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Pence will face off against Harris in VP debate described as 'most consequential' in history
Vice President Mike Pence speaks at an event hosted by The Family Leader Foundation, Oct. 1, 2020, in Des Moines, Iowa. (AP/Charlie Neibergall)
(AP/Charlie Neibergall)
Ultimately, the debate is a chance for voters to decide whether Pence and Harris are in a position to step into the presidency at a moment's notice.
The faceoff Wednesday night in Salt Lake City is the most highly anticipated vice presidential debate in recent memory.
Writing in the Wall Street Journal on Tuesday, William McGurn says it may be "the most consequential vice-presidential debate since they began in 1976, when Republican Bob Dole squared off against Democrat Walter Mondale."
It will unfold while President Donald Trump recovers at the White House after testing positive last week for the coronavirus and spending several days in the hospital, a serious setback for his campaign that adds pressure on Vice President Mike Pence to defend the administration's handling of the pandemic.
His Democratic challenger, California Sen. Kamala Harris, offers a starkly different vision for a country confronting escalating crises.
For Harris, the debate is her highest-profile opportunity to vocalize how the Democratic presidential nominee, Joe Biden, would stabilize the U.S., especially when it comes to resolving the pandemic and addressing racial injustice.
She will be able to explain her views on law enforcement, an area in which she's viewed warily by some progressives, given her past as a prosecutor.
It's hardly a theoretical question, particularly as Democratic presidential candidate Joe Biden, 77, shows clear signs of mental decline.
While the debate will likely cover a range of topics, the virus will be at the forefront.
Pence and Harris will appear on stage exactly 12.25 feet (3.7 meters) apart separated by plexiglass barriers. Anyone in the small audience who refuses to wear a mask will be asked to leave.
Pence, who was with Trump and others last week who have since tested positive, has faced questions about whether he should be at the debate at all. The vice president has repeatedly tested negative for the virus, and his staff and doctors insist he does not need to quarantine under Centers for Disease Control and Prevention guidelines.
Read Classified documents at Pence's home, too, his lawyer says
The CDC defines risky "close contact" as being within 6 feet (1.8 meters) of an infected person for at least 15 minutes starting from two days before the onset of symptoms or a positive test.
Pence's team objected to Harris' request for plexiglass barriers, arguing it was medically unnecessary. But the Commission on Presidential Debates had already agreed to the barriers, and Pence's aides said their presence wouldn't dissuade him from attending the event.
Pence's chief of staff, Marc Short, said there is "zero risk" of the vice president pulling out of the debate over the plexiglass spat. He said Pence "will be there" because it's "too important for the American people."
"The hesitancy seems to be on other side," he added.
Sabrina Singh, a spokesperson for Harris, said the senator "will be at the debate, respecting the protections that the Cleveland Clinic has put in place to promote safety for all concerned." The Cleveland Clinic serves as a health adviser to the Commission on Presidential Debates.
Despite the early sparring, the debate is unlikely to be a repeat of the chaotic debate between Trump and Biden last week.
Pence is eager to seize on Harris and Biden's liberal policies, but it may be difficult to shift the conversation away from the Republican administration's uneven handling of the pandemic. Pence serves as chair of the president's coronavirus task force, which has failed to implement a comprehensive national strategy even as Trump himself recovers from the disease and the national death toll surges past 210,000 with no clear end in sight.
Biden has raised questions about whether he will participate in his next scheduled debate with Trump, telling reporters on Tuesday: "I think if he still has COVID, then we shouldn't have a debate."
Read Kushner had 'knock-down screaming matches' with Trump, warning against election fraud claims
The vice president is a 61-year-old former Indiana governor and ex-radio host, an evangelical Christian known for his folksy charm and unwavering loyalty to Trump.
Harris is a 55-year-old California senator, the daughter of a Jamaican father and an Indian mother. She is also a former prosecutor whose pointed questioning of Trump's appointees and court nominees and cool charm on the campaign trail made her a Democratic star.
She will make history as the first Black woman to appear in a vice presidential debate. Democrats hope the historic nature of her candidacy will help energize key groups of likely Democratic voters — African Americans and young people, in particular — who have shown less excitement for Biden.
It's unclear how aggressive the candidates will be with each other.
Both have adopted a cautious approach on the trail, keeping in line with past running mates who, above all, are tasked with not hurting their party's ticket.
Some Harris allies fear that a conservative approach will prevent her from shining.
"Overly scripting Kamala Harris is tantamount to removing five bullets out of her gun before you walk into a gun fight," said Nathan Barankin, who served as Harris' chief of staff in the Senate and when she was California attorney general.
While some Democrats have set high expectations for the debate, Harris and her allies have been trying to keep them low. Last month, when California's state Senate president told Harris on a Zoom call that home state fans were excited to watch her debate, Harris quickly interjected.
"He's a good debater," she said, laughing. "I'm just, I'm so concerned, like I can only disappoint."
Gender will likely play a role in the debate, Hillary Clinton, the first woman to lead a presidential ticket, said during a recent fundraiser.
She suggested Pence would try to paint Harris as "the inexperienced woman candidate." Harris will have to be mindful of the double standard for women in politics as she responds, Clinton said.
Read Biden compares illegal migrants to Jews fleeing Nazi Europe
"She's got to be firm and effective in rebutting any implication that comes from the other side, but to do it in a way that doesn't, you know, scare or alienate voters," Clinton said.
Harris has been preparing for the debate in Salt Lake City since Saturday. Karen Dunn, a Washington lawyer who helped prepare Clinton for her 2016 debates against Trump, is leading Harris' debate preparations.
Harris plans to focus on failures of leadership by the Trump-Pence administration but avoid personal attacks against Trump, as Biden has done since the president was hospitalized for the virus, according to a campaign aide who wasn't authorized to discuss debate planning publicly and spoke on condition of anonymity.
Pence's team has been reluctant to discuss his strategy in debate preparations except to note that he is "thorough" in everything he does.
Pence has spent the last four years defending the president on a near-daily basis and mastered the art of turning Trump's chaotic rhetoric into more palatable, middle-of-the-road commentary.
Aides note that Pence's criticism has focused almost exclusively on Biden and his record instead of Harris. It's likely to stay that way Wednesday, but he has not completely ignored Harris.
In an interview with a conservative talk show host in Iowa last week, the vice president cast the California senator as a left-wing extremist. Pence also said he was "counting the days to the debate."
"In Kamala Harris, you have someone who actually was the most liberal member of the United States Senate in 2019," Pence said.
"While I'm going to go there and make our case to the American people, and I'm going to take the fight to Joe Biden and his agenda, we're also going to make sure people know the record of his running mate, and the positions that she's taken."
2020 U.S. electionsKamala HarrisMike Pence
I found a very interesting article on World Israel News! Click to read this: --> https://worldisraelnews.com/pence-will-faceoff-against-harris-in-vp-debate-described-as-most-consequential-in-history/?ios_app=true


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VOR: Congrats to Simeon Tienpont (NED) and his AkzoNobel team for their Leg 6 win into AKL, after &#
AUCKLAND – For team AkzoNobel and SHK/Scallywag, the buffer they'd built up over the past week was only just sufficient, and they were able to match race down the coast all the way to the finish line in the Waitemata Harbour, usually sailing within hailing distance of each other. In the end, the margin at the finish was just two minutes between first and second place, with less than a half hour separating the top five teams as they finished in the early morning (Feb 28, local time).
"I've never sailed a race like this in my life," said AkzoNobel skipper Simeon Tienpont. "We've always been in each other's sights. They were always there. It's been neck and neck. Huge respect to Scallywag, they never stopped fighting and we never stopped defending. I'm so proud of our crew. They never flinched."
It's the first leg win for team AkzoNobel and earns the team an important 8 points on the leaderboard.
Congrats also to series leader MAPFRE, who had their own match race into Auckland with arch-rival and second-overall Dongfeng Race Team. "It's been a crazy 20 or 21 days, match racing all the way here," said MAPFRE skipper Xabi Fernández. "We've been fighting with Dongfeng like crazy. We never stopped, we fought so hard, and now we feel very happy of course. We were hoping for opportunities in the doldrums and they never came, but finally they came today, and we did it."
Full story via the VOR website here.
Overall standings after Leg 6 into Auckland.

redpajama_set_name: RedPajamaCommonCrawl


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Obama: Brexit from EU may stifle future UK trade deal
Marilyn Malara
British Prime Minister David Cameron and President of the United States Barack Obama wave to the International media at No.10 Downing St, London on April 22, 2016. President Obama is on a three day trip to the UK where he will meet the Queen and other dignitaries during his stay. Photo by Hugo Philpott/UPI | License Photo
President Barack Obama talks with the Duke of Cambridge while the Duchess of Cambridge plays with Prince George; at left First Lady Michelle Obama talks with Prince Harry of Wales, at Kensington Palace in London, on April 22, 2016. White House Photo by Pete Souza/UPI | License Photo
President Barack Obama and British Prime Minister David Cameron walk the West wing colonnade of the White House in Washington, DC on January 15, 2015.The two leaders will attend a working dinner in the Blue Room. FIle Photo by Olivier Douliery/UPI/Pool | License Photo
LONDON, April 23 (UPI) -- President Barack Obama made clear his opposition to a possible British exit from the European Union on Friday, hinting it would dampen relations between the United Kingdom and the United States
Obama, standing alongside British Prime Minister David Cameron to deliver his remarks, said he was expressing his opinion as a friend.
"Speaking honestly, the outcome of that decision is a matter of deep interest to the United States because it affects our prospects as well," the U.S. leader said. "The United Kingdom is at its best when it is helping to lead a strong Europe. It leverages UK power to be part of the European Union."
In a surprisingly direct answer to a journalists comment at the conference, Obama asserted a British exit, or "Brexit," from the EU would place England at "the back of the queue" in terms of a future trade deal with the United States.
RELATED Obama meets 2-year-old Prince George after tending to business in Britain
"I think it's fair to say that maybe some point down the line, there might be a U.K.-U.S. trade agreement, but it's not going to happen anytime soon, because our focus is in negotiating with a big bloc, the European Union, to get a trade agreement done," Obama said. "And the U.K. is going to be in the back of the queue."
In his response, the Obama likened Britain's involvement in the EU to its involvement alongside the United States in such institutions as the United Nations, NATO, Bretton Woods and others.
"That's what we built after World War II," he explained. "That, to some degree, constrained our freedom to operate...But we knew that by doing so, everybody was going to be better off."
RELATED Obama urges Britain to stay in European Union
Early Friday, Obama published a newspaper essay urging the U.K. to remain in the EU. Published in the Daily Telegraph, Obama's op-ed advised voters ahead of June's "Brexit" referendum to vote to stay.
"I will say, with the candor of a friend, that the outcome of your decision is a matter of deep interest to the United States. The tens of thousands of Americans who rest in Europe's cemeteries are a silent testament to just how intertwined our prosperity and security truly are... The U.S. sees how your powerful voice in Europe ensures that Europe takes a strong stance in the world, and keeps the EU open, outward-looking, and closely linked to its allies on the other side of the Atlantic. So the U.S. and the world need your outsized influence to continue – including within Europe," Obama wrote.
London Mayor Boris Johnson, a proponent for the Brexit, said in his own op-ed the president's stance was "downright hypocritical," due to the United States' intense guardianship of its democracy.
RELATED Kerry, 174 leaders sign 'powerful' Paris climate change treaty amid skepticism, protests
Johnson criticized the president's removal of a bust of former British leader Winston Churchill from the Oval Office, adding he suspects Obama's Kenyan ancestry keeps him from entirely supporting British interest.
During the press conference, however, Obama reacted to the accusation, saying "I love the guy," but he felt it "appropriate" to replace the bust with Rev. Dr. Martin Luther King Jr. instead.


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text:
\section{Introduction}

This paper was motivated by the following

\begin{problem}\label{prob0} Which compact Hausdorff spaces are homeomorphic to the Higson compactifications or the Higson coronas of (finitary) coarse spaces?
\end{problem}

Answering this problem we shall prove the following statements:
\begin{itemize}
\item each first-countable compact Hausdorff space with dense set of isolated points is homeomorphic to the Higson compactification of some finitary coarse space;
\item each perfectly normal compact Hausdorff space is homeomorphic to the Higson corona of a suitable finitary coarse space;
\item under CH, each (zero-dimensional) compact Hausdorff space of weight $\le\w_1$ is homeomorphic to the Higson (resp. binary) corona of some cellular finitary coarse space.
\end{itemize}
Also we present simple examples of  compact Hausdorff spaces with countable dense set of isolated points, which are not homeomorphic to the Higson compactifications of finitary coarse spaces.

It should be mentioned that for compact metrizable spaces the ``corona'' part of Problem~\ref{prob0} was resolved by Mine and Yamashita \cite{MY} who proved that each compact metrizable space is homeomorphic to the Higson corona of a suitable locally compact metric space endowed with the $C_0$ coarse structure of Wright \cite{Wri}.

