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Home Article 2010 SANS VOINOVICH.
2010 SANS VOINOVICH.
by Tim Fernholz
It seems that yet another big-name Republican will be foregoing a senate campaign two years hence, in this case, Ohio's own Senator George Voinovich. The others are Florida's Mel Martinez (where Jeb Bush is declining to run), Missouri's Kit Bond and Kansas' Sam Brownback. Phil Singer's analysis after Bush declined to run is still operative:
The heads of the congressional campaign committees spend the first year of the campaign cycle recruiting candidates. The successful ones keep prospective candidates who are attractive but reluctant to run from making any kind of decision for as long as possible. The logic behind doing so is to keep the candidate open to running in case the political environment shifts or something happens that changes the person's mind and gets him/her into the race.
Jeb Bush's decision to announce that he's not running is a knock on Sen. Cornyn because it means that he failed to keep Bush in the on deck circle. The prospect of a marquee candidate like Bush running in Florida would have been a sign to GOP candidates in other states that 2010 might be a good year for them to run.
So despite worries (or expectations setting) from some Democrats about the electoral challenges Democrats face in 2010, their Senate map is looking cleaner and cleaner; given the potential for competitive races in states like Ohio, Florida, New Hampshire, Missouri and Pennsylvania and not too much worry about defending incumbents -- Colorado has to be a concern, but rumors of Harry Reid's political death have been greatly exaggerated. As Singer notes, there simply aren't a lot of good role models for less-well-known Republicans seeking to enter a Senate race.
But the really interesting aspect of this decision comes in the here-and-now. Voinovich has a tendency to fall into the more moderate camp of senate Republicans, voicing skepticism on everything from tax cuts to John Bolton's nomination as Ambassador to the United Nations. With reelection looming, the need to firm up party and interest group support usually has politicians looking more like ideologues. Now, though, with his political career coming to a close, the 72-year-old Voinovich will be more likely to vote his conscience -- and more receptive to arguments from Harry Reid and Barack Obama on issues like stimulus legislation, health care or even card check. Nothing like a little bipartisan legislative compromise to cement the legacy of a man who's been working in Ohio politics since 1963.
-- Tim Fernholz
Tim Fernholz
Tim Fernholz is a former staff writer for the Prospect. His work has been published by Newsweek, The New Republic, The Nation, The Guardian, and The Daily Beast. He is also a Research Fellow at the New America Foundation.
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K.J. Loerop is the Vice President of Construction Services, for Lakshore Recycling System. Nearly 20 years ago, K.J. Loerop and Ken Hoving began a startup business based out of West Chicago that focused on developing the Construction & Demolition recycling industry. K.J. & Ken Hoving grew the construction and demolition recycling facility to process over 800-1000 tons per day. He also served on the board of directors of the Illinois Recycling Association for six years promoting and increased recycling throughout the state of Illinois.


