text:
\section{Introduction}

A subfamily $\mathcal{I}\subseteq \mathcal{P}(X)$, the power set of some countable set $X$, is called an ideal if \begin{enumerate}
    \item $X\notin\mathcal{I}$,

    \item $A\cup B\in\mathcal{I}$ for all $A,B\in\mathcal{I}$, and

    \item $A\in\mathcal{I}$ whenever $A\subseteq B$ and $B\in\mathcal{I}$.
\end{enumerate} Ideals on countable sets play an important role in infinite combinatorics. In one hand we have the quotients modulo some ideal\footnote{The reader is referred to \cite{handbook} for general concepts and notations on Boolean algebras.}: If $\mathcal{I}\subseteq \mathcal{P}(X)$ is an ideal, for some countable set $X$, define on $\mathcal{P}(X)$ the equivalence relation $\sim_{\mathcal{I}}$, where $A\sim_{\mathcal{I}}B$ if their symmetric difference $A\triangle B\in\mathcal{I}$, for all $A,B\subseteq X$. Then on the set of equivalence classes $\lbrace[A]_{\mathcal{I}}\mid A\subseteq X\rbrace$ define the operations $[A]_{\mathcal{I}}\vee[B]_{\mathcal{I}}:=[A\cup B]_{\mathcal{I}}$, $[A]_{\mathcal{I}}\wedge[B]_{\mathcal{I}}:=[A\cap B]_{\mathcal{I}}$ and $-[A]_{\mathcal{I}}:=[X\setminus A]_{\mathcal{I}}$. It is not hard to see that these operations are well defined and that define a Boolean algebra on the set of equivalence classes of $\sim_\mathcal{I}$, denoted $\mathcal{P}(X)\slash\mathcal{I}$. When studying this structure the notation $[A]_\mathcal{I}$ is mostly avoided and positive elements of $\mathcal{P}(X)\slash\mathcal{I}$ are identified with the elements of $\mathcal{I}^+ :=\mathcal{P}(X)\setminus\mathcal{I}$, as well as the Boolean operations are identified with those of $\mathcal{P}(X)$. The order relation of $\mathcal{P}(X)\slash\mathcal{I}$ being important and not as directly induced by $\subseteq$ as the Boolean operations, its interpretation on $\mathcal{I}^+$ is highlighted: $A\subseteq_{\mathcal{I}}B$ iff $[A]_\mathcal{I} \leq[B]_{\mathcal{I}}$ iff $A\setminus B\in\mathcal{I}$, for all $A,B\in\mathcal{I}^+$.

Capital among these structures is the case when $X=\omega$, $\mathcal{I}=fin:=[\omega]^{<\omega}$, and $\subseteq^* :=\subseteq_{fin}$. Many combinatorial substructures of $\mathcal{P}(\omega)\slash fin$ and their related cardinal invariants have been extrapolated to general quotients. A short list of some of them, relevant to this text, is given here.

\begin{mydef}
\label{cardinal_invariants_quotient}
Take an ideal $\mathcal{I}\subseteq \mathcal{P}(X)$, of some countable set $X$, and $A,B\in\mathcal{I}^+$.

\begin{enumerate}
    \item If $A\cap B\in\mathcal{I}$, it will be said that $A$ and $B$ are $\mathcal{I}$-almost disjoint. A subfamily of $\mathcal{I}^+$ consisting of pairwise $\mathcal{I}$-almost disjoint sets is called an $\mathcal{I}$-ad family. If said family is maximal with this property it will be called an $\mathcal{I}$-mad family or an infinite partition of $\mathcal{P}(X)\slash\mathcal{I}$. The cardinal $\mathfrak{a}(\mathcal{I})$ is defined as the smallest size of an infinite $\mathcal{I}$-mad family.

    \item If $\mathcal{T}\subseteq\mathcal{I}^+$ is a well-ordered substructure of $(\mathcal{I}^+ ,\supseteq_{\mathcal{I}})$ with no lower bound in $\mathcal{I}^+$, it will be called an $\mathcal{I}$-tower. The cardinal $\mathfrak{t}(\mathcal{I})$ is defined as the smallest size of an $\mathcal{I}$-tower.