The paper is organized as follows. In Section~\ref{s:def} we recall the necessary information on coarse spaces and present (and characterize) two important classes of coarse spaces: metrizable and finitary. In Section~\ref{s:H}, given a proper metric space $M$ and a coarse space $(X,\E)$, we introduce two compact Hausdorff spaces $h_M(X,\E)$ and $\hbar_M(X,\E)$, called the $M$-compactification and the $M$-corona of the coarse space $(X,\E)$, respectively. The compact spaces $h_\IR(X,\E)$ and $\hbar_\IR(X,\E)$ (resp. $h_{\mathsf 2}(X,\E)$ and $\hbar_{\mathsf 2}(X,\E)$) coincide with the Higson compactification and the Higson corona (resp.  the binary compactification and the binary corona) of the coarse space $(X,\E)$, introduced by Higson \cite{Higson} (resp. Protasov \cite{Prot_bc}, \cite{Prot05}). In Section~\ref{s:P}, for any compact Hausdorff space $K$ we define  a finitary coarse structure $\E_{X,K}$ on the set $X$ of isolated points of $K$ and call the obtained coarse space $(X,\E_{X,K})$ {\em the permutation coarse space} of the compact space $K$. The main result of Section~\ref{s:M} is Theorem~\ref{t:M} saying each metrizable finitary coarse space $(X,\E)$ coincides with the permutation coarse space $(X,\E_{X,\bar X})$ of the Higson compactification $\bar X$ of $X$. In Section~\ref{s:soft} we define the notion of an $\IR$-soft compactification and prove that a compact Hausdorff space $K$ with dense set of isolated points is $\IR$-soft if and only if $K$ coincides with the Higson compactification of its permutation coarse space. Moreover, in Theorem~\ref{t:mainCH} we prove that under CH any $\IR$-soft separable compact space of weight $\le\w_1$ with dense set of isolated points is homeomorphic to the Higson compactification of some cellular finitary coarse space. In Section~\ref{s:12soft} we provide some sufficient conditions for $\IR$-softness, in particular, we observe that each Fr\'echet-Urysohn compact Hausdorff space with dense set of isolated points is $\IR$-soft.
In Section~\ref{s:nonsoft} we present some examples of compactifications which are not soft. In final Section~\ref{s:Hcor} we address the problem of recognizing the Higson coronas of finitary coarse spaces and prove that a compact Hausdorff space $K$ is homeomorphic to the Higson corona of some finitary coarse space if  $K$ is perfectly normal (more generally, is a corona of character $\chi(K)<\mathfrak p$). Moreover, we prove that under CH, each (zero-dimensional) compact Hausdorff space of weight $\le\w_1$ is homeomorphic to the Higson (resp. binary) corona of some cellular finitary coarse space.
\smallskip

\section{Coarse spaces}\label{s:def}

In this section we recall the necessary definitions from Asymptotic Topology (see \cite{PB}, \cite{PZ} and \cite{Roe} for more details).

For a set $X$, any subset $E\subseteq X\times X$ containing the diagonal $\Delta_X:=\{(x,x):x\in X\}$ of $X\times X$ is called an {\em entourage} on $X$. For entourages $E,F$ on $X$ the sets
$$
\begin{aligned}
&E^{-1} :=\{(y,x):(x,y)\in E\}\mbox{ and }\\
&E\circ F:=\{(x,z):\exists y\in X\mbox{ with } (x,y)\in E\mbox{ and }(y,z)\in F\}
\end{aligned}
$$are entourages on $X$. An entourage $E$ on $X$ is called {\em cellular} if $E$ is an equivalence relation, i.e., $E=E^{-1}=E\circ E$.

For an entourage $E\subseteq X\times X$ and point $x\in X$ the set $E[x]:=\{y\in X:(x,y)\in E\}$ is called the {\em $E$-ball centered at} $x$.  For a subset $A\subseteq X$ the set $E[A]=\bigcup_{a\in A}E[a]$ is the {\em $E$-neighborhood} of $A$.

A {\em ballean} is a pair $(X,\E)$ consisting of a set $X$ and a family $\E$ of entourages on $X$ satisfying two axioms:
\begin{itemize}
\item $\bigcup\E=X\times X$.
\item for any sets $E,F\in\E$ there exists a set $G\in\E$ such that $E\circ F^{-1}\subseteq G$.
\end{itemize}
The family $\E$ is called the {\em ball structure} of the ballean $(X,\E)$.

A ballean $(X,\E)$ is called a {\em coarse space} if for any sets $\Delta_X\subseteq E\subseteq F\subseteq X\times X$ the inclusion $F\in\E$ implies $E\in\E$. In this case the family $\mathcal E$ is called the {\em coarse structure} of the coarse space $(X,\E)$.

For a coarse space $(X,\E)$ a subfamily $\mathcal B\subseteq \mathcal E$ is called a {\em base} of $\E$ if each entourage $E\in\E$ is contained in some entourage $B\in\mathcal B$. It is easy to see that any base $\mathcal B$ of a coarse structure is a ball structure. Conversely, each ball structure $\mathcal B$ on a set $X$ is a base of a unique coarse structure
$${\downarrow}\mathcal B:=\{E:\exists B\in\mathcal B\;\;\Delta_X\subseteq E\subseteq B\}.$$

A coarse space is called {\em cellular} if its coarse structure has a base consisting of cellular entourages. More information on cellular coarse spaces can be found in \cite[\S3.1]{PZ}.

A subset $B\subseteq X$ of a coarse space $(X,\E)$ is called {\em bounded} if $B\subseteq E[x]$ for some $E\in\E$ and some $x\in X$. The family of all bounded subsets is called the {\em bornology} of the coarse space.

Two subsets $A,B\subseteq X$ of a coarse space $(X,\E)$ are called {\em asymptotically separated} if for any entourage $E\in\E$ the intersection $E[A]\cap E[B]$ is bounded.

Now we consider two important examples of coarse spaces.

\begin{example} Each metric space $(X,d)$ carries the canonical coarse structure $\E_d$ generated by the countable base consisting of the entourages
$$E_n:=\{(x,y)\in X\times X:d(x,y)\le n\}\mbox{ \ where $n\in\w$}.$$
\end{example}

\begin{definition} A coarse space $(X,\E)$ is called {\em metrizable} if $\E=\E_d$ for some metric $d$ on $X$.
\end{definition}

The following characterization of metrizable coarse spaces can be found in \cite[2.1.1]{PZ}.

\begin{theorem} A coarse space $X$ is metrizable if and only if its coarse structure has a countable base.
\end{theorem}

A coarse space is called {\em ordinal} if its coarse structure has a well-ordered base.
In particular, each metrizable coarse space is ordinal.
\smallskip

We recall that a {\em $G$-space} is a set $X$ endowed with an action of a group $G$ (which can be identified with a subgroup of the permutation group $S_X$ of $X$).

\begin{example}\label{ex2} Each $G$-space $X$ carries the canonical coarse structure $\E_G$ generated by the base consisting of the entourages
$$E_F:=\{(x,y)\in X\times X:y\in\{x\}\cup\{g(x)\}_{g\in F}\}$$where $F$ runs over finite subsets of the acting group $G$.
\end{example}

Coarse spaces whose coarse structure is generated by the action of some group can be characterized as finitary coarse spaces.

\begin{definition} A coarse space $(X,\E)$ is called {\em finitary} if $$\sup\{|E[x]|:x\in X\}<\w$$ for all $E\in\E$.
\end{definition}

 It is clear that the bornology of any finitary coarse space $(X,\E)$ coincides with the family $[X]^{<\w}$ of all finite subsets of $X$. The following characterization of finitary coarse spaces was proved by Protasov in \cite{Prot} (see also \cite{PP} and \cite{P19}).

\begin{theorem}\label{t:Prot} A coarse space $(X,\E)$ is finitary if and only if $\E=\E_G$ for some subgroup $G\subseteq S_X$.
\end{theorem}

\section{The $M$-compactification and $M$-corona of a coarse space}\label{s:H}

Let  $(M,d_M)$ be a proper metric space. The properness of $M$ means that $|M|\ge 2$ and each closed bounded set in $M$ is compact. For a non-empty subset $A\subseteq M$ let $$\diam(A):=\sup\{d_M(x,y):x,y\in A\}$$be the {\em diameter} of the set $A$ in $M$.

A function $\varphi:X\to M$ on a coarse space $(X,\E)$ is called {\em slowly oscillating} if for every $\e>0$ and $E\in\E$ there exists a bounded set $B\subseteq X$ such that $\diam\, \varphi(E[x])<\e$ for all $x\in X\setminus B$.

A coarse space $(X,\E)$ is called {\em $M$-normal} if for any disjoint and asymptotically separated nonempty sets $A,B\subseteq X$ there exists a slowly oscillating function $\varphi:X\to M$ such that $$\inf\{d_M(\varphi(a),\varphi(b)):a\in A,\;b\in B\}>0.$$
$\IR$-Normal coarse spaces are called {\em normal}. They were introduced by Protasov \cite{Prot03} and studied in \cite{BP}. By \cite{Prot03}, each metrizable (and cellular) coarse space is $\IR$-normal (and $\mathsf 2$-normal). Here by $\mathsf 2$ we denote the doubleton $\{0,1\}$ endowed with the (unique) $\{0,1\}$-valued metric.

Given a coarse space $(X,\E)$, denote by $\so(X,\E;M)$ the family of all bounded slowly oscillating $M$-valued functions on $X$. It can be shown that the function
$$\delta_M:X\to M^{\so(X,\E;M)},\;\;\delta:x\mapsto (f(x))_{f\in\so(X,\E;M)},$$is injective. The properness of the metric space $M$ ensures that the closure $\overline{\delta_M(X)}$ of $\delta_M(X)$ in $M^{\so(X,\E;M)}$ is compact.

The compact Hausdorff space $h_M(X,\E):=\overline{\delta_M(X)}$ is called the {\em $M$-compactification} of the coarse space $(X,\E)$. The injectivity of the map $\delta_M:X\to h_M(X,\E)$ allows us to identify $X$ with its image $\delta_M(X)$ in $h_M(X,\E)$. Observe that a function $f:X\to M$ is slowly oscillating if it is constant outside of some bounded set in $X$. This observation implies that  $X$ coincides with the set of isolated points of the compact space $h_M(X,\E)$, and for any bounded set $B\subseteq X$ its closure $\bar B$ in $h_M(X,\E)$ is open. Consequently, the set
$$\hbar_M(X,\E):=h_M(X,\E)\setminus\bigcup\{\bar B:B\subseteq X\mbox{ is bounded}\},$$
called the {\em $M$-corona} of $(X,\E)$, is compact. If all bounded sets in $X$ are finite, then the $M$-corona $\hbar_M(X,\E)$ coincides with the remainder $h_M(X,\E)\setminus X$ of the $M$-compactification of $X$.

The $\IR$-compactification $h_\IR(X,\E)$ and the $\IR$-corona $\hbar_\IR(X,\E)$ of a coarse space $(X,\E)$ were introduced by Higson \cite{Higson} and are called the {\em Higson compactification} and the {\em Higson corona} of $(X,\E)$, respectively.

For the doubleton $\mathsf 2:=\{0,1\}$, the $\mathsf 2$-compactification $h_{\mathsf 2}(X,\E)$ and the $\mathsf 2$-corona $\hbar_{\mathsf 2}(X,\E)$ of a coarse space $(X,\E)$ were introduced by Protasov \cite{Prot_bc}, \cite{Prot05} and are called the {\em binary compactification} and the {\em binary corona} on $(X,\E)$, respectively.  Observe that the binary compactification (and the binary corona) of a coarse space is zero-dimensional. In Geometric Group Theory, the binary corona is known as the space of ends, see \cite{Yves}.

It is known that for any unbounded metrizable coarse space $(X,\E)$, the Higson corona $\hbar_\IR(X,\E)$ contains a copy of the  remainder $\beta\IN\setminus\IN$ of the Stone-Chech compactification of the discrete space $\IN$ of positive integers, so has rather complicated topological structure.

In the following sections we shall discuss the following

\begin{problem}\label{prob0a} Which compact Hausdorff spaces are homeomorphic to the $M$-compactification (or the $M$-corona) of a finitary coarse space?
\end{problem}

\section{Compactifications and their permutation coarse spaces}\label{s:P}

Let $M$ be a proper metric space.
In order to answer Problem~\ref{prob0a}, observe that for any coarse metric space $X$ its $M$-compactification is a compact Hausdorff space containing $X$ as a dense discrete subspace.

Motivated by this observation, we define a compact Hausdorff space $K$ to be  {\em a compactification} if the set $K'$ of isolated points of $K$ is dense in $K$. So, $K$ is indeed a compactification of the discrete space $K'$. In particular, the $M$-compactification $h_M(X,\E)$ of a coarse space $(X,\E)$ is a compactification (of the discrete space $X$).