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Black Patients Miss Out On Promising Cancer Drugs
Black people and Native Americans are underrepresented in clinical trials of new drugs, even when the treatment targets a type of cancer that disproportionately affects them.
Caroline Chen
Riley Wong
Chiara Morra for ProPublica
This article was originally co-published by Stat and ProPublica.
It's a promising new drug for multiple myeloma, one of the most savage blood cancers. Called Ninlaro, it can be taken as a pill, sparing patients painful injections or cumbersome IV treatments. In a video sponsored by the manufacturer, Takeda Pharmaceutical Co., one patient even hailed Ninlaro as "my savior."
The U.S. Food and Drug Administration approved it in 2015 after patients in a clinical trial gained an average of six months without their cancer spreading. That trial, though, had a major shortcoming: its racial composition. One out of five people diagnosed with multiple myeloma in the U.S. is black, and African Americans are more than twice as likely as white Americans to be diagnosed with the blood cancer.
Yet of the 722 participants in the trial, only 13—or 1.8 percent—were black.
The scarcity of black patients in Ninlaro's testing left unanswered the vital question of whether the drug would work equally well for them. "Meaningful differences may exist" in how multiple myeloma affects black patients, what symptoms they experience and how they respond to medications, FDA scientists wrote in a 2017 journal article.
The racial disparity in the Ninlaro study isn't unusual. Reflecting the reluctance of the FDA to force drugmakers to enroll more minority patients, and the failure of most manufacturers to do so voluntarily, stark under-representation of African Americans is widespread in clinical trials for cancer drugs, even when the type of cancer disproportionately affects them. A ProPublica analysis of data recently made public by the FDA found that in trials for 24 of the 31 cancer drugs approved since 2015, fewer than 5 percent of the patients were black. African-Americans make up 13.4 percent of the U.S. population.
As a result, desperately ill black patients who have exhausted other treatment options aren't getting early access to experimental drugs that could extend their life spans or improve their quality of life. While unapproved treatments also carry a risk of setbacks or side effects, new cancer drugs have dramatically shifted outcomes for some patients.
Recently approved lung cancer treatments are "revolutionary," said Dr. Karen Kelly, associate director for research at UC Davis Comprehensive Cancer Center. Even in the first phase of clinical testing, which is aimed at making sure a drug is safe, 20 percent of cancer patients now see their tumors shrink or disappear, up from 5 percent in the early 1990s, Kelly said.
Dr. Kashif Ali, research head at Maryland Oncology Hematology, has spent seven years recruiting patients for about 30 cancer and blood disease trials a year. He said he's often seen minorities, including African Americans, miss out on trials because of financial hurdles, logistical challenges and their lingering distrust of the medical community due to a history of being victimized by medical experimentation.
"They're potentially losing out on life-extending opportunities because it's one more option they no longer have," Ali said. "Especially when patients are in advanced stages of cancer, treatments are like stepping stones: When one stops working, you move on to the next."
Not joining a trial can mean "you've lost life expectancy," he said.
"If they don't have African Americans to test it on, how will they know it's going to work?"
Pat Conley, a 72-year-old retired business analyst whose multiple myeloma relapsed last year, has never participated in a clinical trial. She was interested several times and didn't meet the criteria. Now she's eligible for one, but worries about the burden of regular hourlong trips from her Fayetteville, Georgia, home to the trial site in Atlanta, as well as the copays for medical tests. "They'll want a new biopsy and Lordy, biopsies are not cheap," she said.
Still, two drug regimens haven't halted her cancer and she wants something positive to come from her illness. "If they don't have African Americans to test it on, how will they know it's going to work?" she asked. "If it doesn't help me, it'll help my children, it'll help somebody else."
Pharmaceutical companies contacted by ProPublica all said diversity in clinical trials is important to ensure that drugs meet patients' needs. The issue "is not elevated high enough in the discussion on clinical studies," said John Maraganore, chair of the industry group Biotechnology Innovation Organization. But he added that enrolling minorities is challenging, often for reasons beyond the manufacturer's control, and that it would require a "public-private partnership, working with the FDA and NIH [National Institutes of Health]."
Black participation reached 10 percent in trials for only two of the 31 cancer drugs: the multiple myeloma drug Darzalex, where the figure was exactly 10 percent, and Yondelis, which treats two types of soft tissue cancer. Twelve percent of patients in the Yondelis trial were black, the highest proportion in the ProPublica study. Both drugs are made by Johnson & Johnson, which said it has an internal working group on trial diversity. The working group trains site leaders in best practices for diverse recruitment and seeks minority physicians to help run trials, since some patients prefer being treated by doctors of their own race.
Not enrolling in clinical trials is just one of many ways that African Americans trail white Americans in the quality of their health care. From diagnosis to death, they often experience inferior care and worse outcomes. Because some black Americans can't afford the health insurance mandated under the Affordable Care Act, they remain less likely to have coverage than non-Hispanic white Americans. African Americans are 30 percent more likely than white Americans to die from heart disease. Black mothers are three to four times as likely as white mothers to die in pregnancy or childbirth, and black children are diagnosed with autism later than white children.
While the scarcity of African Americans stands out in ProPublica's analysis, there appear to be gaps in participation of other minority groups as well. Asians were well represented in trials held in some foreign countries, but they made up only 1.7 percent of patients for drugs for which at least 70 percent of trials were conducted within the U.S. By comparison, about 6 percent of the U.S. population identifies as Asian. Almost two-thirds of the trials didn't report any Native Americans or Alaska Natives, who together make up about 2 percent of the U.S. population. ProPublica's analysis excluded Hispanics, because the FDA reports did not have a separate category for them until 2017 and do not distinguish between white and non-white Hispanics.
The very relationship of race to drug development is fraught with controversy. Race is primarily seen as a social concept, rather than as a product of measurable biological traits. Yet there's growing evidence that, whether for environmental or genetic reasons, drugs may have different effects on different populations.
Inadequate minority representation in drug trials means that "we aren't doing good science."
In 2014, the state of Hawaii sued Bristol-Myers Squibb Co. and Sanofi, the manufacturers of the blood thinner Plavix, accusing them of deceptive marketing for failing to disclose that the drug was less effective for some patients of East Asian or Pacific Islander descent. The drugmakers have denied allegations of misconduct and argue that genetic traits have not been proven to affect how well Plavix works. The case is pending. In the meantime, the FDA has added a warning to the label saying that Chinese patients are more likely to have a gene variation that renders the drug ineffective. Researchers at the University of California, San Francisco have found that a commonly used asthma medication, albuterol, doesn't work as well for African-American and Puerto Rican children as it does for European American or Mexican children.
Inadequate minority representation in drug trials means that "we aren't doing good science,"— said Dr. Jonathan Jackson, founding director of the Community Access, Recruitment and Engagement Research Center at Massachusetts General Hospital in Boston. "If we aren't doing good science and releasing these drugs out into the public, then we are at best being inefficient, at worst being irresponsible."
The National Black Church Initiative, a coalition of 34,000 U.S. churches, urged the FDA in 2017 to mandate diversity in all clinical trials before approving a drug or device. "Simply put, the pharmaceutical community is not going to improve minority participation in clinical trials until the FDA compels them to do so via regulations," it wrote to Commissioner Scott Gottlieb.
The FDA hasn't done so. Although it noted in a 2013 report that a lack of diversity betrays a key ethical principle of medical research—equal justice for all—the agency has shied away from setting quotas or numerical guidelines for participation by race.
Instead, it has relied on persuasion. In its 2014 Action Plan, it said it aims to "share best practices," to "support" industry efforts at improving diversity in clinical trials, and to "encourage" patients to participate in trials via social media, emails and blog posts.
"The FDA believes that enrollment should reflect the patients most likely to use a medical product," spokeswoman Gloria Sanchez-Contreras said. The FDA "does not have the regulatory authority to require specific levels of minority representation in clinical trials," although it may ask drugmakers for additional data, she said.
Dr. Rachel Sherman, the FDA's principal deputy commissioner, said that she's "not entirely satisfied" with minority enrollment but that clinical trials have become more diverse in other ways. Two decades ago, women and children were rarely included in drug studies. Now those groups are better represented and the FDA is working on including more minorities, she said.
"If we spend on one thing, it won't get spent on another. We have to be judicious in what we require and what we demand and what we encourage."
Sherman added that the agency has to balance how much information it demands from drugmakers against the need to get a drug onto the market, where it will be broadly available for all patients.
"When it comes to clinical research in this country, there's a credit card, and there's a limit on the credit card," she said. "If we spend on one thing, it won't get spent on another. We have to be judicious in what we require and what we demand and what we encourage."
Diversity has its trade-offs. Clinical trials already cost hundreds of millions of dollars, and drugmakers say that requiring participants to be racially representative would likely add more time and expense.
"If you have a significant delay in enrollment, that would delay the medication advancing to the whole patient population, hurting everybody including the black population," said Maraganore, who is also CEO of drugmaker Alnylam Pharmaceuticals Inc.
To offset costs caused by these delays, manufacturers might reduce the number of drugs in development, depriving some patients of experimental treatments, or raise prices, which would translate into higher insurance premiums and make new drugs even less affordable for the uninsured. Maraganore favors improving diversity through patient education—"a carrot-based approach"—rather than government regulation.
Despite these short-term expenses, eliminating racial disparities in clinical trials would ultimately save costs through disease prevention and improved treatment, according to a 2015 analysis by researchers at the University of California, San Francisco. Without FDA pressure, though, manufacturers may be unlikely to increase efforts to recruit African Americans, especially if their sights are set on a worldwide market. They can use the same clinical trial data to gain approval in the European Union, or Japan.
Takeda, which is based in Tokyo, Japan, tested Ninlaro in the U.S., Japan and 24 other countries, including Australia, Austria, Denmark, New Zealand, Sweden and others with small black populations. "Clinical trials are reflective of the ethnicity distribution in the population where the study takes place," Phil Rowlands, head of Takeda's oncology unit, said in an email. "While we cannot control enrollment eligibility based on the strict clinical criteria, our recruitment efforts extend to diverse patient populations, including minorities."
Asked why Takeda didn't pick sites with higher black populations, Rowlands said that Takeda does not "select sites with ethnicity as an eligibility criteria unless specific risk factors require that."
Drugmakers are careful not to pay too much because they do not want to appear to be enticing patients to join when it isn't in their medical interests
Income is another reason for sparse African-American representation. Clinical trials are largely a middle-class option. A 2015 study found that patients with an annual household income below $50,000 had 32 percent lower odds of participating in a trial than patients with income above that threshold. The median household income of African Americans in 2016 was $39,490, compared to $65,041 for non-Hispanic white Americans.
Patients who can't afford to travel long distances to a trial, take time off work or find child care are at a disadvantage. They are usually reimbursed for travel and food costs, but drugmakers are careful not to pay too much because they do not want to appear to be enticing patients to join when it isn't in their medical interests, said Laurie Halloran, founder and CEO of a consulting firm for the drug industry.
While the experimental drug used in a study is free, any approved treatments that are part of the trial typically need to be covered by a patient's own insurance, according to Ali, the research director at Maryland Hematology Oncology.
He recalled one black patient who was eligible for a study that entailed taking an already approved drug in combination with a new treatment. The approved drug cost $10,000 a month and the patient's insurance charged 20 percent as his copayment, Ali recalled. The patient decided that joining the trial would be too expensive. He instead underwent chemotherapy, which was cheaper, but he didn't tolerate it well. The patient is now considering hospice, Ali said.
Criteria for admission to clinical trials have become more stringent over the years, Mass. General's Jackson said, and patients of color are increasingly excluded. They are more likely than white people to have other conditions such as stroke, hypertension and diabetes, which could complicate research results. Trials often want to enroll "the healthiest sick people they can find" and preclude patients with other conditions. "The wall basically gets taller and taller," he said.
Aredia Taylor didn't qualify on other grounds. A former U.S. Department of Agriculture supervisor for food safety inspections, she was diagnosed with multiple myeloma in 2014 and has undergone a gamut of treatment including various chemotherapy drugs and a stem cell transplant.
Though Aredia Taylor's cancer is in remission, there is no cure for multiple myeloma. She wants to join a trial so she can contribute to research.
Elroy Johnson for ProPublica
While the transplant drove her cancer into remission, there is no cure for multiple myeloma, and she needs to take Revlimid—a standard treatment—daily to keep the cancer at bay. The drug has given her side effects that she describes as devastating to her daily life, including diarrhea, muscle spasms and an inability to concentrate. She said she wants to stop taking Revlimid and would be willing to try an experimental treatment.
Taylor, 58, said she saw pamphlets at her doctor's office encouraging patients to ask about clinical trials, so she asked her oncologist, Dr. Larry Anderson at the University of Texas Southwestern Medical Center in Dallas, whether she should join a study. He told her that she "wasn't a fit," she said. Anderson told ProPublica that most trials are seeking patients with either newly diagnosed or relapsed multiple myeloma, and since Taylor is currently in remission, she wouldn't be accepted.
Taylor was disappointed. Knowing that African Americans like herself are especially susceptible to multiple myeloma, she'd like to be a part of developing more effective treatments.
"I want to pay it forward and be a blessing to somebody else," she said. "I want to be one of the people that they try to do a clinical trial for, so they find a way to a cure."
Spurred by Congress, the FDA began in January 2015 to regularly publish a "Drug Trials Snapshot" for every new drug approved, delineating the demographic breakdown for clinical trial participants by sex, race and age subgroups. ProPublica's analysis focused on participants in trials for the 31 cancer drugs approved since then, comparing their demographics with data from the National Cancer Institute on the incidence of various cancers by race.
Eighteen of those drugs targeted cancers that are at least as likely to afflict black Americans as white Americans. On average, only 4.1 percent of patients in those trials were black. Trials for four multiple myeloma drugs, including Ninlaro, averaged 5 percent black participation.
An FDA study over a longer time period corroborated ProPublica's findings. In a 2017 article in the journal Blood co-authored by Richard Pazdur, the FDA's cancer chief, the agency reported that black patients on average made up 4.5 percent of participants in trials for multiple myeloma drugs since 2003. Higher enrollment of minorities in myeloma trials would have "multiple benefits," the FDA scientists noted. Not only would "underserved populations" gain access to new therapies, but additional data could help researchers identify subtypes of the blood cancer and develop targeted treatments.
A similar pattern emerges for treatments of non-small cell lung cancer. It occurs in 56 out of 100,000 black Americans, versus 49 out of 100,000 white Americans. Yet ProPublica found that in trials for two recently approved drugs for a type of non-small cell lung cancer driven by a mutation in what is known as the ALK gene, less than 2 percent of participants were black.
Black Americans are twice as likely as white Americans to die from prostate cancer.
Those drugs, Takeda's Alunbrig and Genentech's Alecensa, are approved for patients whose lung cancer has spread to other parts of the body. In trials, both drugs were able to shrink tumors, including lesions in the brain. Patients taking Alecensa lived without the disease progressing for an average of 25.7 months, more than double the 10.4 months for patients on another treatment.
"We believe that we must consider differences across all populations to deliver on the promise of personalized health care," said Meghan Cox, a Genentech spokeswoman, adding that the drugmaker is "continuing to study patient response to Alecensa across populations in the post-marketing setting."
Similarly, prostate cancer affects 178 out of every 100,000 African Americans, more than for any other race in the U.S., compared to 106 out of every 100,000 white Americans. Black Americans are twice as likely as white Americans to die from prostate cancer.
Yet from 2009 to 2015, in seven trials conducted for five new prostate cancer therapies, only 3 percent of participants were black, according to a study conducted by Dr. Daniel Spratt, vice chair of research for the department of radiation oncology at the University of Michigan. More recently, 11 times as many white as black participants—or 66 percent compared to 6 percent—joined trials for Johnson & Johnson's new prostate cancer treatment, Erleada.
The FDA approved Erleada in February after trials showed patients on the drug lived an average of two years longer without their cancer spreading into other organs than if they were taking a placebo. In a J&J press release, physicians hailed Erleada's "impressive clinical benefits" for prostate cancer patients.
J&J is "confident in the efficacy and safety data that have supported the approval" of Erleada, J&J spokeswoman Satu Glawe said.
Like Genentech, J&J and Takeda said they track drugs after approval to see if any racial differences emerge. For example, after another J&J drug for prostate cancer, Zytiga, went on the market in 2011, the company ran a new study of 100 patients, 50 black and 50 white.
"We were aware of the low number of African American men" in the pre-approval drug studies of Zytiga, Glawe said. The post-marketing study was intended "to ensure the medication was also providing clinical benefit to these patients." In fact, it showed that black men responded better than white men to the drug.
The FDA is able to look for signs that an approved drug isn't helping a specific population through a surveillance system launched in 2016, called Sentinel, which provides access to medical claims and other data on 200 million Americans obtained from insurers and other health providers, Sherman, the FDA's principal deputy commissioner, said.
Still, post-marketing surveillance doesn't compensate for lack of diversity in clinical trials. Minorities still miss out on experimental treatments—and, if they take a drug once it's approved, may suffer unanticipated side effects.
Leaving analysis of a drug's effect on minorities until it's already on the market is "a hubristic assumption, unnecessarily arrogant," Jackson said. Drugs should "work for the individuals who are the most vulnerable," he said. "That necessarily includes racial and ethnic minorities."
Like African Americans, Native Americans rarely enroll in clinical trials. Among the drug trials analyzed by ProPublica, 64.5 percent did not report any Native American participants, even for types of cancer that Native Americans get at similar rates as other races. (It's possible that some Native Americans were enrolled but lumped into a generic "other" category.)
Native Americans are at higher risk of colorectal cancer than white or Asian Americans. Yet the drugmaker Taiho Oncology didn't report a single Native American among the 800 participants in its trials for the colorectal cancer treatment Lonsurf. Taiho didn't respond to requests for comment.
To understand how a small minority group responds to a drug, researchers might have to enroll more patients than what would be a nationally representative sample. In a trial of 100 people, two Native American or Alaska Native patients would reflect those groups' proportion of the U.S. population, but wouldn't give doctors enough information about race-related impacts.
"The medical community has to engage in some serious trust-building initiatives."
Dr. Linda Burhansstipanov, founder of the Native American Cancer Research Corporation, estimated that 85 percent of Native Americans want to participate in clinical trials when informed of the opportunity. She knows several Native American patients who drove three hours each way to participate in a trial, despite losing an entire day of work.
Still, trial protocols are rarely designed with minority communities in mind, Burhansstipanov said. It has long been rumored among Native Americans, she said, that a clinical trial in the 1990s required patients to take a medication upon rising in the morning. In many Native American tribes, the first thing people do when they wake up is greet the sun with morning prayers. For some tribes, prayers can take more than half an hour. Because of the delay, the tale goes, Native American patients were kicked out of the clinical trial for violating the protocol. "The story spread and became a barrier for people to take part in clinical trials," she said.
In addition, the history of unscrupulous medical experimentation on minorities feeds wariness of clinical trials. Angela Marshall, an internal medicine doctor and board member of Black Women's Health Imperative, said she sees a "lack of excitement among minority communities for clinical trials, coming from a mistrust of the medical community."
Her black patients often cite the infamous Tuskegee study, conducted from 1932 to 1972 by the U.S. Public Health Service, in which researchers knowingly withheld treatment from African American sharecroppers with syphilis in order to study the progression of the disease. Some Native Americans are similarly suspicious of the medical community, Burhansstipanov said, because the Indian Health Service (a unit of the U.S. Department of Health and Human Services) sterilized thousands of Native American women in the 1970s without their consent.
"The medical community has to engage in some serious trust-building initiatives," Marshall said.
Goode is an active member of the Leukemia and Lymphoma Society and runs a support group for patients. He holds an award from the foundation.
Sarah Elizabeth Borst for ProPublica
That trust is what has helped to keep Thomas Goode alive. An African American diagnosed with multiple myeloma in 2005 when he was 34, Goode has endured three stem cell transplants. In between each transplant, he participated in clinical trials, where he gained access to drug cocktails that helped bridge him to the next procedure. He counted himself lucky that he lived close to Durham, North Carolina, which is a research hub, so he was able to see specialists and take part in trials.
"I never knew anything about clinical trials, but a lot of my trust and faith was in the doctor," he said. "I said, 'I trust you, doctor, whatever you say.'"
Now, Goode has gone six years without a relapse. "I'm in a good place right now," he said, "But if I didn't try a trial, who's to say I would still be here?"
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The Racial Inequality in Cancer Deaths
Megan Jula

redpajama_set_name: RedPajamaCommonCrawl


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Marsella ist eine Gemeinde (municipio) im Departamento Risaralda in Kolumbien.

Geographie
Marsella liegt im Departamento de Risaralda in der kolumbianischen Kaffeezone (Eje Cafetero) 30 km von Pereira entfernt auf einer Höhe von 1575 Metern. Marsella liegt in den kolumbianischen Anden im Gebiet zwischen dem Río Cauca und dem Río San Francisco. Die Gemeinde grenzt im Norden an Belalcázar und Chinchiná im Departamento de Caldas, im Osten an Chinchiná sowie an Santa Rosa de Cabal, im Süden an Pereira und Dosquebradas und im Westen an Belalcázar und La Virginia.

Bevölkerung
Die Gemeinde Marsella hat 24.131 Einwohner, von denen 14.107 im städtischen Teil (cabecera municipal) der Gemeinde leben (Stand: 2019).

Geschichte
Das Gebiet des heutigen Marsella war vor der Ankunft der Spanier von den indigenen Völkern der Quimbaya bewohnt. Der heutige Ort wurde 1860 im Zuge der Besiedlung der Region von aus Antioquia stammenden Siedlern gegründet.

Wirtschaft
Die wichtigsten Wirtschaftszweige von Marsella sind Landwirtschaft (insbesondere werden Kaffee, Bananen, Avocados, Zitrusfrüchte und Zierpflanzen angebaut) sowie in geringerem Maße Rinderproduktion.

Weblinks

Einzelnachweise

Ort in Kolumbien
Geographie (Departamento de Risaralda)
Gegründet 1860

redpajama_set_name: RedPajamaWikipedia


text:
\section{Introduction}

Since its inception, the Szlenk index has proven to be a valuable tool with many applications in Banach space theory, including renorming (\cite{KOS}, \cite{GKL}, \cite{Cpower}, \cite{Cflat}), embedding and universality (\cite{Boss}, \cite{FOSZ}, \cite{C}), and factorization of operators (\cite{Brooker}, \cite{CN2}). See \cite{Lancien} for a survey of the Szlenk index and its applications.