For a compact Hausdorff space $K$ by $C(K;M)$ we denote the family of all continuous functions from $K$ to $M$. For a subset $X\subset K$, let
$$\uc(X,K;M):=\{f{\restriction}X:f\in C(K;M)\}.$$

We shall say that two compactifications $K_1,K_2$ of the same set $X=K'_1=K'_2$ are
\begin{itemize}
\item {\em equivalent} if there exists a homeomorphism $h:K_1\to K_2$ such that $h(x)=x$ for any $x\in X$;
\item {\em $M$-equivalent} if $\uc(X,K_1;M)=\uc(X,K_2;M)$.
\end{itemize}

\begin{proposition} Let $K_1,K_2$ be two compactifications of the same set $X=K_1'=K_2'$.
\begin{enumerate}
\item If $K_1$ and $K_2$ are equivalent, then they are $M$-equivalent;
\item If $K_1$ and $K_2$ are homeomorphic to subspaces of $M^{\kappa}$ for some cardinal $\kappa$, then $K_1$ and $K_2$ are equivalent if and only if they are $M$-equivalent;
\item $K_1,K_2$ are equivalent if and only if they are $\IR$-equivalent;
\item If $K_1,K_2$ are zero-dimensional, then they are equivalent if and only if they are $\mathsf 2$-equivalent.
\end{enumerate}
\end{proposition}

\begin{proof} 1. If $K_1,K_2$ are equivalent, then there exists a homeomorphism $h:K_1\to K_2$ such that $h(x)=x$ for all $x\in X$. Given any function $\varphi\in\uc(X,K_2;M)$, find a continuous function $\bar \varphi:K_2\to M$ such that $\varphi=\bar\varphi{\restriction}X$. It follows that the function $\psi=\bar\varphi\circ h:K_1\to M$ is continuous and its restriction $\varphi=\psi{\restriction}X$ belongs to the family $\uc(X;K_1;M)$. This proves the inclusion $\uc(X,K_2;M)\subseteq  \uc(X,K_1;M)$. By analogy we can prove that $\uc(X,K_1;M)\subseteq \uc(X,K_2;M)$.
So the compactifications $K_1,K_2$ are $M$-equivalent.
\smallskip

2. Now assume that $K_1,K_2$ are $M$-equivalent and for every $i\in\{1,2\}$ the compact space $K_i$ admits a topological embedding $\delta_i:K_i\to M^{\kappa}$, where $\kappa$ is a suitable cardinal. We claim that the map $\delta_1{\restriction}X:X\to K_2\subset M^\kappa$ has a continuous extension $K_1\to K_2$. For every $\alpha\in\kappa$ consider the coordinate projection $\pi_\alpha:M^\kappa\to M$ onto the $\alpha$th coordinate. Observe that the restriction $\pi_\alpha{\restriction}K_2:K_2\to M$ is continuous and  the restriction $\pi_\alpha{\restriction}X\in \uc(X,K_2;M)=\uc(X,K_1;M)$ admits a unique continuous extension $\bar\pi_\alpha:K_1\to M$. Then the map $f=(\bar\pi_\alpha)_{\alpha\in\kappa}:K_1\to M^\kappa$ is a continuous extension of the map $\delta_1{\restriction}X$. The density of $X$ in the compact spaces $K_1,K_2$ ensures that $f(K_1)=K_2$. By analogy we can prove that the identity map $X\to K_1$ extends to a continuous map $g:K_1\to K_2$. Taking into account that $f\circ g(x)=g\circ f(x)=x$ for all $x\in X$, we conclude that $f:K_1\to K_2$ is a homeomorphism witnessing that the compactifications $K_1,K_2$ are equivalent.
\smallskip

3,4. The last two statements follow immediately from the second statement.
\end{proof}

For a compact Hausdorff space $K$ let $X$ be the set of isolated points of $K$ and $\Homeo(K,X)$ be the group of homemorphisms $g$ of $K$ such that $g(x)=x$ for all $x\in K\setminus X$. The group $\Homeo(K,X)$ induces the subgroup
$$G_{X,K}:=\{g{\restriction}X:g\in\Homeo(K,X)\}$$
in the permutation group $S_{X}$ of the set $X$. The group $G_{X,K}$ will be called the {\em permutation group} of the compactification $K$.

The action of the group $G_{X,K}$ on the set $X$ determines the coarse structure $\E_{X,K}:=\E_{G_{X,K}}$ on the set $X$. The coarse space $(X,\E_{X,K})$ will be called the {\em permutation coarse space} of the compact Hausdorff space $K$.

The construction of the permutation coarse space rises two problems, the first of which is related to Problems~\ref{prob0} and \ref{prob0a}.

\begin{problem}\label{prob1} Characterize compactifications  that are equivalent to the Higson compactifications of their permutation coarse spaces.
\end{problem}

\begin{problem}\label{prob2} Characterize coarse spaces that coincide with the permutation coarse spaces of their Higson compactifications.
\end{problem}

\section{The $M$-compactifications of finitary metrizable coarse spaces}\label{s:M}

A partial answer to Problem~\ref{prob2} is given by the following theorem.

\begin{theorem}\label{t:M} Let $M$ be a proper metric space, $(X,\E)$ be a coarse space,  $G:=G_{X,\bar X}$ be the permutation group of the $M$-compactification $\bar X:=h_M(X,\E)$ of $X$, and $\E_G$ be the coarse structure on $X$ generated by the action of the group $G$.\begin{enumerate}
\item If $(X,\E)$ is finitary, then $\E\subseteq\E_G$.
\item If $(X,\E)$ is metrizable and $M$-normal, then $\E_G\subseteq\E$.
\item If $(X,\E)$ is finitary, metrizable and $M$-normal, then $\E=\E_G$.
\end{enumerate}
\end{theorem}

\begin{proof} 1. Assuming that $(X,\E)$ is finitary, we shall prove that $\E\subseteq\E_G$. By Theorem~\ref{t:Prot}, $\E=\E_H$ for some subgroup $H\subseteq S_X$. We claim that $H\subseteq G_{X,\bar X}$. This will follow from the definition of the group $G_{X,\bar X}$ as soon as we prove that every $g\in H$ has a continuous extension $\bar g:\bar X\to\bar X$ such that $\bar g(x)=x$ for all $x\in \bar X\setminus X$. Since $\bar X$ embeds into the product $M^{\so(X,\E;M)}$ it suffices to prove that for every $\varphi\in\so(X,\E;M)$ the function $\psi:=\varphi\circ h:X\to M$ has a continuous extension $\bar\psi:\bar X\to M$ such that $\bar\psi(x)=x$ for all $x\in\bar X\setminus X$. Such extension $\bar\psi$ exists  if and only if for every $z\in \bar X\setminus X$ and $\e>0$ there exists a neighborhood $O_z\subseteq\bar X$ of $z$ such that $\diam\,\psi (X\cap O_x)<\e$. So, fix any point $z\in\bar X\setminus X$ and $\e>0$.

The permuation $g$ determines the entourage  $$E_g=\{(x,y)\in X\times X:y\in\{x,g(x)\}\}\in\E_H=\E.$$ Since the function $\varphi$ is slowly oscillating, there exists a bounded (and hence finite set) $B\subseteq X$ such that $\diam \,\varphi([E_g[x])<\frac14\e$ for all $x\in X\setminus B$.

By definition of $\bar X$, the slowly oscillating function $\varphi$ admits a continuous extension $\bar\varphi:\bar X\to M$. Then for the neighborhood $O_z:=\{x\in \bar X\setminus B:d_M(\bar\varphi(x),\bar\varphi(z))<\frac14\e\}$ of $z$ we have $\diam\, \psi(X\cap O_z)<\e$.  Indeed, for any $x,y\in X\cap O_z$, we get
$$d_M(\psi(x),\bar\varphi(z))\le d_M(\varphi\circ h(x),\varphi(x))+d_M(\varphi(x),\bar\varphi(z))<\tfrac14\e+\tfrac14\e=\tfrac12\e$$
and then
$$d_M(\psi(x),\psi(y))\le d_M(\psi(x),\bar\varphi(z))+d_M(\psi(y),\bar\varphi(z))<\tfrac12\e+\tfrac12\e=\e.$$
This completes the proof of the inclusion $H\subseteq G_{X,\bar X}$. The latter inclusion implies $\E=\E_H\subseteq \E_{G_{X,\bar X}}$.
 \smallskip

2. Assume that the coarse space $(X,\E)$ is metrizable and $M$-normal. By Theorem~\ref{t:M}, the coarse structure $\E$ has a countable base $\{E_n\}_{n\in\w}\subseteq\E$ such that $E_n\subseteq E_{n+1}$ for all $n\in\w$. Assuming that $\E_G\not\subseteq\E$, we can find a finite subset $F\subseteq \Homeo(\bar X,X)$ such that the entourage $$E_F:=\{(x,y)\in X\times X:y\in\{x\}\cup\{g(x)\}_{g\in F}\}$$ does not belong to the coarse structure $\E$.
In particular, $E_F\not\subseteq E_n^{-1}\circ E_n$ for every $n\in\w$. Replacing $F$ by a larger finite set in $\Homeo(\bar X,X)$, we can assume that $F$ contains the identity homeomorphism of the space $\bar X$.

Take any point $x_0\in X$ and inductively construct a sequence of points $(x_n)_{n\in\w}$ in $X$ and bijections $g_n\in F$ such that for every $n\in\w$
$$g_n(x_n)\notin E_n^{-1}\circ E_n[x_{n}]\mbox{ and }x_{n}\notin \bigcup_{k<n}\bigcup_{g\in F}E_n^{-1}\circ E_n[g(x_k)].$$
We claim that the sets $A:=\{x_n\}_{n\in\w}$ and $B:=\{g_n(x_n)\}_{n\in\w}$ are asymptotically separated. Since $\{E_n\}_{n\in\w}$ is a base of the coarse structure $\E$, for every entourage $E\in\E$ there exists $n\in\w$ such that $E\subseteq E_n$. The choice of the points $(x_k)_{k\in\w}$ guarantees that $E[A]\cap E[B]\subseteq  E_n[A]\cap E_n[B]\subseteq \bigcup_{i\le n}E_n[x_i]$ is finite.

By the $M$-normality of the coarse space $(X,\E)$, there exists a slowly oscillating function $\varphi:X\to M$ such that $$\e:=\inf\{d_X(\varphi(a),\varphi(b)):a\in A,\;b\in B\}>0.$$ By the definition of the $M$-compactification, the slowly oscillating function $\varphi$ admits a continuous extension $\bar\varphi:\bar X\to M$.

Since the set $F\subseteq \Homeo(\bar X,X)$ is finite, by the continuity, for every point $z\in \bar X\setminus X$ we can find a neighborhood $O_z\subseteq\bar X$ such that $$d_M(\bar\varphi(g(x)),\bar\varphi(g(z)))<\tfrac12\e$$ for any $g\in F$ and $x\in O_z$. In particular, $d_M(\bar\varphi(x),\bar\varphi(z))<\frac12\e$ for any $x\in O_z$.

Next, choose an open neighborhood $U_z\subseteq O_z$ such that $g(U_z)\subseteq O_z$ for all $g\in F$. It follows that $U:=\bigcup_{z\in\bar X\setminus X}U_z$ is an open set containing $\bar X\setminus X$ and then $F:=\bar X_\E\setminus U$ is a compact (and hence finite) set in the discrete space $X$.

Choose any $n\in\w$ with $x_n\notin F$ and find $z\in \bar X_\E\setminus X$ with $x_n\in U_z\subseteq O_z$. Then $g_n(x_n)\in O_z$ and  $$
\begin{aligned}
d_M(\varphi(g_n(x_n)),\varphi(x_n))&\le d_M(\varphi(g_n(x_n)),\bar\varphi(g_n(z)))+d_M(\bar\varphi(g_n(z)),\bar\varphi(z))+d_M(\bar\varphi(z),\varphi(x_n))\\
&<\frac\e2+0+\frac\e2=\e,
\end{aligned}
$$ which contradicts the definition of $\e$. This contradiction completes the proof of the inclusion $\E_G\subseteq\E$.
\smallskip

3. The last statement follows from the first two statements.
\end{proof}

Since each metrizable coarse space is $\IR$-normal \cite{Prot03}, Theorem~\ref{t:M} implies

\begin{corollary}\label{c:R} Let $(X,\E)$ be a coarse space,  $G:=G_{X,\bar X}$ be the permutation group of its Higson compactification $\bar X:=h_\IR(X,\E)$, and $\E_G$ be the coarse structure on $X$ generated by the action of the group $G$.\begin{enumerate}
\item If $(X,\E)$ is finitary, then $\E\subseteq\E_G$.
\item If $(X,\E)$ is metrizable, then $\E_G\subseteq\E$.
\item If $(X,\E)$ is finitary and metrizable, then $\E=\E_G$.
\end{enumerate}
\end{corollary}

\section{$M$-Soft compactifications}\label{s:soft}

Let $(M,d_M)$ be a proper metric space.
 In order to answer Problems~\ref{prob0} and \ref{prob1}, we shall introduce the notion of an $M$-soft compactification. First we recall the definition of oscillation. Let $\varphi:X\to M$ be a function defined on a subset $X$ of a topological space $K$. For a point $z\in\bar X\subseteq K$ denote by $\tau_z$ the family of all open neighborhoods of $z$ in $K$. The number
$$\osc_z(\varphi):=\inf_{O_x\in\tau_x}\diam\,\varphi(X\cap O_x)$$
is called {\em the oscillation} of $\varphi$ at the point $z$. By Lemma 4.3.16 in \cite{Eng}, $\varphi$ admits a continuous extension $\bar X\to M$ if and only if $\osc_z(\varphi)=0$ for each $x\in\bar X$.

\begin{definition} A compactifiction $K$ is defined to be {\em $M$-soft} if for any bounded function $\varphi:K'\to M$ with $\{z\in K:\osc_z(\varphi)>0\}\ne\emptyset$, there exist a homeomorphism $g\in \Homeo(K,K')$ and $\e>0$ such that $d_M(\varphi(g(x)),\varphi(x))\ge\e$ for infinitely many points $x\in K'$.
\end{definition}

\begin{lemma}\label{l:soft} The $M$-compactification $h_M(X,\E)$ of any finitary coarse space $(X,\E)$ is $M$-soft.
\end{lemma}

\begin{proof} Let $(X,\E)$ be a finitary coarse space. To show that its $M$-compactification $K:=h_M(X,\E)$ is $M$-soft, fix any bounded function $\varphi:K'\to M$ with $\osc_z(\varphi)>0$ for some $z\in K$. This means that $\varphi$ does not  extend to a continuous function on $K$ and hence $\varphi\notin \so(X,\E;M)$. So, there exists an entourage $E\in\E$ and $\e>0$ such that the set $\Omega:=\{x\in X:\diam\,\varphi(E[x])\ge\e\}$ is unbounded and hence infinite in $X$.

By Theorem~\ref{t:Prot}, $\E=\E_G$ for some subgroup $G\subseteq S_X$. The definition of the coarse structure $\E_G$ yields a finite subset $F=F^{-1}\subseteq G$ such that $E\subseteq E_F:=\{(x,y):y\in\{x\}\cup \{g(x)\}_{g\in F}\}$. Then for any $x\in\Omega$ we have
$$\e\le \diam\,
\varphi(E_F[x])\le 2\max_{g\in F}d_M(\varphi{\circ}g(x),\varphi(x)).$$
For every $g\in F$ consider the set $\Omega_g:=\{x\in\Omega:d_M(\varphi\circ g(x),\varphi(x))\ge\frac12\e\}$ and observe that $\Omega=\bigcup_{g\in F}\Omega_g$.
By the Pigeonhole Principle, for some $g\in F$ the set $\Omega_g$ is infinite. So, $d_M(\varphi\circ g(x),\varphi(x))\ge\frac12\e$ for infinitely many points $x$ (in particular, for all points in $\Omega_g$).
\end{proof}

The following theorem answers Problem~\ref{prob1}.

\begin{theorem}\label{t:main2} Let $M$ be a proper metric space, $K$ be a compactification, $X$ be the set of isolated points of $K$ and $(X,\E_{X,K})$ be the permutation coarse space of $K$. The space $K$ is $M$-equivalent to the $M$-compactification $h_M(X,\E)$ of the coarse space $(X,\E_{X,K})$ if and only if $K$ is $M$-soft.
\end{theorem}

\begin{proof} The ``only if'' part follows from Lemma~\ref{l:soft}. To prove the ``if'' part, assume that the compactification $K$ is $M$-soft. Let $X$ be the set of isolated points of $K$. Denote by $C(K;M)$ the family of all continuous $M$-valued functions on $K$ and let $\uc(X,K;M):=\{f{\restriction}X:f\in C(K;M)\}$. The $M$-equivalence of the compactifications $K$ and $\bar X:=h_M(X,\E)$ will follow as soon as we check that $\uc(X,K;M)=\so(X,\E_{X,K};M)$.

First we check that each function $\varphi\in\uc(X,K;M)$ belongs $\so(X,\E_{X,K};M)$. Given any $\e>0$ and a finite subset $F\subseteq G_{X,K}$, we will find a finite set $B\subseteq X$ such that
 $\diam\{\varphi(g(x)):g\in F\}<\e$ for all $x\in X\setminus B$.

Let $\bar\varphi:K\to M$ be a unique continuous function such that $\varphi=\bar\varphi{\restriction}X$. By the uniform continuity of $\bar\varphi$, there exists an open neighborhood $U\subseteq K\times K$ of the diagonal $\Delta_K$ such that $\diam\, \bar\varphi(U[z])<\frac12\e$ for any $z\in K$. By the definition of the group $G_{X,K}$, every permutation $g\in F\subseteq G_{X,K}$ coincides with the restriction $\bar g{\restriction}X$ of some homeomorphism $\bar g$ of $K$ such that $\bar g(z)=z$ for all $z\in K\setminus X$. Then any $z\in K\setminus X$ has an open neighborhood $V_z\subseteq K$ such that $V_z\times \bar g(V_z)\subseteq U$ for all $g\in F$. It follows that $V:=\bigcup_{z\in K\setminus X} V_z$ is an open neighborhood of $K\setminus X$ in $K$. Since the subspace $X$ is discrete, the compact set $B:=K\setminus V\subseteq X$ is finite.

For any point $x\in X\setminus B=V$ there exits a point $z\in K\setminus X$ such $x\in V_z$. Then for any $g\in F$ we have $(x,\bar g(x))\in V_z\times \bar g(V_z)\subseteq U$ and hence $d_M(\varphi\circ \bar g(x),\varphi(x))<\frac12\e$. Finally, $\diam \{\varphi\circ g(x):g\in F\}<\frac12\e+\frac12\e=\e$ for any $x\in X\setminus B$.
 \smallskip

Next, we show that each function $\varphi\in\so(X,\E_{X,K};M)$ belongs to $\uc(X,K;M)$. To derive a contradiction, assume that $\varphi$ does not belong to $\uc(X,K;M)$ and hence admits no continuous extensions to $K$.
This means that $\osc_z(\varphi)>0$ for some $z\in K\setminus X$.
Since $K$ is $M$-soft, there exist a permutation $g\in \Homeo(K,X)$ and $\e>0$ such that $d_M(\varphi(g(x)),\varphi(x))\ge\e$ for infinitely many points $x\in X$. This means that the function $\varphi$ is not slowly oscillating (with respect to the coarse structure $\E_{X,K}=\E_{G_{X,K}}$, which contradicts the choice of $\varphi$.
\end{proof}

\begin{corollary}\label{c:main}  A compactification $K$ is homeomorphic to the Higson compactification of some finitary coarse space if and only if $K$ is $\IR$-soft.
\end{corollary}

Under some additional set-theoretic assumptions, we can prove more. Namely, that every separable $\IR$-soft compactification is homeomorphic to the Higson compactification of some {\em cellular} finitary coarse space. This holds under the equality $\as=\mathfrak c$, where $\as$ is the smallest cardinality of a base of a cellular finitary coarse structure $\E$ on $\w$ containing no infinite $\E$-discrete subsets.  A subset $D$ of a coarse space $(X,\E)$ is called {\em $\E$-discrete} if for any entourage $E\in\E$ there exists a bounded set $B\subset X$ such that $D\cap E[x]=\{x\}$ for any $x\in D\setminus B$. In \cite{LP} $\E$-discrete sets are called thin.

The cardinal $\as$ has been introduced in \cite{j}, where it is shown that $\max\{\mathfrak b,\mathfrak s,\mathrm{cov}(\mathcal N)\}\le\as\le\mathfrak c$ and hence $\as=\mathfrak c$ under Martin's Axiom. On the other hand, we do not know whether $\as=\mathfrak c$ in ZFC. The equality $\as=\mathfrak c$ allows us to prove the following consistent improvement of the ``if'' part in Corollary~\ref{c:main}.

\begin{theorem}\label{t:mainCH} Let $M$ be a proper metric space. Under $\as=\mathfrak c$, each separable $M$-soft compactification $K$ is $M$-equivalent to the $M$-Higson compactification $h_M(K',\E)$ of the set $K'$ of isolated points of $K$ endowed with a suitable cellular finitary coarse structure $\E$.
\end{theorem}

\begin{proof} By the separability of $K$, the set $X$ of isolated point of $K$ is countable. Being proper, the metric space $M$ has cardinality $|M|\le\mathfrak c$. Let $C(K;M)$ be the family of continuous $M$-valued maps on $K$ and $\uc(X,K;M):=\{f{\restriction}X:f\in C(K;M)\}$. Let $\mathsf b(X;M)$ be the set of bounded $M$-valued functions on $X$. If $\uc(X,K;M)=\mathsf b(X;M)$, then each bounded function $\varphi:X\to M$ has a continuous extension $\bar\varphi:K\to M$, which implies that $K$ is a Stone-\v Cech compactification of $X$ and $K$ is $M$-equivalent to $h_M(X,\E_H)$ for the cellular finitary coarse structure $\E_H$,  induced by the group $H=FS_X$ of finitely supported permutations of $X$.

So, we assume that $\uc(X,K;M)\ne\mathsf b(X,K;M)$.  Since $|M^X|\le|\mathfrak c^\w|=\mathfrak c$, the non-empty set $\mathsf b(X;M)\setminus\uc(X,K;M)$ can be written as $\{\varphi_\alpha\}_{\alpha<\mathfrak c}$.

Let $G_{X,K}$ be the permutation group of the compactification $K$. A subgroup $H\subset G_{X,K}$ is defined to be {\em cellular} if the coarse structure $\E_H$ on $X$ is cellular. This happens if and only if for any finitely-generated subgroup $F\subseteq H$ the cardinal $\sup_{x\in X}|F(x)|$ is finite. Here $F(x):=\{f(x):f\in F\}$.

Assume that $\as=\mathfrak c$. We shall inductively construct an increasing transfinite sequence of cellular subgroups $(H_\alpha)_{\alpha\in\mathfrak c}$ of $G_{X,K}$ such that for any $\alpha<\mathfrak c$ the group $H_\alpha$ has cardinality $|H_\alpha|\le |\alpha+\w|$ and there are $\e_\alpha>0$ and $g_\alpha\in H_\alpha$ such that the set $\{x\in X:d_M(\varphi\circ g_\alpha(x),\varphi(x))>\e_\alpha\}$ is infinite.

To start the inductive construction, consider the function $\varphi_0$. By the $M$-softness of $K$, for the function $\varphi_0\notin \uc(X,K;M)$ there exist $\e_0>0$ and a permutation $h_0\in G_{X,K}$ such that the set $\Omega_0=\{x\in K':d_M(\varphi_0(h_0(x)),\varphi_0(x))>\e_0\}$ is infinite. Choose a sequence of points $x_n\in\Omega$ such that $x_{n+1}\notin \{x_k\}_{k\le n}\cup \{h_0(x_k)\}_{k\le n}\cup\{h_0^{-1}(x_k)\}_{k\le n}$ for all $n\in\w$. Consider the permutation $g_0:X\to X$, defined by
$$
g_0(x)=
\begin{cases}h_0(x_n)&\mbox{if $x=x_n$ for some $n\in\w$};\\
x_n&\mbox{if $x=h_0(x_n)$ for some $n\in\w$};\\
x&\mbox{otherwise};
\end{cases}
$$and observe that $g_0$ belongs to the group $G_{X,K}$. Let $H_0$ be the cyclic subgroup of $G_{X,K}$ generated by $g_0$. Observe that $|H_0|=2$ and hence the subgroup $H_0\subset S_X$ is cellular.

Now assume that for some nonzero ordinal $\alpha<\mathfrak c$ we have constructed an increasing transfinite sequence $(H_\beta)_{\beta<\alpha}$ of cellular subgroups of $S_{X,K}$ such that $|H_\beta|\le|\beta+\w|$ for all $\beta<\alpha$. Then the union $H_{<\alpha}=\bigcup_{\beta<\alpha}H_\beta$ is a cellular subgroup of $S_X$ with $|H_{<\alpha}|\le\sum_{\beta<\alpha}|H_\beta|\le |\alpha|\cdot |\alpha+\w|=|\alpha+\w|$ and $\E_{H_{<\alpha}}$ is a cellular finitary coarse structure on $\w$ having a base of cardinality $|[H_{<\alpha}]^{<\w}|\le|\alpha+\w|<\mathfrak c=\as$. Consider the function $\varphi_\alpha\in \mathsf b(X;M)\setminus \uc(X,K;M)$. By the $M$-softness of $K$, there exist a permutation $h_\alpha\in G_{X,K}$ and $\e_\alpha>0$ such that the set $\Omega_\alpha=\{x\in X:d_M(\varphi_\alpha\circ h_\alpha(x),\varphi_\alpha(x))>\e_\alpha\}$ is infinite.

Two cases are possible.