In this paper, we say that a property which a Banach space may or may not possess is a \emph{three space property} if whenever $X$ is a Banach space and $Y$ is a closed subspace of $X$ such that two of the spaces $X, Y, X/Y$ has the property, then the third space also has the property. It is well known that for a Banach space $X$ and a closed subspace $Y$ of $X$, $Sz(Y), Sz(X/Y)\leqslant Sz(X)$. Therefore in order to decide whether the property `$Sz(\cdot)\leqslant \omega^\xi$' is a three space property, it is necessary and sufficient to answer the question of whether $Sz(Y), Sz(X/Y)\leqslant \omega^\xi$ implies $Sz(X)\leqslant \omega^\xi$.  We recall that the Szlenk index of an Asplund Banach space is always $\omega^\xi$ for some ordinal $\xi$ (\cite{Lancien}), so that we lose no generality by considering only such ordinals.  In the case that $\xi=\omega^\gamma$ for some $\gamma$, Brooker and Lancien gave an affirmative answer.

\begin{theorem}\cite{BL} Let $\gamma$ be any ordinal. Let $X$ be a Banach space and let $Y$ be a closed subspace. If two of the three spaces $X,Y,X/Y$ have Szlenk index not exceeding $\omega^{\omega^\gamma}$, then the third space also has Szlenk index not exceeding $\omega^{\omega^\gamma}$.

\label{BLth}
\end{theorem}

We note that except for the case $\gamma=0$, this result was already shown by Lancien in \cite{Lancien1}. Theorem \ref{BLth} result was also shown in \cite{CAlt} by a method dual to that in \cite{BL}.

We also recall that an ordinal $\xi$ is said to be a \emph{gamma number} if it is not the sum of two smaller ordinals.  It is a standard fact about ordinals that  $\xi$ is a gamma number if and only if $\xi$ is equal to $0$ or $\omega^\gamma$ for some $\gamma$. We recall that an ordinal is said to be \emph{additively indecomposable} if it is a non-zero gamma number. Since a Banach space $X$ has $Sz(X)\leqslant 1=\omega^0$ if and only if it is finite dimensional, combining this trivial fact with Theorem \ref{BLth} yields that `$Sz(\cdot)\leqslant \omega^\xi$' is a three space property whenever $\xi$ is a gamma number. To the best of our knowledge, prior to this writing, there were no ordinals $\xi$ for which it was known that `$Sz(\cdot)\leqslant \omega^\xi$' is not a three space property.  The main result of this paper is to show that for countable ordinals, the countable gamma numbers form a complete list of countable ordinals for which `$Sz(\cdot)\leqslant \omega^\xi$' is a three space property.

\begin{theorem} For $\xi\in (0,\omega_1)\setminus \{\omega^\gamma: \gamma<\omega_1\}$, the property of having Szlenk index not exceeding $\omega^\xi$ is not a three space property.

\label{main1}
\end{theorem}

Also, it was shown in \cite{BCM} that the property of not admitting an $\ell_1^\xi$-spreading model is a three space property whenever $\xi=\omega^\gamma$ for some countable $\gamma$.   As a matter of convention, this result is also true when $\xi=0$.  Again, prior to this writing, we are not aware of an ordinals $\xi$ for which it is known that not admitting an $\ell_1^\xi$-spreading model fails to be a three space property.  The spaces which we use to prove Theorem \ref{main1} will also give a complete characterization of the collection of $\xi$ such that not admitting an $\ell_1^\xi$-spreading model is a three space property.

\begin{theorem} For $\xi\in (0, \omega_1)\setminus \{\omega^\gamma: \gamma<\omega_1\}$, the property of admitting no $\ell_1^\xi$-spreading model is not a three space property.

\label{main2}
\end{theorem}

\section{Szlenk index and spreading models}

For a Banach space $X$, a weak$^*$-compact subset $K$ of $X^*$, and $\ee>0$, we let $s_\ee(K)$ denote the subset of $K$ consisting of those $x^*\in K$ such that for any weak$^*$-neighborhood $V$ of $x^*$, $\text{diam}(V\cap K)>\ee$.   We define the transfinite derivations as $$s^0_\ee(K)=K,$$ $$s^{\xi+1}(K)=s_\ee(s^\xi_\ee(K)),$$ and if $\xi$ is a limit ordinal, $$s^\xi_\ee(K)=\bigcap_{\zeta<\xi}s^\zeta_\ee(K).$$  If there exists an ordinal $\xi$ such that $s^\xi_\ee(K)=\varnothing$, then we let $Sz(K,\ee)$ denote the minimum such $\xi$. If no such $\xi$ exists, we agree to the convention that $Sz(K, \ee)=\infty$.   If for each $\ee>0$, there exists an ordinal $\xi$ such that $s^\xi_\ee(K)=\varnothing$, then we let $Sz(K)=\sup_{\ee>0}Sz(K,\ee)$. If $Sz(K, \ee)=\infty$ for some $\ee>0$, then we establish the convention $Sz(K)=\infty$. We also establish the convention that $\xi<\infty$ for any ordinal $\xi$. For a Banach space $X$, we define $Sz(X, \ee)=Sz(B_{X^*}, \ee)$ for each $\ee>0$ and $Sz(X)=Sz(B_{X^*})$.

We recall that if $Y$ is a closed subspace of $X$, then $Sz(Y)\leqslant Sz(X)$. Furthermore, $Sz(\ell_1)=\infty$, since the set $K=\{\pm 1\}^\nn$ is a $2$-separated subset of $B_{\ell_1^*}$ with no weak$^*$-isolated points, from which it follows that $K\subset s^\xi_1(B_{\ell_1^*})$ for all $\xi$. Therefore if the Banach space $X$ contains an isomorphic copy $\ell_1$, then $Sz(X)=\infty$.

Throughout, we identify subsets of $\nn$ with strictly increasing sequences of natural numbers in the usual way.  For an infinite subset $M$ of $\nn$, we let $[M]$ denote the set of infinite subsets of $M$. For non-empty sets $E,F$, we let $E<F$ denote the relation $\max E<\min F$. For a non-empty subset $E$ of $\nn$ and $n\in\nn$, we let $n\leqslant E$ denote the relation that $n\leqslant \min E$.  The relation $E<n$ is defined similarly.  If $E<F$, we let $E\smallfrown F$ denote the concatenation of $E$ with $F$. Given our identification of sets with increasing sequences, if $E<F$, $E\smallfrown F$ is identified with $E\cup F$. We let $E\prec F$ denote the relation that $E$ is a proper initial segment of $F$. That is, $E\prec F$ if either $E=\varnothing\neq F$ or $F=(n_i)_{i=1}^t$ and $E=(n_i)_{i=1}^s$ for some $1\leqslant s< t$.   We recall the \emph{Schreier families} $\mathcal{S}_\xi$, $\xi<\omega_1$. We let $$\mathcal{S}_0=\{\varnothing\}\cup \{(n): n\in\nn\},$$ $$\mathcal{S}_{\xi+1}=\{\varnothing\}\cup \Bigl\{\bigcup_{n=1}^t E_n: t\leqslant E_1<\ldots <E_t, \varnothing\neq E_n\in\mathcal{S}_\xi\Bigr\},$$ and if $\xi<\omega_1$ is a limit ordinal, we fix $\xi_n\uparrow \xi$ and let $$\mathcal{S}_\xi=\{\varnothing\}\cup \{E: \exists n\leqslant E\in\mathcal{S}_{\xi_n}\}.$$ We let $MAX(\mathcal{S}_\xi)$ denote the set of $E\in \mathcal{S}_\xi$ which are maximal in $\mathcal{S}_\xi$ with respect to inclusion.

 We also recall that for $0\leqslant \alpha, \beta<\omega_1$, $$\mathcal{S}_\beta[\mathcal{S}_\alpha]:=\{\varnothing\}\cup \Bigl\{\bigcup_{n=1}^t E_n: E_1<\ldots <E_t, \varnothing \neq E_n\in \mathcal{S}_\alpha, (\min E_n)_{n=1}^t\in \mathcal{S}_\beta\Bigr\}.$$

We recall that for two sets $(m_i)_{i=1}^l$, $(n_i)_{i=1}^l$, $(n_i)_{i=1}^l$ is said to be a \emph{spread} of $(m_i)_{i=1}^l$ if $m_i\leqslant n_i$ for all $1\leqslant i\leqslant l$.  We now collect the necessary facts concerning these families.

\begin{proposition}\cite[Proposition $3.2$]{OTW}\cite[Propositions $3.1$,$3.2$]{CConcerning} \begin{enumerate}[(i)]\item For each $\xi<\omega_1$ and $E\in \mathcal{S}_\xi$, either $E\in MAX(\mathcal{S}_\xi)$ or $E\smallfrown (n)\in \mathcal{S}_\xi$ for all $E<n$. \item $\mathcal{S}_\xi$ contains no infinite $\preceq$-ascending chains. \item $\mathcal{S}_\xi$ contains all subsets of its members. \item $\mathcal{S}_\xi$ is \emph{spreading}. That is, $\mathcal{S}_\xi$ contains all spreads of its members.  \item For any $\alpha, \beta<\omega_1$, there exists $(m_n)_{n=1}^\infty\in[\nn]$ such that for any $E\in \mathcal{S}_{\alpha+\beta}$, $(m_n)_{n\in E}\in \mathcal{S}_\beta[\mathcal{S}_\alpha]$. \end{enumerate}

\label{trivlist}

\end{proposition}

Throughout, $c_{00}$ will denote the space of eventually zero scalar sequences and $(e_i)_{i=1}^\infty$ will denote the canonical basis of $c_{00}$. Given $x=\sum_{i=1}^\infty a_ie_i\in c_{00}$, we let $\ran(x)$ be the smallest interval in $\nn$ which contains $\{i\in \nn: a_i\neq 0\}$. Given $E\subset \nn$ and $x=\sum_{i=1}^\infty a_ie_i\in c_{00}$, we let $Ex=\sum_{i\in E}a_ie_i$.    For $0<\xi<\omega_1$, a sequence $(x_n)_{n=1}^\infty$ in some Banach space $X$ is an $\ell_1^\xi$-\emph{spreading model} if it is bounded and $$0<\inf\Bigl\{\|x\|: E\in\mathcal{S}_\xi, x=\sum_{n\in E}a_nx_n, \sum_{n\in E}|a_n|=1\Bigr\}.$$  For completeness, we say $(x_n)_{n=1}^\infty$ is an $\ell_1^0$-\emph{spreading model} if it is a seminormalized basic sequence.  We note here that by the almost monotone property of the  Schreier families (see \cite{OTW}), if $X$ admits an $\ell_1^\xi$-spreading model, then it admits an $\ell_1^\zeta$-spreading model for each $\zeta<\xi$.

\begin{rem}\upshape Let $\xi<\omega_1$ be an ordinal. \begin{enumerate}[(i)]\item If the Banach space $X$ admits a weakly null $\ell_1^\xi$-spreading model, then $Sz(X)>\omega^\xi$.  Indeed, if $(x_n)_{n=1}^\infty \subset B_X$ is weakly null and $$0<\inf\Bigl\{\|x\|: E\in\mathcal{S}_\xi, x=\sum_{n\in E}a_nx_n, \sum_{n\in E}|a_n|=1\Bigr\},$$ then by \cite[Proposition $5$]{OSZ} combined with the main theorem of \cite{AJO} , the collection $(x_E)_{E\in \mathcal{S}_\xi\setminus \{\varnothing\}}$ given by $x_E=x_{\max E}$ witnesses that $Sz(X)>\omega^\xi$.