1. For some finitely-generated subgroup $F\subseteq H_{<\alpha}$, the set $\{x\in\Omega_\alpha:h_\alpha(x)\in F(x)\}$ is infinite.
 Let $S=S^{-1}$ be a finite set of generators of the group $F$. Let $e$ be the unit of the group $F$, $S^0=\{e\}$ and $S^{k+1}=S^k\circ S=\{f\circ g:f\in S^k,\;g\in S\}$ for $k\in\w$. Since the subgroup $F\subseteq H_{<\alpha}\subset S_X$ is cellular, the cardinal $m=\sup_{x\in X}|F(x)|$ is finite. We claim that for some $k<m$ the set  $\{x\in \Omega_\alpha:S^k(x)=S^{k+1}(x)\}$ is infinite. Assuming the converse, we conclude that the set $\Omega_\alpha'=\bigcup_{k< m}\{x\in \Omega_\alpha:S^k(x)=S^{k+1}(x)\}$ is finite and then for any $x\in\Omega_\alpha\setminus\Omega_\alpha'$ the sequence $(S^k(x))_{k=0}^m$ is strictly increasing which is not possible as $\bigcup_{k\in\w}S^k(x)=F(x)$ has cardinality $\le m$. This contradiction shows that for some $k\in\w$ the set $$\widetilde\Omega_\alpha:=\{x\in\Omega_\alpha:S^k(x)=S^{k+1}(x)\}=\{x\in\Omega_\alpha:S^k(x)=F(x)\}$$ is infinite. Then for every $x\in \widetilde\Omega_{\alpha}$ we have $h_\alpha(x)\in F(x)=S^k(x)$. By the Pigeonhole Principle,  for some permutation $s_\alpha\in S^k$, the set $\Lambda_\alpha=\{x\in \widetilde \Omega_{\alpha}:h_\alpha(x)=s_\alpha(x)\}$ is infinite.

Put $H_\alpha:=H_{<\alpha}$ and $g_\alpha:=s_\alpha$, and observe that the subgroup $H_\alpha$ of $S_X$ is cellular and the set
$\{x\in X:d_M(\varphi_\alpha\circ g_\alpha(x),\varphi(x))>\e_\alpha\}$ contains the set $\Lambda_\alpha$ and hence is infinite.
\smallskip

2. For every finitely generated subgroup $F\subseteq H_{<\alpha}$ the set $\{x\in\Omega_\alpha:h_\alpha(x)\in F(x)\}$ is finite. Since the cellular finitary coarse structure $\E_{H_{<\alpha}}$ on $X$ has base of cardinality $<\mathfrak c=\as$, the definition of the cardinal $\as$ ensures that the infinite set $\Omega_\alpha$ contains an infinite $\E_{H_{<\alpha}}$-discrete subset $T_\alpha\subseteq\Omega_\alpha$. Find a sequence $\{x_n\}_{n\in\w}$ of pairwise distinct points of $T_\alpha$ such that the sets $\{x_n\}_{n\in\w}$ and $\{h_\alpha(x_n)\}_{n\in\w}$ are disjoint.

 Consider the permutation $g_\alpha:X\to X$, defined by
$$
g_\alpha(x)=
\begin{cases}h_\alpha(x_n)&\mbox{if $x=x_n$ for some $n\in\w$};\\
x_n&\mbox{if $x=h_\alpha(x_n)$ for some $n\in\w$};\\
x&\mbox{otherwise;}
\end{cases}
$$and observe that $g_\alpha$ belongs to the group $G_{X,K}$. Let $H_\alpha$ be the subgroup of $G_{X,K}$ generated by the set $\{g_\alpha\}\cup H_{<\alpha}$. Using Lemma~5.3 from \cite{j}, one can show that the subgroup $H_\alpha$ of $S_X$ is cellular and
the set $\{x\in X:d_M(\varphi_\alpha\circ g_\alpha(x)),\varphi_\alpha(x))>\e_\alpha\}\supseteq\{x_n\}_{n\in\w}$ is infinite. This completes the inductive step.

After completing the inductive construction, consider the cellular subgroup $H=\bigcup_{\alpha<\mathfrak c}H_\alpha$ of $G_{X,K}$.
The property of the transfinite sequence $(H_\alpha)_{\alpha<\mathfrak c}$ ensures that any function $\varphi\in \mathbf b(X;M)\setminus \uc(X,K;M)$ is not slowly oscillating with respect to the coarse structure $\E_H$. Consequently, $\uc(K,h_M(X,\E_H);M)=\so(X,\E_H;M)\subseteq\uc(X,K;M)$. On the other hand,  Theorem~\ref{t:main2} ensures that
$\uc(X,K;M)=\so(X,\E_{G_{X,K}};M)\subseteq \so(X,\E_H;M)$, where the latter inclusion folows from $H\subseteq G_{X,K}$. Therefore, $\uc(X,K;M)=\uc(X,h_M(X,\E_H);M)$ and the compactifications $K$ and $h_M(X,\E_H)$ are $M$-equivalent.
\end{proof}

Applying Theorem~\ref{t:mainCH} to the proper metric spaces $\IR$ and $\mathsf 2$, we obtain the following corollaries.

\begin{corollary}\label{c:CHR} Under $\as=\mathfrak c$, each separable $\IR$-soft compactification $K$ is homeomorphic to the Higson compactification of some cellular finitary coarse space.\end{corollary}

\begin{corollary}\label{c:CH0} Under $\as=\mathfrak c$, each zero-dimensional separable $\mathsf 2$-soft compactification $K$ is homeomorphic to the binary compactification of some cellular finitary coarse space.
\end{corollary}

\section{Soft compactifications}\label{s:12soft}

The results of the preceding section motivate a deeper study of $M$-soft compactifications. For this purpose, we introduce a stronger version of the $M$-softness.

\begin{definition} A compactification $K$  is defined to be  {\em soft} if for any disjoint sets $A,B\subseteq K'$ with $\bar A\cap\bar B\ne\emptyset$ there exists a homeomorpism $h\in\Homeo(K,K')$ such that the set $\{x\in A:h(x)\in B\}$ is infinite.
\end{definition}

\begin{proposition}\label{p:sss} Let $M$ be a proper metric space and $K$ be a compactification.
\begin{enumerate}
\item If $K$ is soft, then $K$ is $\IR$-soft.
\item If $K$ is $\IR$-soft, then $K$ is $M$-soft.
\item If $K$ is $M$-soft, then $K$ is $\mathsf 2$-soft.
\end{enumerate}
\end{proposition}

\begin{proof} 1. Assume that a compactification $K$ is soft. To show that $K$ is $\IR$-soft, fix any bounded function $\varphi:K'\to\IR$ with $\osc_z\varphi>0$ for some $z\in  K\setminus K'$. Let $\tau_z$ be the family of open neighborhoods of $z$ in $K$. By definition of the oscillation $\e:=\osc_z\varphi$, for every $U\in\tau_z$ the closed bounded set $\overline{\varphi(U\cap K')}$ in $\IR$ has diameter $\ge\e$. Then the intersection $I:=\bigcap_{U\in\tau_z}\overline{\varphi(U\cap K')}$ also has diameter $\ge \e$ and we can find two points $a,b\in I$ with $b-a=\e$. It follows that the sets
$$
A:=\{x\in X:\varphi(x)<\tfrac23a+\tfrac13b\}\mbox{ \ and \ }B:=\{x\in X:\varphi(x)>\tfrac13a+\tfrac23b\}
$$contain the point $z$ in their closures. Since the compactification $K$ is soft, there exists a homeomorphism $g\in\Homeo(K,K')$ such that the set $\Omega:=\{x\in A:g(x)\in B\}$ is infinite. Then for any $x\in\Omega$ we have
$$|\varphi(g(x))-\varphi(x)|\ge\varphi(g(x))-\varphi(x)>\tfrac13a+\tfrac23b-(\tfrac23a+\tfrac13b)=\tfrac13(b-a)=\tfrac13\e>0,$$
witnessing that the compactification $K$ is $\IR$-soft.
\smallskip

2. Assume that a compactification $K$ is $\IR$-soft. To prove that it is $M$-soft, fix any bounded function $\varphi:K'\to M$ such that $\e:=\osc_z(\varphi)>0$ for some $z\in K$. Let $\tau_z$ be the family of neighborhoods of $z$ in $K$. It follows that for any $U\in\tau_z$ the image $\varphi(U\cap K')$ is a bounded set of diameter $\ge\e$ in $M$. The properness of $M$ ensures that the intersection $I=\bigcap_{U\in\tau_z}\overline{\varphi(U\cap K')}$ is a compact set of diameter $\ge\e$ in $M$. By the compactness of $I$, there are points $a,b\in I$ with $d_M(a,b)\ge\e$. Observe that $f:M\to\IR$, $f:x\mapsto d_M(a,x)$, is a non-expanding function with  $|f(a)-f(b)|=d_M(a,b)\ge\e$. It follows that the function $f\circ\varphi:K'\to \IR$ has $\osc_z(f\circ\varphi)\ge\e>0$. By the $\IR$-softness of $K$, there exists a permutation $g\in G_{X,K}$ and a positive real number $\delta$ such that the set $\{x\in K':|f\circ\varphi(g(x))-f\circ\varphi(x)|>\delta\}$ is infinite. Since the function $f$ is non-expanding, the set $\{x\in K':d_M(\varphi(g(x)),\varphi(x))>\delta\}\supseteq  \{x\in K':|f\circ\varphi(g(x))-f\circ\varphi(x)|>\delta\}$ is infinite, too.
\smallskip

3. Assume that a compactification $K$ is $M$-soft. To show that $K$ is $\mathsf 2$-soft, fix any function $\varphi:K'\to \mathsf 2$ with $\osc_z(\varphi)>0$ for some $z\in K$. By definition, the proper space $M$ contains more than one point and hene admts an injective map $f:\mathsf 2\to M$. Then the map $f\circ\varphi:K'\to M$ has $\osc_z(f\circ\varphi)>0$ and by the $M$-softness of $K$, there exists a permutation $g\in G_{X,K}$ and $\e>0$ such that the set $\{x\in K':d_M(f\circ\varphi\circ g(x),f\circ \varphi(x))>\delta\}$ is infinite. Taking into account that for two points $x,y\in \mathsf 2$ the inequality $d_M(f(x),f(y))>0$ is equivalent to $|x-y|>\frac12$, we conlcude that the set   $\{x\in K':|\varphi\circ g(x)-\varphi(x)|>\frac12\}$ contains $\{x\in K':d_M(f\circ\varphi\circ g(x),f\circ \varphi(x))>\delta\}$ and hence is infinite.
\end{proof}

Proposition~\ref{p:sss} implies that for any proper metric space $M$ and any compactification we have the implications
$$\mbox{soft $\Ra$ $\IR$-soft $\Ra$ $M$-soft $\Ra$ $\mathsf 2$-soft}.$$

\begin{example} The Stone-\v Cech compactification $\beta\w$ of the countable discrete space $\w$ is soft.
\end{example}

\begin{proof} The softness of $\beta\w$ follows from the well-known fact \cite[3.6.2]{Eng} that disjoint sets $A,B\subseteq\w$ have disjoint closures in $\beta\w$.
\end{proof}

\begin{proposition}\label{p:ss} A compactification $K$ is soft if for any disjoint sets $A,B\subseteq K'$ with $\bar A\cap\bar B\ne\emptyset$ there are sequences $\{a_n\}_{n\in\w}\subseteq A$ and $\{b_n\}_{n\in\w}\subseteq B$ that converge to some point $z\in K$.\end{proposition}

\begin{proof} Given disjoint sets $A,B\subseteq K'$ with $\bar A\cap\bar B\ne\emptyset$ find sequences $\{a_n\}_{n\in\w}\subseteq A$ and $\{b_n\}_{n\in\w}\subseteq B$ that converge to some point $z\in K$. Since the sets $A,B$ are disjoint, the point $z$ is not isolated. So, we can assume that $a_n\ne a_m$ and $b_n\ne b_m$ for all $n\ne m$. It can be shown that the bijective function $g:K\to K$ defined by
$$g(x)=\begin{cases}
b_n&\mbox{if $x=a_n$};\\
a_n&\mbox{if $x=b_n$};\\
x&\mbox{otherwise};
\end{cases}
$$ is a homeomorphism of $K$ and $g\in\Homeo(K,K')$. For this homeomorphism the set $\{x\in A:g(x)\in B\}$ contains the set $\{a_n\}_{n\in\w}$ and hence is infinite.
\end{proof}

We recall that a topological space $X$ is {\em Fr\'echet-Urysohn} if for any set $A\subseteq X$ and point $x\in \bar A\setminus A$ there exists a sequence $\{a_n\}_{n\in\w}$ that converges to $x$. Proposition~\ref{p:ss} implies:

\begin{corollary}\label{c:fu} A compactification $K$ is soft if  the space $K$ is Fr\'echet-Urysohn.
\end{corollary}

Since each first-countable space is Fr\'echet-Urysohn, Corollary~\ref{c:fu} implies another corollary.

\begin{corollary} Each first-countable compactification  is soft.
\end{corollary}

By $\mathfrak p$ we denote the smallest cardinality $|\F|$ of a family $\F$ of infinite subsets of $\w$ such that
\begin{itemize}
\item for any $F_1,\dots,F_n\in\F$ the intersection $\bigcap_{i=1}^nF_i$ is infinite;
\item for any infinite subset $I\subseteq\w$ there exists a set $F\in\F$ such that $I\setminus F$ is infinite.
\end{itemize}
It is known  that $\w_1\le\mathfrak p\le\mathfrak b\le\mathfrak c$, and $\mathfrak p=\mathfrak c$ under Martin's Axiom (see, \cite{Blass}, \cite{vD}, \cite{Vau}).

We recall that the {\em character} $\chi(x;X)$ of a point $x$ in a topological space $X$ is the smallest cardinality of a neighborhood base at $x$. The cardinal $$\chi(X):=\sup_{x\in X}\chi(x;X)$$ is called the {\em character} of the space $X$. The definition of the cardinal $\mathfrak p$ implies the following known fact.

\begin{proposition}\label{p:p2} Let $A$ be a countable set in a topological space $X$ and $x\in\bar A$. If $\chi(x;X)<\mathfrak p$, then the set $A$ contains a sequence that converges to $x$.
\end{proposition}

Proposition~\ref{p:p2} and \ref{p:ss} imply the following corollary.

\begin{corollary}\label{c:p1} Each separable compactification of character $<\mathfrak p$ is  soft.
\end{corollary}

Unifying Corollaries~\ref{c:main} and \ref{c:fu}, we obtain

\begin{corollary}\label{c:FU2} A compact Hausdorff space $K$ is homeomorphic to the Higson compactification of some finitary coarse space if $K$ is Fr\'echet-Urysohn and has dense set of isolated points.
\end{corollary}

Also, Corollaries~\ref{c:CHR}, \ref{c:CH0} and \ref{c:fu} imply the following CH-improvements of Corollary~\ref{c:FU2}.

\begin{corollary} Under $\as=\mathfrak c$, a separable compact Hausdorff space $K$ is homeomorphic to the Higson compactification of some cellular finitary coarse space if the space $K$ is Fr\'echet-Urysohn and has dense set of isolated points.
\end{corollary}

\begin{corollary} Under $\as=\mathfrak c$, a zero-dimensional separable compact Hausdorff space $K$ is homeomorphic to the binary compactification of some cellular finitary coarse space if the space $K$ is Fr\'echet-Urysohn and has dense set of isolated points.
\end{corollary}

\section{Examples of compactifications which are not soft}\label{s:nonsoft}

In this section we present some examples of compactifications which are not soft and hence are not homeomorphic to the Higson compactifications of finitary coarse spaces.

There are two sources of such examples. One source comes from compactifications that are composed with two overlapping  pieces with different topological properties.

\begin{proposition}\label{p:not1} Let $K$ be a compactification containing two disjoint sets $A,B\subseteq K'$ such that $\bar A\cap \bar B\ne\emptyset$ and for any infinite sets $I\subseteq A$ and $J\subseteq B$ with $\bar I\setminus I=\bar J\setminus J$ the spaces $\bar I$ and $\bar J$ are not homeomorphic. Then the compactification $K$ is not soft. If $A\cup B=K'$, then $K$ is not $\mathsf 2$-soft and hence is not homeomorphic to the Higson compactification of a finitary coarse space.
\end{proposition}

\begin{proof} Assuming that $K$ is soft, we can find a homeomorphism $h\in\Homeo(K,K')$ such that the set $I=\{x\in A:h(x)\in B\}\subseteq A$ is infinite and  so is the set $J=h(I)$. Take any accumulation point $x\in\bar I\setminus I$ and observe that $x=h(x)\in\bar J\setminus J$. So, $\bar I\setminus I\subseteq\bar J\setminus J$. By analogy we can prove that $\bar J\setminus J\subseteq \bar I\setminus I$. Since $h(\bar I)=\bar J$, the spaces $\bar I,\bar J$ are homeomorphic, but this contradicts our assumption.

If $A\cup B=X$, then we can consider the characteristic function $\chi_A:X\to\mathsf 2$. Assuming that $K$ is $\mathsf 2$-soft, we can find a homeomorphism $h\in\Homeo(K,X)$ such that the set $I=\{x\in A:h(x)\notin A\}$ is infinite and  so is the set $J=h(I)$. Then $\bar I\setminus I=\bar J\setminus J$ and the spaces  $\bar J\setminus J\subseteq \bar I\setminus I$. Since $h(\bar I)=\bar J$, the spaces $\bar I,\bar J$ are homeomorphic, which contradicts our assumption. This contradiction shows that the compactification $K$ is not $\mathsf 2$-soft. By Proposition~\ref{p:sss}, $K$ is not $\IR$-soft and by Corollary~\ref{c:main},  $K$  is not homeomorphic to the Higson compactification of a finitary coarse space.
\end{proof}

Proposition~\ref{p:not1} allows us to construct a more concrete example of a non-soft compactification.

\begin{example} Let $K$ be a compactification containing two disjoint sets $A,B\subseteq K'$ such that $\bar A\cap \bar B\ne\emptyset$, $\bar A$ is homeomorphic to $\beta\w$ and $\bar B$ is first-countable.
Then the compactification $K$ is not soft. If $A\cup B=K'$, then $K$ is not $\mathsf 2$-soft and hence is not homeomorphic to the Higson compactification of a finitary coarse space.
\end{example}

The other source of non-soft compactifications are compactifications $K$ with small permutation group $G_{K',K}$. For a set $X$ by $FS_{X}$ we denote the group of permutations $g:X\to X$ that have finite support $\supp(g):=\{x\in X:g(x)\ne x\}$.
It is clear that $FS_{K'}\subseteq G_{K',K}$ for any compactification $K$.
The Stone-\v Cech compactification $\beta X$ of a discrete space $X$ has $G_{X,\beta X}=FS_X$.

\begin{proposition} A compactification $K$ with $G_{K',K}=FS_{K'}$ is equal to the Higson compactification of its permutation coarse space $(K',\E_{K',K})$ if and only if $K=\beta K'$.
\end{proposition}

\begin{proof} Let $X$ be the set of isolated points of the compactification $K$. By our assumption, $G_{X,K}=FS_X$. Then for any entourage $E\in E_{FS_X}$ the complement $E\setminus \Delta_X$ is finite, which implies that each bounded function $\varphi:K'\to\IR$ is slowly oscillating and then the Higson compactification $\bar X_{\E_{X,K}}$ of the coarse space $(X,\E_{X,K})$ is equivalent to  the Stone-\v Cech compactification of $X$. Now we see that $K=\beta X$ if and only if $\bar X_{\E_{X,K}}=\beta X$.
\end{proof}

We recall (see \cite{Blass}, \cite{vD}, \cite{Vau}) that $\mathfrak s$ is the smallest cardinality $|\F|$ of a family $\F$ of subsets of $\w$ such that for any infinite set $I\subseteq \w$ there exists a set $F\in\F$ such that both sets $I\cap F$ and $I\setminus F$ are infinite. By Theorem 6.1 in   \cite{vD}, each compact space of weight $<\mathfrak s$ is sequentially compact (so contains many non-trivial convergent sequences).

\begin{proposition} A compactification $K$ has $G_{K',K}\ne FS_{K'}$ if $K$ is infinite and one of the following conditions is satisfied:
\begin{enumerate}
\item $K$ contains a non-trivial convergent sequence of isolated points;
\item $K$ has weight $w(K)<\mathfrak s$;
\item $K$ is separable and has character $\chi(X)<\mathfrak p$.
\end{enumerate}
\end{proposition}

\begin{proof} 1. Assume that the compactification $K$ contains a convergent sequence $(x_n)_{n\in\w}$ consisting of pairwise distinct isolated points of $K$. It can be shown that the homeomorphism $h\in \mathcal H(K,K')$ defined by the formula
$$h(x)=\begin{cases}x_{2n+1}&\mbox{if $x=x_{2n}$ for some $n\in\w$};\\
x_{2n}&\mbox{if $x=x_{2n+1}$ for some $n\in\w$};\\
x&\mbox{otherwise};
\end{cases}
$$
determines the bijection $h{\restriction}K'\in G_{K',K}\setminus FS_{K'}$, which implies that $G_{K',K}\ne FS_{K'}$.
\smallskip

2. If $K$ has weight $w(K)<\mathfrak s$, then each sequence in $K$ has a convergent subsequence. Since $K$ is infinite and the set $K'$ of isolated points is dense in $K$, it contains a non-trivial convergent sequence. Now the preceding item applies.
\smallskip

3. If $K$ is separable and has $\chi(K)<\mathfrak p$, then the set $K'$ of isolated points is countable. Applying Proposition~\ref{p:p2}, we conclude that $K'$ contains a non-trivial sequence, convergent to some non-isolated point of $K$. The first item of this proposition implies that  $G_{K',K}\ne FS_{K'}$.
\end{proof}

By $\non(\M)$ we denote the smallest cardinality of a non-meager set in the Cantor cube $\{0,1\}^\w$.  It is known (see \cite{Blass}, \cite{Vau}) that the strict inequality $\non(\M)<\mathfrak c$ is consistent.

\begin{proposition}
 There exists a zero-dimensional compactification $c\IN$ of the discrete countable space $\IN$ such that $w(c\IN)\le\non(\mathcal M)$ and $G_{\IN,c\IN}=FS_{\IN}$. If $\non(\mathcal M)<\mathfrak c$, then the compactification $c\IN$ is not homeomorphic to the Higson compactification of a finitary coarse space.
\end{proposition}

\begin{proof} Endow the family $[\IN]^{\le\w}$ of all subsets of $\IN$ with the natural compact metrizable topology, making $[\IN]^{\le\w}$ homeomorphic to the Cantor cube $\{0,1\}^\w$. A base of this topology consists of the sets $[F,E]=\{A\subseteq\IN:A\cap E=F\}$ where $F\subseteq E$ are finite subsets of $\IN$.

 Take any non-meager set $\A\subseteq [\IN]^{\le\w}$ of cardinality $|\A|=\non(\mathcal M)$. Replacing $\A$ by a larger family of the same cardinality, we can assume that $\A$ contains all finite subsets of $\IN$. For any set $A\in\A$ consider its characteristic function $\chi_A:\IN\to\{0,1\}$ (i.e., the unique function such that $\chi^{-1}_A(1)=A$). The functions $\chi_A$, $A\in\A$, determine the function $\chi_\A:\IN\to\{0,1\}^{\A}$, $\chi_\A:x\mapsto(\chi_A(x))_{A\in\A}$.

Let $c\IN$ denote the closure of the set $\chi_\A(\IN)$  in the cube $\{0,1\}^\A$.
 Since the family $\A$ contains all finite subsets of $\IN$, the function $\chi_\A$ is injective, so we can identify $\IN$ with its image $\chi_\A(\IN)$ in $c\IN$ and consider $c\IN$ as a compactification of $\IN$.

 It is clear that $c\IN$ has weight $w(c\IN)\le|\A|=\non(\M)$. We claim that $G_{\IN,c\IN}=FS_{\IN}$. To derive a contradiction, assume that the compact space $c\IN$ admits a homeomorphism $h\in\Homeo(c\IN,\IN)$ such that the set $\{x\in\IN: h(x)\ne x\}$ is infinite.

In the power-set $[\IN]^{\le\w}$ consider the subfamily
$$\A_h:=\big\{A\subseteq \IN:|\{x\in A:h(x)\notin A\}|=\w\big\}.$$
We claim that $\A_{h}$ is a dense $G_\delta$-set in $[\IN]^{\le\w}$.
For every $n\in\IN$ denote by $[\IN]^n$ the family of $n$-element subsets of $\IN$. For any sets $I,J\in[\IN]^n$ consider the open family
$$\A_{I,J}:=\{A\subseteq\IN: I\subseteq A,\;J\subseteq \IN\setminus A,\;h(I)=J\}$$in $[\IN]^{\le\w}$ and observe that the union $\bigcup_{I,J\in[\IN]^n}\A_{I,J}$ is dense in $[\IN]^{\le\w}$. Then the set
$$\A_{h}=\bigcap_{n\in\IN} \bigcup_{I,J\in[\IN]^n}\A_{I,J}$$is a dense $G_\delta$ in $[\IN]^{\le\w}$. Since the family $\A$ is not meager in $[\IN]^{\le\w}$, there exists a set $A\in\A\cap \A_{h}$.  By the definition of the topology of the space $c\IN$, the function $\chi_A:\IN\to\{0,1\}$ admits a continuous extension $\bar\chi_A:c\IN\to\{0,1\}$.

The definition of the family $\A_h$ guarantees that the set $\Omega=\{x\in A:h(x)\notin A\}$ is infinite. By the compactness of $c\IN$, there exists a point $z\in\bar\Omega\setminus\Omega\subseteq c\IN\setminus\IN$.
It follows that $\bar\chi_A(z)\subseteq \bar\chi_A(\bar \Omega)\subseteq\overline{\chi_A(\Omega)}=\{1\}$ and $\bar\chi_A(h(z))\in \bar\chi_A(\overline{h(\Omega)})\subseteq\bar \chi_A(\IN\setminus A)\subseteq\overline{\chi_A(\IN\setminus A)}=\{0\}$. But this contradicts the equality $h(z)=z$ following from $h\in\Homeo(c\IN,\IN)$. This contradiction shows that $G_{\IN,c\IN}=FS_{\IN}$.