\item The Banach space $X$ admits an $\ell_1^0$-spreading model if and only if it is infinite dimensional if and only if $Sz(X)>1=\omega^0$. If $0<\xi$ and if the Banach space $X$ admits an $\ell_1^\xi$-spreading model, then either it contains an isomorhpic copy of $\ell_1$, in which case  $Sz(X)=\infty>\omega^\xi$, or $X$ admits a weakly null $\ell_1^\xi$-spreading model (see \cite[Remark $5$]{BCM}), and $Sz(X)>\omega^\xi$ by $(i)$.  Therefore any Banach space which admits an $\ell_1^\xi$-spreading model (weakly null or otherwise) must have Szlenk index exceeding $\omega^\xi$. \item For $\alpha, \beta<\omega_1$,  if a Banach space $X$ admits a bounded sequence $(x_n)_{n=1}^\infty$ such that $$0<\inf\Bigl\{\|\sum_{i\in E}a_ix_i\|_Y: E\in \mathcal{S}_\beta[\mathcal{S}_\alpha], \sum_{i\in E}|a_i|=1\Bigr\},$$ then the sequence $(x_n)_{n=1}^\infty$ admits a subsequence which is an $\ell_1^{\alpha+\beta}$-spreading model, and by $(ii)$,  $Sz(X)>\omega^{\alpha+\beta}$.  Indeed, the indicated subsequence can be taken to be $(x_{m_n})_{n=1}^\infty$, where $(m_n)_{n=1}^\infty\in [\nn]$ is as in Proposition \ref{trivlist}$(v)$.   \end{enumerate}

\label{remark}
\end{rem}

For $\gamma<\omega_1$, the \emph{Schreier space of order} $\gamma$, denoted by $X_\gamma$, is the completion of $c_{00}$ with respect to the norm $$\|x\|_{X_\gamma}=\sup \{\|Ex\|_{\ell_1}: E\in \mathcal{S}_\gamma\}.$$ The \emph{Baernstein} $2$ \emph{space of order} $\gamma$, denoted by $X_\gamma^2$,  is the completion of $c_{00}$ with respect to the norm $$\|x\|_{X^2_\gamma}= \sup\Bigl\{\bigl(\sum_{n=1}^\infty \|E_nx\|_{X_\gamma}^2\bigr)^{1/2}: E_1<E_2<\ldots\Bigr\}.$$  We note that $X_0^2=\ell_2$ isometrically.

We recall the relevant properties of the Schreier and Baernstein spaces.

\begin{proposition} Fix $\gamma<\omega_1$, \begin{enumerate}[(i)] \item The canonical $c_{00}$ basis is normalized and $1$-unconditional in $X_\gamma$ and $X_\gamma^2$. \item $Sz(X_\gamma)=\omega^{\gamma+1}$ and $Sz(X_\gamma^2)=\omega^{\gamma+1}$. In particular, neither $X_\gamma$ nor $X_\gamma^2$ admits an $\ell_1^{\gamma+1}$-spreading model.\item The canonical $c_{00}$ basis is shrinking in both $X_\gamma$ and $X_\gamma^2$. \item The canonical $c_{00}$ basis of $X_\gamma^2$ satisfies a $1$-lower $\ell_2$ estimate. That is, for any $n\in\nn$ and any integers $0=p_0<\ldots <p_n$ and scalars $(a_i)_{i=1}^{p_n}$, $$\Bigl\|\sum_{i=1}^{p_n} a_ie_i\Bigl\|_{X_\gamma^2}^2 \geqslant \sum_{j=1}^n \Bigl\|\sum_{i=p_{j-1}+1}^{p_j} a_ie_i\Bigl\|_{X_\gamma^2}^2.$$ \item  For any $t\in\nn$ and integers $1\leqslant p_1<\ldots<p_{t+1}$ and $x_1, \ldots, x_t$ such that for each $1\leqslant n\leqslant t$,  $x_n\in B_{X_\gamma^2}\cap \text{\emph{span}}\{e_i: p_n \leqslant i< p_{n+1}\}$, and for any scalars  $(a_n)_{n=1}^t$, $$\Bigl\|\sum_{n=1}^t a_nx_n\Bigr\|_{X_\gamma^2} \leqslant 4\Bigl\|\sum_{n=1}^t a_ne_{p_n}\Bigr\|_{X_\gamma^2}.$$ \item The canonical basis of $X_\gamma^2$ is $1$-\emph{right dominant}. That is, for any sequences of positive integers $l_1<l_2<\ldots$, $m_1<m_2<\ldots$ such that $l_n\leqslant m_n$ for all $n\in\nn$, and for any $(a_n)_{n=1}^\infty\in c_{00}$, $$\Bigl\|\sum_{n=1}^\infty a_ne_{l_n}\Bigr\|_{X_\gamma^2} \leqslant \Bigl\|\sum_{n=1}^\infty a_ne_{m_n}\Bigr\|_{X_\gamma^2}.$$    \end{enumerate}

\label{important}
\end{proposition}

\begin{proof} Item $(i)$ is clear. The part of  $(ii)$ concerning the Szlenk index can be found in \cite[Proposition $4.6$]{C} and \cite[Proposition $4.2$]{CF}. The part concerning spreading models can be deduced by combining the part of $(ii)$ concerning Szlenk index with Remark \ref{remark}$(ii)$.   Item $(iii)$ can be deduced from the fact that since $Sz(X_\gamma), Sz(X_\gamma^2)<\infty$, neither of these spaces  can contain an isomorphic copy of $\ell_1$. Since the canonical $c_{00}$ basis is unconditional in each of these and neither space contains an isomorphic copy of $\ell_1$, the canonical $c_{00}$ basis is shrinking in $X_\gamma$ and in $X_\gamma^2$.

For item $(iv)$, we first note that for any interval $I\subset \nn$,  $x\in \text{span}\{e_i:i\in I\}$, and $E_1<E_2<\ldots$, if $J=\{n\in\nn: E_n\cap I\neq 0\}$, then $$\sum_{n=1}^\infty \|E_nx\|_{X_\gamma}^2 = \sum_{n\in J} \|(I\cap E_n) x\|_{X_\gamma}^2.$$  Therefore $$\|x\|_{X^2_\gamma}^2 = \sup\Bigl\{\Bigl(\sum_{n=1}^m \|E_nx\|^2_{X_\gamma}\Bigr)^{1/2}:E_1<\ldots <E_m, E_n\subset \ran(x)\Bigr\}.$$   Now if $n$, $p_0, \ldots, p_n$, $(a_i)_{i=1}^{p_n}$ are as in the proposition, then for each $1\leqslant j\leqslant n$, we can choose   $E^j_1<\ldots <E^j_{k_j}$, $E^j_i\subset (p_{j-1}, p_j]$, such that $$\Bigl\|\sum_{i=p_{j-1}+1}^{p_j} a_ie_i\Bigr\|^2_{X^2_\gamma} = \sum_{l=1}^{k_n} \Bigl\|E_l\sum_{i=p_{j-1}+1}^{p_j} a_ie_i\Bigr\|_{X_\gamma}^2.$$   Enumerate $E^1_1, \ldots, E^1_{k_1}, E^2_1, \ldots, E^n_{k_n}$ as $E_1, \ldots, E_m$ and note that  $$\Bigl\|\sum_{i=1}^{p_n} a_ie_i\Bigr\|_{X_\gamma^2}^2 \geqslant \sum_{l=1}^m \Bigl\|E_l\sum_{j=1}^n\sum_{i=p_{j-1}+1}^{p_j} a_ie_i\Bigr\|_{X_\gamma}^2 = \sum_{j=1}^n  \Bigl\|\sum_{i=p_{j-1}+1}^{p_j} a_ie_i\Bigr\|_{X_\gamma}^2.$$

Items $(v)$ and $(vi)$ were shown in \cite[Lemma $3.2$]{CF}.

\end{proof}

We recall that a (finite or infinite) sequence $(g_n)_{n=1}^N\subset c_{00}$ of non-zero vectors is said to be a \emph{block sequence with respect to the canonical} $c_{00}$ \emph{basis} provided there exist integers $0=p_0<\ldots <p_N$ (resp. $0=p_0<p_1<\ldots$ if $N=\infty$) such that for each $n\leqslant N$ (resp. $n<N$ if $N=\infty$) such that $g_n\in \text{span}\{e_i: p_{n-1}<i\leqslant p_n\}$.

We last recall that for a Banach space $G$ for which the canonical $c_{00}$ basis is a Schauder basis, $0<\xi<\omega_1$, and a collection $(g_E)_{E\in \mathcal{S}_\xi\setminus \{\varnothing\}}\subset G$, $(g_E)_{E\in \mathcal{S}_\xi\setminus \{\varnothing\}}$ is said to be a \begin{enumerate}[(i)]\item \emph{weakly null tree} if for any $E\in \mathcal{S}_\xi\setminus MAX(\mathcal{S}_\xi)$, $(g_{E\cup (n)})_{n> \max E}$ is a weakly null sequence in $G$,  \item \emph{block tree} provided that for each $\varnothing \prec E_1\prec E_2\prec \ldots \prec E_n\in \mathcal{S}_\xi$, $(g_{E_i})_{i=1}^n$ is a block sequence with respect to the canonical $c_{00}$ basis, and for each $E\in \mathcal{S}_\xi\setminus MAX(\mathcal{S}_\xi)$, $(g_{E\cup (n)})_{n>\max E}$ is a block sequence with respect to the canonical $c_{00}$ basis. \end{enumerate}

We note that if $(g_E)_{E\in \mathcal{S}_\xi\setminus \{\varnothing\}}$ is a block tree, then for any $\varnothing\neq E\in\mathcal{S}_\xi$, $\ran(g_E)\geqslant  \max E.$

\section{The example}

Our example is a modification, and in some cases coincides with, spaces constructed by Lindenstrauss \cite{L}. Fix two countable ordinals $\alpha, \beta<\omega_1$.  Fix a dense sequence $(f_i)_{i=1}^\infty$ in the unit sphere of the Schreier space $X_\alpha$. Define a norm $\|\cdot\|_{G_{\alpha,\beta}}$ on $c_{00}$ by $$\Bigl\|\sum_{n=1}^\infty a_ne_n\Bigr\|_{G_{\alpha,\beta}} = \sup\Bigl\{   \Bigl\|\sum_{n=1}^\infty\|\sum_{i\in I_n} a_if_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X_\beta^2}: I_1<I_2<\ldots, I_n\text{\ an interval}\Bigr\}.$$    Let $G_{\alpha,\beta}$ be the completion of $c_{00}$ with respect to this norm. In the case $\beta=0$, since $X_0^2=\ell_2$,  this construction coincides with that of Lindenstrauss in \cite{L}.

\begin{rem}\upshape
Note that by  $1$-right dominance of the $X_\beta^2$ basis, we can compute the $\|\cdot\|_{G_{\alpha,\beta}}$ norm of some non-zero $g\in c_{00}$ as $$\|g\|_{G_{\alpha,\beta}}=\sup\Bigl\{   \Bigl\|\sum_{n=1}^m\|\sum_{i\in I_n} a_if_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X_\beta^2}: I_1<\ldots <I_m, I_n\text{\ an interval}, I_n\subset \ran(g)\Bigr\}.$$    Indeed, for any intervals $J_1<J_2<\ldots$, if $K_1<\ldots <K_m$ is a list of those $J_i$ such that $J_i\cap \ran(g)\neq \varnothing$ and if $I_n= K_n\cap \ran(g)$ for $n=1, \ldots, m$, then by $1$-right dominance of the canonical $X_\beta^2$ basis together with the fact that $\sum_{i\in K_n}a_if_i=\sum_{i\in I_n}a_if_i$ and $\min I_n\geqslant \min K_n$ for each $1\leqslant n\leqslant m$,  $$  \Bigl\|\sum_{n=1}^\infty\|\sum_{i\in J_n} a_if_i\|_{X_\alpha}e_{\min J_n}\Bigr\|_{X_\beta^2} =\Bigl\|\sum_{n=1}^m\|\sum_{i\in K_n} a_if_i\|_{X_\alpha}e_{\min K_n}\Bigr\|_{X_\beta^2} \leqslant \Bigl\|\sum_{n=1}^m\|\sum_{i\in I_n} a_if_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X_\beta^2} .$$

\label{easyrem}
\end{rem}

We collect the following easy properties of $G_{\alpha,\beta}$.

\begin{proposition} \begin{enumerate}[(i)]\item The canonical $c_{00}$ basis is a boundedly-complete, normalized, bimonotone basis for $G_{\alpha,\beta}$.\item If $g=\sum_{i=1}^\infty a_ie_i\in G_{\alpha,\beta}$, then $Qg:=\sum_{i=1}^\infty a_if_i$ is convergent in $X_\alpha$, and $Q:G_{\alpha,\beta}\to X_\alpha$ is  a quotient map. \end{enumerate}

\label{tech1}
\end{proposition}

\begin{proof}$(i)$ It is obvious that $\|e_i\|_{G_{\alpha, \beta}}=1$ for each $i\in\nn$.  Fix $(a_i)_{i=1}^\infty\in c_{00}$ and an interval $J$.  Let $b_i=a_i$ for $i\in J$ and $b_i=0$ for $i\in\nn\setminus J$.   As noted in Remark \ref{easyrem}, there exist intervals $I_1<\ldots <I_k$, $I_i\subset J$, such that $$\Bigl\|\sum_{i=1}^\infty b_ie_i\Bigr\|_{G_{\alpha, \beta}}  =\Bigl\|\sum_{n=1}^k \|\sum_{i\in I_n} b_if_i\|_{X_\alpha} e_{\min J_n} \Bigr\|_{X_\beta^2}=\Bigl\|\sum_{n=1}^k \|\sum_{i\in I_n} a_if_i\|_{X_\alpha} e_{\min J_n} \Bigr\|_{X_\beta^2} \leqslant \Bigl\|\sum_{i=1}^\infty a_ie_i\Bigr\|_{G_{\alpha, \beta}}.  $$ Therefore $(e_i)_{i=1}^\infty$ is bimonotone in $G_{\alpha, \beta}$.