Now assuming that $\non(\M)<\mathfrak c$, we prove that the compactification $c\IN$ is  not homeomorphic to the Higson corona of a finitary coarse space. To derive a contradiction, assume that $c\IN$ is  homeomorphic to the Higson corona of a finitary coarse space. By Lemma~\ref{l:soft}, the compactification $c\IN$ is $\IR$-soft.
Since $w(c\IN)\le\non(\M)<\mathfrak c=w(\beta\IN)$, the compactification  $c\IN$ is not homeomorphic to $\beta\IN$. Then there exists a bounded function $\varphi:\IN\to\IR$ that has no continuous extension to $c\IN$ and hence $\osc_z(\varphi)>0$ for some $z\in c\IN$. Since the compactification $c\IN$ is $\IR$-soft, there exist a homeomorphism $g\in \mathcal H(c\IN,\IN)$ and $\e>0$ such that the set $\{x\in \IN:|\varphi(g(x))-\varphi(x)|>\e\}$ is infinite. The homeomorphism $g$ determines a permutation $g{\restriction}\in G_{\IN,c\IN}\setminus FS_\IN$, which contradicts the equality $G_{\IN,c\IN}=FS_\IN$.
\end{proof}

\section{Constructing a finitary coarse space with a given Higson or binary corona}\label{s:Hcor}

In this section we apply Theorem~\ref{t:main2} for constructing a finitary coarse space with a given Higson or binary corona. All compact spaces considered in this section are Hausdorff.

\begin{definition}\label{d:p} A compact Hausdorff space $K$ is called a {\em corona} if $K$ is homeomorphic to the remainder $c\mathbb N\setminus \mathbb N$ of suitable compactification $c\mathbb N$ of the discrete space $\mathbb N$ of positive integers. If the compactification $c\IN$ can be chosen to be soft (or $M$-soft for some proper metric space $M$), then the corona $K$ is called a {\em soft} (resp. {\em $M$-soft\/}).
\end{definition}

\begin{remark} Coronas can be equivalently defined as  continuous images of the remainder $\beta\IN\setminus\IN$ of the Stone-\v Cech compactification of $\IN$. Each corona $K$ has weight $w(K)\le\mathfrak c$ and cardinality $|K|\le 2^{\mathfrak c}$. It is known that the class of coronas  includes all separable compact spaces, all perfectly normal compacta (see, \cite{Przym}), and all compact spaces of weight $\le \w_1$ (see, \cite{Par} or \cite[1.3.4]{vM}). By a recent result of Hart \cite{Hart},  under CH each compact space of weight $\le\w_1$ is a soft corona. On the other hand, Bell \cite{Bell} constructed a consistent example of a separable compact space which is not a corona, and Dow observed that under his  principle (NT) introduced in \cite{Dow}, the compact Hausdorff space $K=\w_1+1+\w_1^*$ is not a soft corona, see Example in \cite{Hart}.
\end{remark}

The above remark and Proposition~\ref{p:sss} imply that for any compact Hausdorff space $K$ and any proper metric space $M$ we have the implications:
$$
\xymatrix{
\mbox{$\IR$-soft corona}\ar@{=>}[r]&\mbox{$M$-soft corona}\ar@{=>}[r]&\mbox{$\mathsf 2$-soft corona}\ar@{=>}[d]\\
\mbox{soft corona}\ar@{=>}[u]&\mbox{weight $\le\w_1$}\ar@{=>}[r]\ar_{\;\;+CH}[l]&\mbox{corona}\ar@{=>}[r]&\mbox{weight $\le\mathfrak c$}
}
$$

Definition~\ref{d:p} and Theorem~\ref{t:main2} imply the following characterizations.

\begin{corollary}\label{c:corP} A compact Hausdorff space $K$ is homeomorphic to the Higson corona of a finitary coarse space $(X,\E)$ with $|X|=\w$ if and only if $K$ is an $\IR$-soft corona.
\end{corollary}

\begin{corollary}\label{c:corP} A zero-dimensional compact Hausdorff space $K$ is homeomorphic to the binary corona of a finitary coarse space $(X,\E)$ with $|X|=\w$ if and only if $K$ is a $\mathsf 2$-soft corona.
\end{corollary}

These corollaries rise the problem of recognizing $\IR$-soft coronas among coronas.
We recall that a topological space $X$ is {\em perfectly normal} if it is normal and each closed subset of $X$ is of type $G_\delta$. By \cite{Przym}, perfectly normal compact spaces are coronas.

\begin{theorem}\label{t:Par} A compact space $K$ is a soft corona if one of the following conditions is satisfied:
\begin{enumerate}
\item $K$ is a corona of character $\chi(K)<\mathfrak p$;
\item $K$ is perfectly normal;
\item $K$ has weight $w(K)\le\w_1<\mathfrak p$;
\item $K$ has weight $w(K)\le\w_1=\mathfrak c$.
\end{enumerate}
\end{theorem}

\begin{proof} 1. Assume that $K$ is a corona with $\chi(K)<\mathfrak p$. Then  $K$ can be identified with the remainder $c\IN\setminus\IN$ of some compactification $c\IN$ of $\IN$. Taking into account that $\chi(K)<\mathfrak p$ and  $c\IN\setminus K=\IN$ is countable, we conclude that $c\IN$ has (pseudo)character $<\mathfrak p$. By Corollary~\ref{c:p1}, the compactification $c\IN$ is soft. Consequently, $K$ is a soft corona.
\smallskip

2. If $K$ is perfectly normal, then $K$ is a corona according to \cite{Przym}. Being perfectly normal, the space $K$ is first-countable. By the first statement, the space $K$ is a soft corona.
\smallskip

3. If $K$ has weight $w(K)\le\w_1<\mathfrak p$, then $K$ is a corona by \cite{Par} or \cite[1.3.4]{vM} and has character $\chi(K)\le w(K)<\mathfrak p$. Applying the first statement, we conclude that $K$ is a soft corona.
\smallskip

4. If $w(K)\le\w_1=\mathfrak c$, then $K$ is a soft corona by a result of Hart  \cite{Hart}.
\end{proof}

Corollary~\ref{c:corP} and Theorems~\ref{t:Par}, \ref{t:mainCH} imply the following corollaries.

\begin{corollary}\label{c:Higson} (Under $\as=\mathfrak c$) a compact space $K$ is homeomorphic to the Higson corona of a (cellular) finitary coarse space $(X,\E)$ with $|X|=\w$ if one of the following conditions is satisfied:
\begin{enumerate}
\item $K$ is a corona of character $\chi(K)<\mathfrak p$;
\item $K$ is perfectly normal;
\item $w(K)\le\w_1<\mathfrak p$;
\item $w(K)\le\w_1=\mathfrak c$.
\end{enumerate}
\end{corollary}

\begin{corollary}\label{c:Higson0} (Under $\as=\mathfrak c$) a zero-dimensional compact space $K$ is homeomorphic to the binary corona of a (cellular) finitary coarse space $(X,\E)$ with $|X|=\w$ if one of the following conditions is satisfied:
\begin{enumerate}
\item $K$ is a corona of character $\chi(K)<\mathfrak p$;
\item $K$ is perfectly normal;
\item $w(K)\le\w_1<\mathfrak p$;
\item $w(K)\le\w_1=\mathfrak c$.
\end{enumerate}
\end{corollary}

\begin{remark} Corollary~\ref{c:Higson} can be considered as a generalization of the result of Mine and Yamashita \cite{MY} who proved that each compact metric space is homeomorphic to the Higson corona of a suitable coarse space (namely, a totally bounded locally compact metric space carrying the $C_0$ coarse structure of Wright \cite{Wri}).
\end{remark}

With the help of Theorems~\ref{t:mainCH} and \ref{t:Par}(4), we can improve Corollary~\ref{c:Higson}(4) as follows.

\begin{corollary}\label{c:CH-H} Under CH, each compact space of weight $\le\w_1$ is homeomorphic to the Higson corona of some cellular finitary coarse space $(X,\E)$ with $|X|=\w$.
\end{corollary}

\begin{corollary}\label{c:CH-0} Under CH, each zero-dimensional compact space of weight $\le\w_1$ is homeomorphic to the binary corona of some cellular finitary coarse space $(X,\E)$ with $|X|=\w$.
\end{corollary}

\begin{remark} Corollary~\ref{c:CH-H} should be compared with Theorem 4 \cite{Prot_bc} (see also Theorem~11 in \cite{DKU}) saying that the Higson corona of any ordinal cellular coarse space is zero-dimensional.
\end{remark}

\begin{remark} We do not know whether the Continuum Hypothesis can be removed from Corollaries~\ref{c:CH-H} and \ref{c:CH-0}.
\end{remark}

By \cite{Hart}, under CH, each compact space of weight $\le\w_1$ is a soft corona and hence an $\IR$-soft corona. We do not know if the latter fact remains true in ZFC.

\begin{problem} Is each compact Hausdorff space of weight $\le\w_1$ an $\IR$-soft corona? A $\mathsf 2$-soft corona?
\end{problem}

Another problem is motivated by Corollary~\ref{c:fu}.

\begin{problem} Is each (Fr\'echet-Urysohn) corona $\IR$-soft?
\end{problem}

By \cite{Hart}, the Dow Principle (NT) implies that the linearly ordered compact space $\w_1+1+\w_1^*$ is not a soft corona. Here $\w_1^*$ stands for the space $\w_1$ with the reverse linear order. On the other hand, the space $\w_1+1+\w_1^*$ has the following ZFC-property.

\begin{proposition}\label{p:sPar} The compact space $\w_1+1+\w_1^*$ is an $\IR$-soft corona.
\end{proposition}

\begin{proof} If $\w_1<\mathfrak p$, then the compact space $\w_1+1+\w_1^*$ is a  soft corona by Theorem~\ref{t:Par}(3) and Proposition~\ref{p:sss}. It remains to consider the case $\w_1=\mathfrak p$.
Write the linearly ordered space $L=\w_1+1+\w_1^*$ as $$L=\{x_\alpha\}_{\alpha\in\w_1}\cup\{z\}\cup\{y_\alpha\}_{\alpha\in\w_1}$$ where $$x_\alpha<x_\beta<z<y_\beta<x_\alpha$$for any $\alpha<\beta<\w_1$. For two elements $a\le b$ in $L$, consider the order intervals
$$
\begin{aligned}
&L[a,b]:=\{x\in L:a\le x\le b\},\;L[a,b):=\{x\in L:a\le x<b\},\\
&L(a,b]:=\{x\in L:a<x\le b\},\;L(a,b):=\{x\in L:a<x<b\}
\end{aligned}
$$ with end-points $a,b$.

Choose any countable set $D$, which is disjoint with $L$.
According to Theorem 1.2 in \cite{NV}, the equality $\w_1=\mathfrak p$ implies the existence of a tight Hausdorff gap on $D$, which is a transfinite sequence $\langle (X_\alpha,Y_\alpha)\rangle_{\alpha\in\w_1}$ of pairs of infinite subsets of $D$ satisfying the following conditions:
\begin{enumerate}
\item $X_\alpha\subset^* X_\beta$ and $Y_\alpha\subset^* Y_\beta$ for any countable ordinals $\alpha<\beta$;
\item $X_\alpha\cap Y_\alpha$ is finite for every $\alpha\in\w_1$;
\item for any disjoint sets $X,Y\subseteq D$ there exists $\alpha\in\w_1$ such that $X_\alpha\not\subseteq^* X$ or $Y_\alpha\not\subseteq^* Y$;
\item for any infinite set $I\subseteq D$ there exists $\alpha\in\w_1$ such that $I\cap(X_\alpha\cup Y_\alpha)$ is infinite.
\end{enumerate}
Here given sets $A,B$, we write $A\subseteq^* B$ (and $A\subset^* B$)  if the complement $A\setminus B$ is finite (and the complement $B\setminus A$ is infinite).

Consider the space $K=L\cup D$ endowed with the topology $\tau$ consisting of the sets $U\subseteq K$ satisfying the following conditions:
\begin{itemize}
\item if $x_\alpha\in U$ for some ordinal $\alpha\in\w_1$, then there exists an ordinal $\beta<\alpha$ such that $L(x_\beta,x_\alpha]\cup (X_\alpha\setminus X_\beta)\subseteq^* U$;
\item if $y_\alpha\in U$ for  some ordinal $\alpha\in\w_1$, then there exists an ordinal $\beta<\alpha$ such that $L[y_\alpha,y_\beta)\cup (Y_\alpha\setminus Y_\beta)\subseteq^* U$;
\item if $z\in U$, then there exists an ordinal $\alpha\in\w_1$ such that $L(x_\alpha,y_\alpha)\cup (D\setminus (X_\alpha\cup Y_\alpha))\subseteq^* U$.
\end{itemize}

\begin{claim} $K$ is a compact Hausdorff space.
\end{claim}

\begin{proof} The Hausdorff property of $K$ follows from the definition of the topology $\tau$. Also the definition of the topology $\tau$ implies that the induced topology on $L$ coincides with the topology generated by the linear order, which implies that the subspace $L$ of $K$ is homeomorphic to the compact linearly ordered space $\w_1+1+\w_1^*$. By \cite[3.12.1]{Eng}, the compactness of $K$ will follows as soon as we check that each infinite subset $S\subseteq K$ has a complete accumulation point, i.e., a point $x\in K$ such that for any neighborhood $O_x\subseteq K$ we have $|O_x\cap S|=|S|$. If $S$ is uncountable, then $|S|=\w_1=|S\cap L|$ and $S\cap L$ has a complete accumulation point by the compactness of $L$. So, we assume that $S$ is countable. If $S\cap L$ is infinite, then $S\cap L$ has a complete accumulation point by the compactness of $L$. So, we assume that $S\cap L$ is finite and hence $S\cap D$ is infinite. Using the condition (5), we can find the smallest cardinal $\alpha$ such that $S\cap D\cap (X_\alpha\cup Y_\alpha)$ is infinite. Then $x_\alpha$ or $y_\alpha$ is a (complete) accumulation point of $S$.
\end{proof}

The definition of the topology $\tau$ ensures that $D$ coincides with the set $K'$ of isolated points of the compact space $K$.

\begin{claim}\label{cl:cont} Any continuous function $\varphi:K\setminus \{z\}\to\IR$ has a continuous extension $\bar\varphi:K\to\IR$.
\end{claim}

\begin{proof}  It suffices to prove that $\osc_z(\varphi)=0$. To derive a contradiction, assume that $\e:=\osc_z(\varphi)>0$.