We next show that $(e_i)_{i=1}^\infty$ is boundedly-complete. Suppose $(a_i)_{i=1}^\infty$, $\ee>0$, and $p_0<p_1<\ldots$ are such that $$\inf_n \Bigl\|\sum_{i=p_{n-1}+1}^{p_n} a_ie_i\Bigr\|_{G_{\alpha, \beta}}\geqslant\ee.$$  Then by Remark \ref{easyrem}, for each $n\in\nn$,  there exist intervals $I^n_1<\ldots <I^n_{k_n}$ such that $I^n_i\subset (p_{n-1}, p_n]$ and $$\ee\leqslant \Bigl\|\sum_{j=1}^{k_n} \|\sum_{i\in I^n_j} a_if_i\|_{X_\alpha} e_{\min I^n_j}\Bigr\|_{X_\beta^2}.$$  Then with $u_n=\sum_{j=1}^{k_n} \|\sum_{i\in I^n_j} a_if_i\|_{X_\alpha} e_{\min I^n_j}$, it follows that  $\ran(u_n)\subset (p_{n-1}, p_n]$ and $\|u_n\|_{X_\beta^2}\geqslant \ee$.   Let $I_1<I_2<\ldots $ be an enumeration of $I^1_1, I^1_2, \ldots, I^1_{k_1}, I^2_1, I^2_2 ,\ldots$. Using the $1$-lower $\ell_2$ estimate of the canonical $X^2_\beta$ basis, for any $m\in\nn$, \begin{align*} \Bigl\|\sum_{i=1}^{p_m} a_ie_i\Bigr\|_{G_{\alpha, \beta}} & \geqslant \Bigl\|\sum_{n=1}^\infty \|\sum_{i\in I_n} a_if_i\|_{X_\alpha} e_{\min I_i}\Bigr\|_{X^2_\beta} = \Bigl\|\sum_{n=1}^m\sum_{j=1}^{k_n} \|\sum_{i\in I^n_j} a_if_i\|_{X_\alpha}\Bigr\|_{X^2_\beta} = \Bigl\|\sum_{n=1}^m u_n\Bigr\|_{X^2_\beta} \geqslant \ee m^{1/2}. \end{align*}   By contraposition, if $\sup_m \|\sum_{i=1}^m a_ie_i\|_{G_{\alpha, \beta}}<\infty$, then $\sum_{i=1}^\infty a_ie_i$ is convergent in $G_{\alpha, \beta}$.

$(ii)$ Note that for any scalars $(a_i)_{i=1}^\infty\in c_{00}$ and any finite interval $I$, $$\Bigl\|\sum_{i=1}^\infty a_ie_i\Bigr\|_{G_{\alpha, \beta}}\geqslant \Bigl\|\|\sum_{i\in I}a_if_i\|_{X_\alpha}e_{\min I}\Bigr\|_{X^2_\beta}= \Bigl\|\sum_{i\in I} a_if_i\Bigr\|_{X_\alpha}.$$   Fix $g=\sum_{i=1}^\infty a_ie_i\in G_{\alpha, \beta}$.   If there exist $\ee>0$ and $p_0<p_1<\ldots$ such that $\inf_n \|\sum_{i=p_{n-1}+1}^{p_n}a_if_i\|_{X_\alpha} \geqslant \ee$, then the inequalities $$\inf_n \Bigl\|\sum_{i=p_{n-1}+1}^{p_n} a_ie_i\Bigr\|_{G_{\alpha, \beta}}\geqslant \Bigl\|\sum_{i=p_{n-1}+1}^{p_n}a_if_i\Bigr\|_{X_\alpha}\geqslant \ee$$ and $$\sup_m \Bigl\|\sum_{n=1}^{p_m} a_if_i\Bigr\|_{X_\alpha} \leqslant \sup_m \Bigl\|\sum_{i=1}^{p_m} a_ie_i\Bigr\|_{G_{\alpha, \beta}}=\|g\|_{G_{\alpha, \beta}}<\infty$$ contradict $(i)$.   Therefore $\sum_{i=1}^\infty a_if_i$ is convergent in $X_\alpha$, and $Q\sum_{i=1}^\infty a_ie_i=\sum_{i=1}^\infty a_if_i$ is well-defined.   By the first line of $(ii)$, $\|Q\|\leqslant 1$. Of course, $\|Q\|\geqslant 1$ by considering its action on any basis vector.   Furthermore, since $QB_{G_{\alpha, \beta}}$ contains $(f_n)_{n=1}^\infty$, which is dense in $S_{X_\alpha}$, $Q$ is a quotient map.

\end{proof}

Let $Q$ be as defined in Proposition \ref{tech1}$(ii)$ and let $H_{\alpha,\beta}=\ker(Q)$.  We recall the notation that for an interval $I\subset \nn$, $I$ also denotes the projection on $c_{00}$ given by $I\sum_{i=1}^\infty a_ie_i=\sum_{i\in I}a_ie_i$. Then for an interval $I\subset \nn$, $QI\sum_{i=1}^\infty a_ie_i=\sum_{i\in I}a_if_i$, which allows for a more convenient expression of the $\|\cdot\|_{G_{\alpha, \beta}}$ norm: $$\|g\|_{G_{\alpha, \beta}}= \sup\Bigl\{\Bigl\|\sum_{n=1}^\infty \|QI_ng\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X_\beta^2}:I_1 <I_2<\ldots, I_n\subset\nn\text{\ an interval}\Bigr\}.$$

\begin{lemma} \begin{enumerate}[(i)]\item If $(g_E)_{E\in \mathcal{S}_{\beta+1}\setminus \{\varnothing\}}\subset B_{G_{\alpha, \beta}}$ is a block tree such that $\|Qg_E\|_{X_\alpha}\leqslant \ee$ for all $\varnothing \neq E\in \mathcal{S}_{\beta+1}$, then $\inf\{\|g\|_{G_{\alpha, \beta}}: E\in \mathcal{S}_{\beta+1}, g\in \text{\emph{co}}(g_F: \varnothing\prec F\preceq E)\} \leqslant \ee.$  \item $Sz(H_{\alpha, \beta})\leqslant \omega^{\beta+1}$.  \item $Sz(H_{\alpha, \beta})= \omega^{\beta+1}$ and $H_{\alpha, \beta}$ contains an $\ell_1^\beta$-spreading model. \end{enumerate}

\label{finish}

\end{lemma}

In the proof, we will use the \emph{repeated averages hierarchy} from \cite{AMT}, but with the notation from \cite{CN}. A complete presentation of the repeated averages hierarchy is unnecessary for our purposes. For readability, we recall here only the properties necessary for the proof below. Proofs of these properties can be found in \cite{AMT}. For each ordinal $\xi<\omega_1$ and each infinite subset $N$ of $\nn$, $\mathbb{S}^\xi_{N,1}:\nn\to [0,1]$ is a non-negative function such that $\sum_{i=1}^\infty \mathbb{S}^\xi_{N,1}(i)=1$ and, if $(n_i)_{i=1}^t$ is the maximal initial segment of $N$ which lies in $\mathcal{S}_\xi$, $(n_i)_{i=1}^t=\{i\in\nn: \mathbb{S}^\xi_{N,1}(i)\neq 0\}$.

\begin{proof}[Proof of Lemma \ref{finish}]$(i)$ We will repeatedly use facts from Proposition \ref{important}. Fix a block tree $(g_E)_{E\in\mathcal{S}_{\beta+1}\setminus \{\varnothing\}}\subset B_{G_{\alpha, \beta}}$ such that $\|Qg_E\|_{X_\alpha}\leqslant \ee$ for all $\varnothing\neq E\in\mathcal{S}_{\beta+1}$.   Combining the fact that  $Sz(X^2_\beta)=\omega^{\beta+1}$ with Remark \ref{remark}, the canonical basis of $X_\beta^2$ has no subsequence which is an $\ell_1^{\beta+1}$ spreading model.  Since the canonical basis of $X_\beta^2$ is shrinking and normalized, it is weakly null. Since the basis is weakly null and has no subsequence which is an $\ell_1^{\beta+1}$-spreading model, for any $\delta>0$, it follows from \cite[Theorem $4.12$]{CN} that there exists $L\in[\nn]$ such that for any $N\in[L]$, $$\Bigl\|\sum_{i=1}^\infty \mathbb{S}^{\beta+1}_{N, 1}(i)e_i\Bigr\|_{X_\beta^2}<\delta.$$  Let $n_1=\min L$ and recursively choose $n_1<n_2<\ldots$, $n_i\in L$, such that if  $1<t\in\nn$ is such that $(n_1, \ldots, n_t)\in \mathcal{S}_{\beta+1}$, then $\max \ran(g_{(n_1, \ldots, n_{t-1})})<n_t$.   Since $\mathcal{S}_{\beta+1}$ contains no infinite, $\preceq$-ascending chains, there exists $t\in \nn$ such that $E:=(n_i)_{i=1}^t$ is the maximal initial segment of $N$ which lies in $\mathcal{S}_{\beta+1}$. By the properties of the repeated averages hierarchy, $E=\{i: \mathbb{S}^{\beta+1}_{N,1}(i)\neq 0\}$ and $\sum_{i=1}^t \mathbb{S}^{\beta+1}_{N,1}(n_i)=1$.     Let $$g=\sum_{i=1}^t \mathbb{S}^{\beta+1}_{N,1}(n_i)g_{(n_1, \ldots, n_i)}\in \text{co}(g_F: \varnothing\prec F\preceq E).$$  Our goal is to estimate $\|g\|_{G_{\alpha, \beta}}$. For convenience, for $i=1, \ldots, t$,  let $g_i=g_{(n_1, \ldots, n_i)}$  and $w_i=\mathbb{S}^{\beta+1}_{N,1}(n_i)$. Then $g=\sum_{i=1}^t w_ig_i$.

We fix intervals $I_1<I_2<\ldots$ such that $$\|g\|_{G_{\alpha, \beta}} = \Bigl\|\sum_{n=1}^\infty \|QI_ng\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X^2_\beta}.$$  By including additional $I_n$ if necessary and using $1$-unconditionality of the canonical $X_\beta^2$ basis, we may assume that $\nn=\cup_{n=1}^\infty I_n$.    For each $n\in\nn$, let $A_n=\{i\in \{1, \ldots, t\}: \ran(g_i)\subset I_n\}$ and $A=\cup_{n=1}^\infty A_n$, $B=\{1, \ldots, t\}\setminus A$.   Then using the pairwise disjointness of $A_1, A_2, \ldots$ together with the fact that $\|Qg_i\|_{X_\alpha}\leqslant \ee$ for each $1\leqslant i\leqslant t$ and $I_ng_i=g_i$ for all $n\in\nn$ and $i\in A_n$, \begin{align*} \|g\|_{G_{\alpha, \beta}} & \leqslant \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in A}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} + \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} \\ & \leqslant \sum_{n=1}^\infty \sum_{i\in A_n} w_i\|QI_ng_i\|_{X_\alpha}+ \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} \\ & = \sum_{n=1}^\infty \sum_{i\in A_n} w_i\|Qg_i\|_{X_\alpha}+ \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} \\ &\leqslant \ee+ \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta}. \end{align*} If $B=\varnothing$, then the second sum here is zero and we are done.  Assume $B\neq \varnothing$. Since $\nn=\cup_{n=1}^\infty I_n$, it follows that for each $i\in B$, there exist at least two values of $n\in\nn$ such that $I_ng_i\neq 0$. Enumerate $B=(b_1, \ldots, b_s)$ and let $B_1=(b_i: i\leqslant s, i\text{\ odd})$ and $B_2=(b_i: i\leqslant s, i\text{\ even})$.    For $i\in B$, let  $$C_i=\{n\in\nn: I_n g_i\neq 0\}.$$  Note that the sets $(C_i)_{i\in B_1}$ are pairwise disjoint, as are the sets $(C_i)_{i\in B_2}$. To see this, note that if $b_i, b_j\in B_1$ with $i<j$ and $n\in C_{b_i}\cap C_{b_j}$, then since $i,j$ are both odd, $b_i<b_{i+1}<b_j$, and  $$\ran(g_{b_{i+1}})\subset (\max \ran(g_{b_i}), \min \ran(g_{b_j}))\subset I_n.$$ But this means $b_{i+1}\in A_n$,  contradicting the fact that $b_{i+1}\in B=\{1, \ldots, t\}\setminus \cup_{m=1}^\infty A_m$. A similar argument yields that $(C_i)_{i\in B_2}$  are pairwise disjoint.

We now turn to estimating $$\Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B_1} w_ig_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X^2_\beta} = \Bigl\|\sum_{i\in B_1} w_i \sum_{n\in C_i}\|QI_ng_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X^2_\beta}.$$   For each $i\in B_1$ and $n\in C_i$, let $J_n=\ran(g_i)\cap I_n\neq \varnothing$.  Let $h_i=\sum_{n\in C_i} \|QJ_n g_i\|_{X_\alpha} e_{\min J_n}$ and $h'_i=\sum_{n\in C_i}\|QI_n g_i\|_{X_\alpha}e_{\min I_n}$. It follows from this definition that $\|h_i\|_{X_\beta^2}\leqslant \|g_i\|_{G_{\alpha, \beta}}\leqslant 1$. Moreover,  $\ran(h_i)\subset \ran(g_i)$. For each $i\in B_1$, $\|QJ_ng_i\|_{X_\alpha}=\|QI_ng_i\|_{X_\alpha}$ and $\min J_n\geqslant \min I_n$ for each $n\in C_i$. Also, by our choice of $n_1, \ldots, n_t$,  \begin{align*} n_1 & \leqslant \min \ran(g_1)=\min \ran(g_{(n_1)}) < n_2 \leqslant \min \ran(g_2)=\min \ran(g_{(n_1, n_2)}) <\ldots \\ & <n_t \leqslant \min \ran(g_t)=\min \ran(g_{(n_1, \ldots, n_t)}).\end{align*}  Therefore by the definition of $h_i'$, $1$-right dominance, Proposition \ref{important}$(v)$, the properties of $\mathbb{S}^{\beta+1}_{N,1}$, and our choice of $N$, \begin{align*}\Bigl\|\sum_{i\in B_1} w_i \sum_{n\in C_i}\|QI_ng_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X^2_\beta} & = \Bigl\|\sum_{i\in B_1} w_i h'_i\Bigr\|_{X_\beta^2} \leqslant \Bigl\|\sum_{i\in B_1} w_ih_i\Bigr\|_{X^2_\beta} \leqslant 4\Bigl\|\sum_{i=1}^t w_ie_{n_i} \Bigr\|_{X^2_\beta} \\ & = 4\Bigl\|\sum_{i=1}^t \mathbb{S}^{\beta+1}_{N,1}(n_i)e_{n_i}\Bigr\|_{X^2_\beta} = 4\Bigl\|\sum_{i=1}^\infty \mathbb{S}^{\beta+1}_{N,1}(i)e_i\Bigr\|_{X^2_\beta} \\ & <4\delta. \end{align*}

An identical argument yields that \begin{align*} \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B_2} w_ig_i\|_{X_\alpha}e_{\min I_n}\Bigr\|_{X^2_\beta} <4\delta. \end{align*} Therefore \begin{align*} \|g\|_{G_{\alpha, \beta}} & \leqslant \ee+ \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} \\ &  \leqslant  \ee+ \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B_1}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} + \Bigl\|\sum_{n=1}^\infty \|QI_n \sum_{i\in B_2}w_ig_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X^2_\beta} \\ & < \ee+8\delta.\end{align*} Since $\delta>0$ was arbitrary, we are done.

$(ii)$ By \cite[Proposition $5$]{OSZ}, it is sufficient to show that for any $\ee>0$ and any weakly null tree $(h_E)_{E\in \mathcal{S}_{\beta+1}\setminus \{\varnothing\}}\subset B_{H_{\alpha, \beta}}$, there exist $E\in \mathcal{S}_{\beta+1}$ and $h\in \text{co}(h_F: \varnothing \prec F\preceq E)$ such that $\|h\|_{G_{\alpha,\beta}}<3\ee$.     Fix such a weakly null tree. By standard perturbation and pruning arguments, we may assume there is a block tree $(g_E)_{E\in \mathcal{S}_{\beta+1}}\subset B_{G_{\alpha, \beta}}$ such that $\|g_E-h_E\|_{G_{\alpha, \beta}}<\ee$ for all $E\in \mathcal{S}_{\beta+1}\setminus \{\varnothing\}$.  This yields that for each $E\in \mathcal{S}_{\beta+1}\setminus \{\varnothing\}$, $$\|Qg_E\|_{X_\alpha} =\|Qg_E-Qh_E\|_{X_\alpha} \leqslant \|g_E-h_E\|_{X_\alpha} \leqslant \ee.$$   By $(i)$, we may choose $\varnothing \neq E\in \mathcal{S}_{\beta+1}$ and non-negative numbers $(w_F)_{\varnothing\prec F\preceq E}$ such that $\sum_{\varnothing\prec F\preceq E} w_F=1$ and $\|\sum_{\varnothing\prec F\preceq E}w_F g_F\|_{G_{\alpha, \beta}}<2\ee$.   Then $$\Bigl\|\sum_{\varnothing \prec F\preceq E}w_Fh_F\Bigr\|_{G_{\alpha, \beta}} \leqslant \Bigl\|\sum_{\varnothing \prec F\preceq E}w_Fg_F\Bigr\|_{G_{\alpha, \beta}} + \sum_{\varnothing\prec F\preceq E}w_F\|h_F-g_F\|_{G_{\alpha, \beta}} <3\ee.$$   This finishes $(ii)$.

$(iii)$  Fix $0<\ee<1$.   We can recursively select positive integers $p_1<q_1<p_2<q_2<\ldots$ such that $\|f_{p_i}-f_{q_i}\|_{X_\alpha}<1-\ee$ for all $i\in\nn$. Since $Q$ is a quotient map, we can choose $(g_i)_{i=1}^\infty\subset B_{G_{\alpha, \beta}}$ such that $\|g_i\|_{G_{\alpha, \beta}}<1-\ee$ and $Qg_i=f_{q_i}-f_{p_i}$. Therefore  $h_i := e_{p_i}-e_{q_i}-g_i\in H_{\alpha, \beta}$.  By the triangle inequality, $\|h_i\|_{G_{\alpha, \beta}}\leqslant 3$.           Fix  $E\in \mathcal{S}_\beta$ and scalars $(b_i)_{i\in E}$ such that $\sum_{i\in E}|b_i|=1$ and note that  $(p_i:i\in E)\in \mathcal{S}_\beta$, since $\mathcal{S}_\beta$ is spreading. Let  \begin{displaymath}
   a_j = \left\{
     \begin{array}{lr}
       b_i & : j=p_i \text{\ for some\ }i\in E\\
       -b_i & : j=q_i \text{\ for some\ }i\in E\\
			0 & : \text{\ otherwise}.
     \end{array}
   \right.
\end{displaymath}  Let $I_i=\{p_i\}$ for $i\in E$ and note that \begin{align*} \Bigl\|\sum_{i\in E} b_i (e_{p_i}-e_{q_i}-g_i)\Bigl\|_{G_{\alpha, \beta}} & \geqslant \Bigl\|\sum_{i\in E} b_i (e_{p_i}-e_{q_i})\Bigl\|_{G_{\alpha, \beta}}-(1-\ee) \geqslant \Bigl\|\sum_{i \in E} \|\sum_{j\in I_i}a_jf_j\|_{X_\alpha} e_{p_i}\Bigr\|_{X_\beta^2} -(1-\ee) \\ & = \Bigl\|\sum_{j\in E} |b_i|e_{p_i}\Bigl\|_{X^2_\beta}-(1-\ee)= 1-(1-\ee)=\ee.\end{align*}   This yields that $(h_i)_{i=1}^\infty$ is an  $\ell_1^\beta$-spreading model.  By Remark \ref{remark}, $Sz(H_{\alpha, \beta})>\omega^\beta$, and $Sz(H_{\alpha, \beta})\geqslant \omega^{\beta+1}$ by \cite[Proposition $3.3$]{Lancien}.   Combining this estimate with  $(ii)$, we deduce that  $Sz(H_{\alpha, \beta})=\omega^{\beta+1}$.

\end{proof}

\begin{proof}[Proof of Theorems \ref{main1} and \ref{main2}] Fix $0<\xi<\omega_1$ such that $$\xi\notin \{\omega^\gamma: \gamma<\omega_1\}.$$  By standard properties of ordinals, there exist $\alpha, \beta<\xi$ such that $\alpha+\beta=\xi$. Let $G=G_{\alpha,\beta}$ and $H=H_{\alpha,\beta}$.

 Note that $G/H\approx X_\alpha$, so $Sz(G/H)=Sz(X_\alpha)=\omega^{\alpha+1}\leqslant \omega^\xi$. Here we are using the fact that  the Szlenk index is an isomorphic invariant (see \cite{Lancien}).  Since $G/H\approx X_\alpha$, $G/H$ admits no $\ell_1^{\alpha+1}$-spreading model. Since $\alpha+1\leqslant \xi$, as noted prior to Remark \ref{remark}, $G/H$ does not admit an $\ell_1^\xi$-spreading model.   It follows from Lemma \ref{finish} that $Sz(H)\leqslant \omega^{\beta+1}\leqslant \omega^\xi$ and that $Sz(H)$ admits no $\ell_1^{\beta+1}$-spreading model.  Since $\beta+1\leqslant \xi$, $H$ admits no $\ell_1^\xi$-spreading model.  By Remark \ref{remark}, we will be done once we show that $G$ admits an $\ell_1^\xi$-spreading model. Since $\alpha+\beta=\xi$, we will be done once we show that $G$ admits an $\ell_1^{\alpha+\beta}$-spreading model.

Fix $0<\ee<1$ and choose positive integers $m_1<m_2<\ldots$ such that for all $i\in\nn$, $\|f_{m_i}-e_i\|_{X_\alpha}<1-\ee$. Fix $E\in\mathcal{S}_\beta [\mathcal{S}_\alpha]$ and write $E=\cup_{n=1}^m E_n$ with $E_1<\ldots <E_m$, $\varnothing \neq E_n\in \mathcal{S}_\alpha$, $(\min E_n)_{n=1}^m \in \mathcal{S}_\beta$.  Let $F=\{m_i: i\in E\}$.    For each $1\leqslant n\leqslant m$, let $I_n$ be the smallest interval containing $\{m_i: i\in E_n\}$. Note that $(\min I_n)_{n=1}^m$ is a spread of $(\min E_n)_{n=1}^m$, since $\min I_n=m_{\min E_n}$, from which it follows that $(\min I_n)_{n=1}^m\in \mathcal{S}_\beta$.   Fix  scalars $(a_i)_{i\in E}$ such that $\sum_{i\in E}|a_i|=1$ and let $b_{m_i}=a_i$ for $i\in E$ and $b_i=0$ for $i\in\nn\setminus F$. Then  \begin{align*} \Bigl\|\sum_{i\in E} a_ie_{m_i}\Bigr\|_{G_{\alpha,\beta}}  & = \Bigl\|\sum_{i=1}^\infty b_ie_i\Bigr\|_{G_{\alpha,\beta}} \geqslant \Bigl\|\sum_{n=1}^m \|\sum_{i\in I_n} b_if_i\|_{X_\alpha} e_{\min I_n}\Bigr\|_{X_\beta^2} =\sum_{n=1}^m \Bigl\|\sum_{i\in I_n} b_if_i\Bigr\|_{X_\alpha} \\ & = \sum_{n=1}^m \Bigl\|\sum_{i\in E_n} a_if_{m_i}\Bigr\|_{X_\alpha}   \geqslant  \sum_{n=1}^m \Bigl\|\sum_{i\in E_n} a_ie_i\Bigr\|_{X_\alpha} -\sum_{i\in E} |a_i|\|f_{m_i}-e_i\|_{X_\alpha} \\ & = \sum_{n=1}^m \sum_{i\in E_n}|a_i| - \sum_{i\in E} |a_i|\|f_{m_i}-e_i\|_{X_\alpha}  > 1-(1-\ee)=\ee. \end{align*} Therefore by Remark \ref{remark}, $(e_{m_i})_{i=1}^\infty$ has a subsequence which is a an $\ell_1^{\alpha+\beta}$-spreading model.  By the criteria established at the end of the preceding paragraph, we are done.