Since any closed unbounded sets in $[0,\w_1)$ have non-empty intersection, the continuous function $\varphi$ is eventually constant on the sets $L[x_0,z)$ and $L(z,y_0]$. This means that for some $\gamma\in\w_1$  and real numbers $a,b$ the sets $\varphi(L[x_\alpha,z))$ and $\varphi(L(z,y_\alpha])$ are equal to the singletons $\{a\}$ and $\{b\}$, respectively. By the normality and zero-dimensionality of the compact space $K$, the compact set $L[x_0,x_\gamma]$ has a clopen neighborhood $U_\gamma$ in $K$ which is disjoint with the compact set $L[x_{\gamma+1},y_0]$. By analogy, the compact set $L[y_\gamma,y_0]$ has a clopen neighborhood $W_\gamma$ in $K$ which is disjoint with the compact set $L[x_0,y_{\gamma+1}]$.

We claim that $a=b$. Assuming that $a\ne b$, put $\delta:=\frac12|a-b|$ and consider the open neighborhood $V=\{x\in K\setminus\{z\}:|\varphi(x)-a|<\delta\}$ of $L[x_\gamma,z)$ in $K$.

 For every $\alpha\in\w_1$, the inclusion $L[x_0,x_\alpha]\subseteq (U_\gamma\setminus W_\gamma)\cup V$ implies $X_\alpha\subseteq^*(U_\gamma\setminus W_\gamma)\cup V$.

The condition (3) yields a countable ordinal $\alpha>\gamma$ such that the set $Y_\alpha\cap(U_\gamma\setminus W_\gamma)\cup V$ is infinite and hence has  an accumulation point $y$ in $K$. The point $y$ belongs to $L\cap \overline{Y_\alpha}\cap (U_\gamma\setminus W_\gamma)\subseteq L[y_\alpha,y_0]\setminus W_\gamma\subseteq L[y_\alpha,y_\gamma)$ and hence has $\bar\varphi(y)=b$. Taking into account that $y\in L[y_\alpha,y_\gamma)$ and $L[y_\alpha,y_\gamma)\cap U_\gamma=\emptyset$, we conclude that $y\in\overline{V}$ and hence $b=\varphi(y)\in\overline{\varphi(V)}\subseteq [a-\delta,a+\delta]$, which contradicts the choice of $\delta=\frac12|a-b|$. This contradiction shows that $a=b$.

Now consider the open neighborhood $W=V_\gamma\cup W_\gamma\cup \{x\in K\setminus \{z\}:|\bar\varphi(x)-a|<\frac13\e\}$ of $L\setminus\{z\}$ in $K\setminus\{z\}$. Then the set $K\setminus W$ is closed and accumulates at $z$ as $\diam(\varphi(W\setminus(V_\gamma\cup W_\gamma)))\le\frac23\e<\e=\osc_z(\varphi)$. Moreover, the continuity of $\varphi$ ensures that $z$ is a unique accumulation point of the set $D\setminus W$. But this contradicts the condition (4). This contradiction shows that $\osc_z(\varphi)=0$ and hence $\varphi$ has a  continuous extension $\bar\varphi:K\to\IR$.
\end{proof}

\begin{claim}\label{cl:soft} The compactification $K$ is $\IR$-soft.
\end{claim}

\begin{proof} Given any bounded function $\varphi:D\to \IR$ with non-empty set $\Omega=\{x\in K:\osc_x(\varphi)>0\}$, we should find a homeomorphism $h\in\mathcal H(K,D)$ and $\e>0$ such that the set $\{y\in D:|\varphi(h(y))-\varphi(y)|>\e\}$ is infinite. By Claim~\ref{cl:cont}, the set $\Omega$ contains some point $x\in L\setminus\{z\}$. Using the first countability of $K$ at $x$, we can repeat the argument of the proof of Proposition~\ref{p:ss} and find a homeomorphism $h\in\mathcal H(K,D)$ and $\e>0$ such that the set $\{y\in D:|\varphi(h(y))-\varphi(y)|>\e\}$ is infinite.
\end{proof}

Since the remainder $L$ of the compactification $K$ is homeomorphic to $\w_1+1+\w_1^*$, Claim~\ref{cl:soft} implies that the space $\w_1+1+\w_1^*$ is an $\IR$-soft corona.
\end{proof}

 \section{Acknowledgements}

 The authors express their sincere thanks to KP Hart and Todd Eisworth for their answers to the MO-questions\footnote{\tt https://mathoverflow.net/q/309583, https://mathoverflow.net/q/351615} of the first author. Those answers were used in the proofs of Theorem~\ref{t:Par}(4) and Proposition~\ref{p:sPar}, respectively.

redpajama_set_name: RedPajamaArXiv


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text:
Simon Grayson Looking For Catapulting Run
North End manager Simon Grayson is looking for a big run of victories to catapult his side into contention for a Play-Off place come May.
The Lilywhites currently sit in tenth place, nine points behind the sixth-placed Sheffield Wednesday and the gaffer recognises that it will require an impressive run if his side are to be contenders for a top six finish.
However, he feels that with the working environment being a happy one, that anything is possible for a squad of players that in many people's eyes are already exceeding expectations.
"I think we have tried to create an atmosphere where people enjoy coming into work since we have been at the football club and results do make it easier, but there is a fine line between being complacent and doing what you need to do.
"We have 13 big games to go and we will approach them all in a positive manner with a big work ethic amongst ourselves and see where that takes us. The lads are enjoying what they are doing and that is a big part of what you need to do as a manager.
"We want to win as many games as possible. We are at that stage of the season where it all about results and not about performances. If you can put the two together, which we have had recently with the Brentford and the Birmingham games then that is even better. Saturday's game was a bit below par and a bit frustrating, but if you can win football matches between now and the end of the season then it can move you up the table.
"If we can keep winning at home and picking something up away then it is a good sign for us. We have done well recently at home, but it is a tough game on Saturday against QPR.
"We have had two hard fought away draws, so we are progressing in the right manner. If you look back, it is now four defeats in 25, which is very encouraging.
"The only disappointment is that if you really want to sustain a push for the Play-Offs, we have been short a couple of times in the last few months where some teams have gone and got four, five or six wins on the bounce, which can really catapult you up the division; for whatever reason we have not done that, but we will look to do it and win as many as possible.
"Everybody will be playing each other and there will some strange results as well where the teams at the bottom will get results against the teams around us and we have plenty of fixtures of teams who are around us, the likes of Newcastle, Leeds, Huddersfield, Reading, Norwich, Derby and Fulham, so we will have plenty to say in this battle."
The PNE chief also feels that his side's relative success thus far has been overlooked, despite sitting well above some of the Championship's other much-vaunted members.
"We are still under the radar to an extent," he continued. "Everybody is still talking about the big teams, what they are going to do and what sort of runs they can go on to get into the Play-Offs and that comes with the territory with the so-called bigger teams in the division and what they have spent.
"Teams are certainly aware of us and that has shown in recent games – Wigan last week, for example, worked extremely hard to stop our better players from affecting the game and you have to be aware of that; the more success you have as a team, the more you will come into focus and the more people will want to do to stop you from hurting them and that's what we look at times."
This weekend's visitors to PR1 have had excellent back-to-back victories and Ian Holloway's Queens Park Rangers will also be looking upwards rather than over their shoulders after they moved above the 40 point threshold on Tuesday night.
"You know what Ollie is like. He is a bubbly character; he is positive about what he is going to do. He got off to an average start, but has had two very good results in their last two games, winning the way they did at Birmingham and then it was a different type of dogged win on Tuesday night against Wigan, which bodes well for them.
"They have spent a bit of money, Ollie has turned the squad around, has pace and ability in the team and he will be looking forward to this fixture to keep them going in the right direction.
"It just shows you how well we are doing that we are above teams who have bigger budgets and who have spent a lot of money on individuals, it shows we have done really well. But it is not just related to money, it is about the players you have got and fight, quality and spirit you have got and all those ingredients can help you along the way.
"Money can get you so far, but it doesn't get you success and we are proving that."
pnefc


text:
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Twenty-Second Sunday in Ordinary Time
2017 Homilies
by Paul A. Magnano Pastor
We're led into this challenging gospel by the first reading from the prophet Jeremiah, "You duped me, O Lord, and I let myself be duped." Sometimes it's translated, "You have seduced me, Lord, and I've let myself be seduced." These are both softenings of the original text. What Jeremiah is really saying is closer to, "You have raped me, God, and I have let myself be raped."
Jeremiah seems to be saying, "I set out on this journey with you that I thought would be wonderful and consoling and gratifying and self-promoting; instead it's turned out to be a life with an awful lot of suffering. I don't know why I ever followed you!"
This lament prepares us for the gospel. Today's gospel reading is quite a reversal from last week's, an earlier part of Matthew 16 where Jesus is praising Peter, "You are a rock, Peter. You're the leader. You've got the keys of the Kingdom of Heaven. You did it right." You can just see Peter strutting around with immense self-confidence, thinking, "I'm the best."
And now, just a few verses later, the infallible pope shows himself, in fact, to be very fallible when Jesus goes so far as to call Peter a devil. How can one who got it so right now get it so wrong? Well, I hope that is comforting, because it's true of all of us – both sides can coexist in the same person.
I was ordained in Rome in the great basilica of St. Peter: there, written in Latin and Greek around the top of the cupola, it says, "Thou are Peter, and upon this rock I will build my church." This is as it should be, but I wonder why there wasn't a second ring of words saying, "Get behind me, Satan! You are an obstacle to me. You are thinking not as God does, but as human beings do," as it says in Matthew today. We acknowledge the positive, but we conveniently forget the negative – it's human nature, I guess.
When we start on our journey toward God, maybe we don't realize it, but the motivation is almost all about ourselves. I'm reflecting on my own spiritual life journey. During high school, at Seattle Prep, I'm buying my eternal fire insurance policy. I'm looking for personal advancement toward heaven. This isn't love of God yet. It's very well-disguised self-interest. I'm not really in it for God. I'm in it for myself, but it takes a while in the seminary to recognize that.
That's why Jesus says it so strongly and I put it on my book shelf in Rome: "If you don't lose your life, if you're in this to gain your life, you haven't yet understood what I'm talking about. You're missing the point." Jesus isn't saying this just to Peter or to myself. He's speaking to all who seek to follow him.
We start the journey largely out of self-interest, taking care of ourselves while thinking we're worshiping God. But we're doing the "work of the devil," as it were, precisely because it is passing for love of God, but it isn't. Whoever wishes to save their own life must lose it, because it's still all about "me" and my private gateway to heaven. It's not about God, truth, love, others, or the bigger world.
How could such smallness be God's great salvation? Why would that show any transformation, which is what the second reading talks about? "Be transformed by the renewal of your mind." This is Peter's moment of initiation, when he turns from earlier, selfish motivations to being drawn by actual desire for God and God's big picture.
It's probably the ultimate challenge, a complete reversal of direction for each of us when somewhere in the middle of life we have to ask the question, "Why am I doing what I'm doing? Why am I coming here for Mass on Sunday? Does God need me to come to church?" I don't think so. God is doing quite well without you and without me, and yet it's God who invites us into this great assembly.
We're really not coming here for God. We're here, I believe, for ourselves, to find out what life's all about. What does it all mean? It's we who have to change. It's we who have to reverse direction and recognize that it's in letting go of what we think is our identity that we actually find our true self. That's the great paradox.
Let me say it plainly: our lives are not about ourselves. We are one little instance of this mystery of life that's been going on for billions of years. All we can do is surrender to that Mystery that is larger than ourselves, bigger than me, more wonderful than me, and which will draw me forward, almost in spite of myself. Despite our struggles, we are persistent, especially in our search for God.
St. Benedict says in the Rule that seeking God is often "harsh and bitter." All we can do is trust it and allow it and suffer it, and surely enjoy it too. Being holy and being real always go together. It's not a matter of doing it right, it's a matter of "doing it!" It is a matter of letting go and losing who we think we are, to fall into who we really are, and always will be. Amen? Amen.
Paul A. Magnano
Christ Our Hope Catholic Church
mail@christourhopeseattle.org
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