\end{proof}

redpajama_set_name: RedPajamaArXiv


With all the pureed pumpkin I have in the freezer, I'm always looking for new and easy ways to use it. My goal is that by the end of November, my skin will have a nice orange quality to it. I just love that orange skin look—hearkens me back to the wonderful tanning booth days. Mmm…what a high point that was in my life.
One Thanksgiving I was given a jar of the most delicious pumpkin butter. OH, was it ever wonderful. I spent an entire week toasting english muffins and sliced sandwich bread, just so I could keep eating the pumpkin butter. It made my whole holiday extra wonderful.
This is a play on that creation, with a couple of twists. First, this pumpkin butter is not heat processed, so the pumpkin maintains its bright orange hue. Second, it contains the wonderful, the marvelous plain, nonfat yogurt, which gives it a nice injection of calcium-rich dairy, as well as a little bit of tangy creaminess. Finally, it takes about a minute to mix together. What could be wrong with that?
Add in 1 cup of the pumpkin you just pureed.
1 teaspoon (more if you desire) pumpkin pie spice.
And 1 heaping cup of powdered sugar.
Stir together well and give it a taste. If you like things spicier, add more spice. If it tastes like it could use more sweetness, add more powdered sugar. And if you need more pumpkin flavor…you know what to do.
I like to store my pumpkin butter in a pretty jar. It makes me feel more special, which gets me through the day.
Pretty jars make everything better.
Of course this needs to be stored in the fridge.
But it'll keep in there for a good five or six days. And if you have a houseful of holiday guests, there's no way it's lasting that long.
Really slather it on, too. And here's a tip: the warmer the bread, the more delicious it is. The pumpkin butter starts to seep into the melting butter and warm bread and ohhhhhhhh, is it ever good.
If you don't believe me, just look at the evidence.
Stir all ingredients together well. Store in the refrigerator; will keep for five to six days.
This looks delicious! I always used to get amazing pumpkin butter from a cider mill in VT (where I went to college). I'll definitely try this recipe.
Ree, you always have the prettiest plates! Where are these from?
And, where do you store all your China!?
That looks interesting. Never had it, but I'm thinking I need to give it a try! Thanks for sharing!
k j/k…. not #1… haha looks yummy!
Oh my word. I need that and I need it now.
Mmm that looks delicious. I'm definitely planning on making some pumpkin puree soon and I will definitely have to make this too.
Question: Is it possible to freeze this? I know there is yogurt in it and i've haven't had any luck with freezing yogurt.
Yum, Yum….I gotta get busy. Really digging the plate you used in the post. So pretty.
that looks wonderful – i have 2 pumpkins left from those i've grown this year – i'll have to give this a go.
Can't wait to try this. Looks super yummy!
That looks great! I'd love to try this!!!!
Ree, there are people that believe that grass is alive and screams when you cut it (something about gases being released).
How's that for making your hiney cringe?
Yum yum yum. Never made my own pumpkin butter, but will give it a go! One of my favorite ways to use pumpkin puree is to mix it with yogurt or fat free sugar free pudding and a little whipped cream if I'm feeling naughty. Delicious and easy dessert!


text:
What's On In Sydney This Week, This Month And Beyond
By Olivia Gee
Dial your mates because Casa Aperol is taking over the Promenade at Bondi Beach every day for summer sips, Mediterranean fare and live tunes—no need to book. Click here for details.
Sydney's 2023 summer of cultural rejuvenation is sure to be a tonic for anyone whose party battery is running on red after a manic festive season. January's artistic highlight is, of course, Sydney Festival, while two weeks of Lunar New Year celebrations will bring hope, prosperity, and joyous community feasts with plenty of pop-ups, exhibitions and parties in between.
Check out the best things to do in Sydney this week, this month, and beyond.
Keep scrolling for:
What's On In Sydney This Week
What's On In Sydney In January
What's Coming Up In Sydney In 2023 And Beyond
Stargazing Dinner
BLAQ Restaurant, Blackheath | Tuesday 31 January, 7pm–10:30pm
Get celestial with Blue Mountains Stargazing's latest curated evening. You'll hit gorgeous hotel Kyah's restaurant, BLAQ, for an astronomical cocktail on arrival followed by a three-course meal by talented chef Mate Herceg. Keep your fingers crossed for clear skies, as the evening will include a naked-eye stargazing tour and professional-grade telescope experience guided by a local astrophysicist to catch glimpses of the rings of Saturn, the moons of Jupiter, and the Orion nebula. Bookings are $160pp (or $80 for children) and can be made here.
Sydney Lunar Festival
Around Sydney | Till Sunday 5 February
According to the Lunar calendar, 2023 is predicted to be a year of hope. And if Sydney's packed calendar of Lunar New Year celebrations is anything to go by, we've certainly got a lot to look forward to. Sydney Lunar Streets in Haymarket will kick off the festivities with a joyful sunset street party with live entertainment and every snack imaginable before the renowned dragon boat races sluice through the waters of Cockle Bay Wharf. Community groups will share traditional performances and contemporary acts for the Lunar spectacular show (4 February) at Centennial Hall, and there are loads of art pieces, workshops, and pop-ups to explore. Check out all the free and ticketed events for the Sydney Lunar Festival.
Disney At The Grounds
The Grounds of Alexandria | Till 30 April
Celebrate 100 years of Disney magic with The Grounds' latest magical walk-through pop-up. Disney at The Grounds will see the huge site's laneways decked out with montages, photo opportunities, art pieces and large-scale installations sharing the rich legacy of storytelling that Disney has fostered since 1923. You do have to be visiting The Grounds venues The Cafe (open 7am–4pm) or The Potting Shed (open 11:30am–late), or attending a function to visit the activation, and you can enjoy Disney-themed treats like an iridescent lime zest sponge filled with caramelised pineapple and topped with white chocolate ganache and vanilla buttercream. Read more about the pop-up and plan a visit here.
WORD MADE FLESH
Carriageworks | Till Saturday 26 February
Paul Yore creates a new world on a foundation of queer experience in WORD MADE FLESH. In an almost overpowering cacophony of found objects, pulsing neon lighting, sculpture, collage, painting, and video, the stirring multidisciplinary artist explores the interaction of religious and artistic histories with queer identity, pop culture, and neo-liberal capitalism. The outcome is an immersive installation on an architectural scale that celebrates the fluidity of identity and sexuality, while juggling political contradictions. You can visit this alternate reality kaleidoscope for free throughout the Sydney Festival and beyond.
Year Of The Rabbit With Lotus
All Lotus Venues | Till Saturday 5 February
For those celebrating the Lunar New Year, it is a time to honour ancestors and come together with family to celebrate the abundance of life, including the great cultural definer: food. Lotus Dining Group is adding to the 15 days of feasting and celebrations with specialised dishes to recognise the holiday and the Year of the Rabbit. All seven of their Sydney venus will serve up a delightful rabbit-shaped dumpling packed with texture and flavour led by mud crab, prawn and ginger. Other new dishes and special banquets will appear on the menus of select Lotus venues, and you'll also be able to watch lion dancing while you dine at Lotus Barangaroo, Lotus The Galeries, or The Gardens by Lotus. Find menus and pricing here.
Westpac OpenAir Cinema
Mrs Macquaries Point | Till Tuesday 21 February
Favourite summertime pop-up Westpac OpenAir is returning in 2023 with a packed season of must-see films, launching with Steven Spielberg's coming-of-age drama The Fablemans on Sunday 8 January. This year, chef Mike McEnearney will be bringing an elevated menu of fish and chips, burgers, wraps and salads to the pop-up for general admission diners, with premium dining upgrades offering full table service available alongside a cocktail bar and DJ tunes. Films on the lineup include Tar starring Cate Blanchett, She Said with Carey Mulligan, and Don't Worry Darling. Tickets are on sale from $36 here.
Entertainment Quarter | Till Sunday 26 February
Van Gogh Alive made a huge splash when it debuted in Sydney in 2020. An immersive exhibition where Gogh's seminal swirls, stars, sunflowers, snowflakes and moonshine drift and dance across walls, just as you imagine the painter would have seen them. If you missed it last time, you have the op to catch a new and improved version of Van Gogh Alive in all its swirly, moving glory, with this year's version housed in a purpose-built venue that offers extra rooms and experiences to the original installation. The multi-sensory exhibition features some 3,000 images beamed across walls, floors, ceilings, and more. The projections move in time to music and specifically designed scents fill each space—it's an experience of timeless artwork akin to diving into Van Gogh's paintings. Learn more and grab tickets from $39 here.
Casa Aperol At The Promenade, Bondi Beach
Bondi Promenade, Bondi Beach | Till Friday 31 March
Casa Aperol is back and this time it's bringing good vibes every day, all summer long. This year's iteration is the result of Aperol and House Made Hospitality (founders of Hinchcliff House) coming together to create an immersive summer pop-up experience that will transport you straight to the Amalfi Coast. Taking over the Promenade at Bondi Beach right out the front of the Pav, the Mediterranean oasis is the ultimate hang for you and your pals to sip a refreshing spritz at, while blissing out to live beats from a range of acts. Head over here for all the details.
Sunset Sessions
Cockatoo Island | Saturdays 5pm–8pm Till 1 April
Stunning historic site Cockatoo Island is hosting a weekly live music series on its gorgeous elevated Biloela Lawn, providing the vibes for some stunning summer sunsets. The picnic-style gigs welcome Aussie musicians from a wide range of genres and include First Nations and LGBTQIA+ talent—you can see the weekly lineup here. Grazing boxes can be pre-ordered with ticket purchases, and the bar will be serving drinks and snacks. Learn more and book tickets for $35pp here.
Art Gallery Cinema: Badlands
Art Gallery Of NSW | Till Sunday 2 April
AGNSW's Art Gallery Cinema is back for 2023, presenting a program of rebellious films from around the world taking inspiration from our new exhibition Outlaw. The film series, called Badlands, explores cinema's outlaw zones, celebrating unruly characters, boundary riders and rule breakers (both on- and off-screen). The lineup, curated by AGNSW curator of film Ruby Arrowsmith-Todd, includes Yojimbo (1961, Japan), Bonnie and Clyde (1967, US), and martial arts films Dragon Inn (1967, Taiwan), The 36th chamber of Shaolin (1978, Hong Kong), and Echo 8 (2022, Australia)—Australia's first female-led martial arts film.
Badlands will take place on Wednesdays and Sundays in the South Building's Domain Theatre—tickets are free and can be reserved here, and an allocation of door tickets will be available for each session. See the full program here.
The Wharf Theatre | Till Thursday 4 March
While Jane Austen's stirring tales of love and yearning lead the base narrative of Hubris & Humiliation, this cackling stage comedy's etiquette is more reminiscent of Kath & Kim than the Bennets and Darcys. Sydney Theatre replaces Austen's sweeping country estates with the pumping queer scene of Sydney, where country Queensland boy Elliott has been sent to source a wealthy husband and rebuild the family fortune after his mother had a costly run-in with an online scammer. Following instructions from his Kirribilli mansion-owning uncle and his sister who snuck along on the mission, Elliot navigates family, obligation, and love while whisking through a rainbow mile of madcap adventures that take him to Berlin and back again. Grab tickets from $54 to see this fresh take.
Siteworks 2022: From A Deep Valley At Bundanon
Bundanon Museum | Till Sunday 12 March
Collaborations between 25 artists and 10 researchers will address climate change, critical thinking, First Nations knowledge, technology, and creative digital spaces in Siteworks 2022: From a Deep Valley at Bundanon. This will be presented as a major exhibition at Bundanon Art Gallery and a program of outdoor installations, performances, workshops, and digital artworks throughout the site's 1,000 hectares of bush and parkland that overlooks the Shoalhaven River. Expect mixed mediums and ideas incorporating natural materials like oyster shells and algae, artist walk-and-talks, outdoor songs, and rituals all paying homage to the earth. Program tickets will be available from late October, but you can see what else is on at Bundanon here.
Belvedere Amphitheatre, Centennial Park | Till 26 March
Pack the picnic rug for a night under the stars catching your fave flicks, or a new release. Moonlight Cinema is back for another year (and, yep, still dog-friendly), with the option to BYO all your food and drinks (booze included), or pick up some snacks from the pop-ups on site. As well as massive must-see releases like Babylon and Avatar: The Way of Water, there's 'A Week of Romance' running 8–15 February for Valentine's Day (or Galentine's), plus Aussie favourites under the stars including The Castle and Muriel's Wedding, thanks to Oz Lotto's Big Aussie Fun screenings. Check out the full February and March program and book your tickets from $22 here.
Australian Geographic: Our Country
ICC Sydney Exhibition Centre | From Saturday 19 November
Marvel at a multi-sensory showcase of Australia's vast wilderness at this immersive cinematic experience. Our Country brings together natural history filmmakers alongside the stewardship of Elders, celebrated Indigenous musicians, and creative technologists to explore the natural landscapes and forces of Australia. Their work is displayed on over 40 screens up to six metres tall, with a powerful 360-degree soundscape for complete immersion, and will tour globally when its Sydney stint wraps up. Tickets to Our Country are $34.90 for adults, and are available here.
Sunset Cinema North Sydney
North Sydney Oval | Till Saturday 1 April
Sunset Cinema is back for its tenth year, making the most of our long, warm summer evenings. This season's screenings include recent creations like the fantastically unreal yet touching sci-fi Everything Everywhere, All at Once, the less-than-ideal dining experience of The Menu, and the Whitney Houston biopic, I Wanna Dance With Somebody. There are also classic rom-coms for date night like Something About Mary, which pair perfectly with a BYO picnic basket and a tipple from the on-site bar. Nab your tickets, pack a blanket, and settle in for movie night.
Powerhouse Museum | From Thursday 27 October
If the words Beastie Boys and Nirvana still get your head bopping, head over to the Powerhouse Museum for a major throwback to the 90s with its upcoming exhibition, Unpopular. Sifting through the goldmine of music promoter, record company founder, and entrepreneur Stephen "Pav" Pavlovic's archive, the exhibition takes you on a rockin' trip through the behind-the-scenes of alternative music. Feast your eyes (and ears) on footage of live performances, unpublished music demos, and original art pieces including new drops by artist and filmmaker Julian Klincewicz and artist Lillian O'Neil. Kurt Cobain fans will even get to see the legend's very own Martin guitar. But wait—there's more. Fancy yourself an artist yourself? Try your hand at the music photography masterclass with Sophie Howarth and analogue art workshops headed by designers Ben Brown and Paul Curtis. For the full program, head over here.
MCA Sydney International Art Series—Do Ho Suh
Museum Of Contemporary Art | From Friday 4 November
Internationally acclaimed South Korean-born and London-based artist Do Ho Suh is celebrating his first large-scale solo exhibition in the Southern Hemisphere at the Museum of Contemporary Art. The renowned artist's survey spans three decades of his work, which typically explores complex relationships between the body, memory, and space through a wide range of mediums. Many of these artworks can be walked around, through, and within. Through these conceptions, notions of identity, belonging, and representations of home will be provocatively evoked and encourage personal introspection. Find more on the exhibition and artist here.
Zampatti Powerhouse
Powerhouse Museum Ultimo | Till Sunday 11 June
Marvel at the work and professional ascension of Australia's longest-working fashion designer, Carla Zampatti. More than 100 pieces, including a host of garments from Zampatti's fashion archive, personal items from her estate, and pieces donated by well-known women reflect the life of a design legend who constantly produced and adapted an elegant aesthetic for more than 50 years. Among the decades of fashion moving from Zampatti's youthful zest in the '60s and '70s through to more recent, understated pieces, you'll find two audio and video installations. These feature the star in intimate conversations with friends, clients, and business leaders, as well as interviews with those who have strutted Zampatti's style around Australia. Entry to this fashion time capsule is free.
Sydney Observatory Nights
Sydney Observatory | Various Dates Until 22 June
Sydney Observatory is gearing up to bring the stars and sky to you. The city's star-studded venue has assembled a line-up of programs featuring renowned astronomers, live tunes, and shows by leading NSW producers. For more information, head over here.
Coming Up In Sydney In 2022 And Beyond
Roslyn Packer Theatre | Friday 3 February–Saturday 18 February
Eryn Jean Norvill will again grace Sydney with her epic shape-shifting performance in this widely-adored contemporary production of The Picture of Dorian Gray. Sydney Theatre Company Artistic Director Kip Williams truly masters the melding of cinema and theatre (coining the 'cine-theatre' genre) to enable Norvill's career-defining portrayal of all 26 characters that appear in the Gothic tale. Live and pre-recorded audio visuals build a world around the metamorph as she and an immensely talented tech crew weave the audience and our modern-day struggles into Oscar Wilde's fable of beauty, projected perfection, and devil's deals. Secure your seats for the limited February run before this modern marvel heads out to conjure theatre magic on international stages.
Greek Summer Festival
Carss Bush Park, Carss Park | Sunday 12 February, 11am–9pm
Immerse yourself in Greek culture (and, more importantly, flavours) at the annual Greek Summer Festival. Expect authentic entertainment including live bouzouki and instrumental shows, traditional dancing, and cooking demonstrations, including from noted Greek chef David Tsikeras who'll be tending to Australia's largest Greek barbecue. Hundreds of food stalls will be serving up the best and most beloved Greek eats including loukoumades, souvlaki and sheftalies, and Greek sweets like kouloria and creamy risogalo. There's more: there'll be donkey rides for kids and a massive fireworks display to traditional Greek music to round out the day's celebrations. Entry is completely free, and you can check out the full program here.
KPMG Australia Sail Grand Prix
Sydney Harbour | Saturday 18 and Sunday 19 February
Don't miss the close-to-shore action of the ninth event of SailGP Season 3, which will see Team Australia, with Olympic medallist Tom Slingsby in the driver's seat, defend a significant lead over Switzerland, Canada, Denmark, France, Great Britain, New Zealand, Spain and the United States. The close-quarters racing around the harbour can be observed from a range of free and ticketed spectator spots on land and water, including the free SailGP Village in Barangaroo with a pop-up bar and DJs, Genesis Island (Shark Island) with gourmet picnic hampers and a premium open bar, or luxury yachts in the public exclusion zone. You can check out the range of ticketing options and get in on the action here.
Bell Shakespeare's Macbeth
Sydney Opera House | Saturday 25 February–Sunday 2 April
Australia's esteemed company dedicated to all things The Bard, Bell Shakespeare is tackling Macbeth in its signature style in 2023. This staging of the gripping tale of humanity, loyalty and murder features Logie award-winning Hazem Shammas in the title role with Jessica Tovey (The Miser, The Merchant of Venice) as the formidable Lady Macbeth. Tickets are available here from $45.
All About Women
Sydney Opera House | Saturday 11–Monday 13 March
Celebrate extraordinary thinkers, writers and artists while pondering vital questions about gender, equality and justice at All About Women's 11th program. Co-curated by Jamila Rizvi (she/her), Dr Amy Thunig (she/they), Jane Caro (she/her), and the Sydney Opera House Talks and Ideas team, led by Chip Rolley (he/him), the festival will honour women and girls who fight to be heard – from protesters in Iran, to pro-choicers in the US, to advocates everywhere who defend the silenced. The international lineup includes Jennette McCurdy, Chloe Hayden, Brooke Blurton and Grace Tame (among so many other notable names), and can be experienced in person at the Opera House, or by tuning into live streams. Check out the full program and ticketing options here.
Tina—The Tina Turner Musical
Theatre Royal Sydney | Thursday 4 May–Sunday 3 September 2023
Celebrate the queen of rock'n'roll in this musical extravaganza that's flying in direct from Broadway. It explores Turner's ongoing resilience in the face of inequalities tied to race, gender, and age, while displaying her talent in full force across her repertoire of hits. The Theatre Royal Sydney will be rocking to songs that have become ingrained in popular culture around the globe like 'Proud Mary', 'Nutbush City Limits', and 'What's Love Got to Do With It'. Tickets are available now for simply the best 2023 experience.
Sydney Lyric | From May 2023
Here we go again! The poptastic musical smash Mamma Mia! is returning to Sydney stages in 2023. Dancing queens and super troupers have shimmied to the box office in droves around the globe to watch the cinematic tale play out live thanks to an Aussie creative team. In this re-staging of the 2017 production, we head back to the idyllic Greek island to watch the wedding, the father-finding mission, the mother-daughter bonding, and the 22 ABBA classics belted out to a glowing backdrop of '70s nostalgia. Tickets are on sale now, so get your money, money, money and book a seat.
Sydney Lyric | From August 2023
What is this feeling? It's the exhilarating news that Wicked is returning to our Emerald City. It's been almost two decades since the musical's Broadway premiere and ten years since the beloved witches Glenda and Elphaba belted out ballads as rivals and then friends in Australia, so gravity certainly isn't keeping fans down with this magic in the air. The production lands in the Lyric Theatre in August 2023, featuring musical staging by Wayne Cilento with original direction by Joe Mantello and production in Australia by John Frost for Crossroads Live. Join the waitlist for news and ticket offers.
For even more things to do in Sydney, head over here.
Image credit: Sydney Festival, Blue Mountains Stargazing, The Grounds of Alexandria, Lotus Dining Group (Alana Dimou), Cockatoo Island, Blue Mountains Stargazing, Yia Yia's Honey Balls, SailGP, Casa Aperol, Van Gogh Alive, MCA


BV-107II-28 Canadian Army basic transport version, similar to CH-46A; 12 built. Two lost in accidents. Ten survivors transferred to Canadian Armed Forces upon unification, reserialled and converted to SAR role. Two further machines lost in accidents. Eight modified under SARCUP program, served until retirement in 2002, when stored. Four sold to Columbia Helicopters International, Oregon, USA.


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text:
MATCH REPORT: Thirteen-man Carlow fail to get past Waterford hurlers in Fraher Field
Waterford 3-15 Carlow 0-10
Source: Carlow GAA
Thirteen-man Carlow failed to get past Waterford on their away trip to Fraher Field after the Carlovians were beaten by fourteen points in Round Three of the Allianz National Hurling League.
A goal for each member of Waterford's full-forward line of Thomas Ryan, Stephen Bennett and Shane Bennett laid the foundations for the Déise men's victory.
Carlow, who grabbed the headlines with a home draw to Galway in the previous round, struggled to get going and suffered a blow to their chances just before half-time with Martin Kavanagh limping out of action after 33 minutes.
Waterford players held their feet better and were in the lead 2-07 to 0-07 at the break.
Shane Bennett turned goalscorer in the 44th minute and there were more woe for Carlow after substitute Diarmuid Byrne was sent off for a reckless pull after 52 minutes.
With the advantage of the extra man, Waterford eased their way to the full-time whistle.
It took Carlow twenty five minutes to register their first score of the second when substitute Chris Nolan landed a point after 60 minutes.
There was further misery for Colm Bonnar's side when they had a second player, Ger Coady, sent to the line in the 62nd minute for a second bookable offence.
It ended on a full time score of: Waterford 3-15 Carlow 0-10
Waterford: Billy Nolan, Shane McNulty, Conor Prunty, Noel Connors, Austin Gleeson , Michael Walsh, Jordan Henley, Jamie Barron (0-02), Colin Dunford (0-01), D.J. Foran (0-01), Mikey Kearney (0-01), Jack Prendergast, Thomas Ryan (1-01), Stephen Bennett (1-08, 0-07f), Shane Bennett (1-00).
Subs: Callum Lyons (0-1) for Jordan Henley (48), Shane Ryan for Thomas Ryan (52), Michéal Harney for Colin Dunford (59), Stephen Roche for Shane Bennett (64m).
Carlow: Damien Jordan, Michael Doyle, Alan Corcoran, Gary Bennett, Ger Coady, David English, Richard Coady (0-01), Jack Kavanagh, Jack Murphy, John Michael Nolan, Ross Smithers, Seamus Murphy, James Doyle, Denis Murphy (0-02,0-01f), Martin Kavanagh (0-04,0-03f).
Subs: Edward Byrne for Martin Kavanagh (33), Chris Nolan (0-1) for John Michael Nolan (37m), Diarmuid Byrne for Seamus Murphy (45m), Ted Joyce for Gary Bennett (48), Sean Whelan for Jack Murphy (53).
Referee: John Keenan (Wicklow)


text:
Q: Sorting DataGridView by Column.DisplayMember I have a DataGridView with a few DataGridViewComboBoxColumn's where the actual value is tied to an ID but the DisplayMember is the string counterpart in a lookup table. I'm trying to make it so when I sort by that column then the sort is done based on the DisplayMember, not the ValueMember. I know this was addressed in this question but the answer was less than in depth nor did I understand it.
What I've tried thus far

*

*Binding to the SortCompare event but discovered it isn't fired on a databound column.

*Manually sorting on the ColumnHeaderMouseClick event but rows in a DataGridViewRowCollection are read-only and I can't programatically insert rows (while swapping) on a databound collection.

*Creating a hidden DataGridViewTextBoxColumn where the cells are automatically set to the DisplayMember of the original column then attempting to sort that column instead. However, a databound collection cannot be sorted based on an unbounded column.

Edit: To further clarify: I'm attempting to sort the entire DataGridView based on the DisplayMember of the comboboxes, not sort the comboboxes themselves.
How can I sort a DataGridView based on the DisplayMember of a databound DataGridViewComboBoxColumn?

A: Your third attempt is almost right: What you need to do is create an extra column in your DataSet to store the value of the DisplayMember. Then, you create an invisible bound DataGridViewColumn, and bind it to this extra DataSet column. Then it's a bound column, and you can programmatically sort on it.
The question you've linked to is doing the same, only that solution is generating the display member text in SQL, before it's returned to the application.
You'd think this'd be simple, wouldn't you? :)

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