    \item If $A\cap B\in\mathcal{I}^+$ and $B\setminus A\in\mathcal{I}^+$, it will be said that $A$ splits $B$. The cardinal $\mathfrak{s}(\mathcal{I})$ is defined as the smallest size of a family $\mathcal{S}\subseteq\mathcal{I}^+$ such that for all $X\in\mathcal{I}^+$ there exists $Y\in\mathcal{S}$ such that $Y$ splits $X$.

    \item The cardinal $\mathfrak{r}(\mathcal{I})$ is defined as the smallest size of a subfamily $\mathcal{R}\subseteq\mathcal{I}^+$ such that no element of $\mathcal{I}^+$ splits all elements of $\mathcal{R}$. In this case $\mathcal{R}$ is called a reaping family.
\end{enumerate}
\end{mydef}

All these cardinals are generalizations of the well-studied case when $\mathcal{I}=fin$. In that case no ideal is specified and the cardinal is identified by the corresponding fraktur font letter alone. For a general and basic understanding of this case the reader is referred to \cite{blass}. For results of both general and specific cases of these cardinal invariants, though mostly of $\mathfrak{a}(\mathcal{I})$, the reader is referred to \cite{steprans}, \cite{farkas} and \cite{el_juris_y_el_dilip}. For more general cases of infinite Boolean algebras a great resource is \cite{monk}.

Another source of study for an ideal $\mathcal{I}$ on a countable set $X$ is seeing it as a combinatorial substructure of $\mathcal{P}(X)\slash[X]^{<\omega}$. There the following cardinal invariants arise.

\begin{mydef}
\label{cichon}

  Let $\mathcal{I}$ be an ideal on the countable set $X$. Define

    \begin{itemize}
        \item (Additivity) $add^* (\mathcal{I}):=\min\lbrace\vert\mathcal{F}\vert\mid\mathcal{F}\subseteq\mathcal{I}~\neg\exists I\in\mathcal{I}~\forall A\in\mathcal{F}~A\nsubseteq^* I\rbrace$

        \item (Covering) $cov^* (\mathcal{I}):=\min\lbrace\vert\mathcal{F}\vert\mid\mathcal{F}\subseteq\mathcal{I}~\forall B\in[X]^\omega ~\exists A\in\mathcal{F}~\vert B\cap A\vert=\omega\rbrace$

        \item (Uniformity) $non^* (\mathcal{I}):=\min\lbrace\vert\mathcal{X}\vert\mid\mathcal{X}\subseteq[X]^\omega ~\neg\exists A\in\mathcal{I}~\forall B\in\mathcal{X}~\vert A\cap B\vert=\omega\rbrace$

        \item (Cofinality) $cof^* (\mathcal{I}):=\min\lbrace\vert\mathcal{F}\vert\mid\mathcal{F}\subseteq\mathcal{I}~\forall A\in\mathcal{I}~\exists B\in\mathcal{F}~A\subseteq^* B\rbrace$
    \end{itemize}
\end{mydef}

In Section \ref{quotient} of this text an ideal on $\omega\times\omega$, the \textit{Nowhere Centered ideal} $\mathcal{NC}$ is introduced and the structures of its quotient, as defined in Definition \ref{cardinal_invariants_quotient}, are studied, with great emphasis on $\mathfrak{a}(\mathcal{I})$ and infinite partitions. In section \ref{third_dimension} hints are given to higher dimensional generalizations of $\mathcal{NC}$, alongside an important result on infinite partitions on the corresponding quotients. Finally in Section \ref{add_cov_non_cof} the cardinal invariants of Definition \ref{cichon} relative to the ideal $\mathcal{NC}$ are studied. The reader is assumed to have a basic knowledge on concepts and notation of set theory and forcing.

Since most of this work is set in $\omega\times\omega$, for $A\subseteq\omega\times\omega$ and $n<\omega$ set the notation $$A(n):=\lbrace m<\omega\mid(n,m)\in A\rbrace.$$ Also, for $X\subseteq\omega$ and $\lbrace Y_n \mid n\in X\rbrace\subseteq \mathcal{P}(\omega)$ define $$\coprod_{n\in X}Y_n :=\bigcup_{n\in X}\lbrace n\rbrace\times Y_n .$$

\section{The Nowhere Centered Ideal and its quotient}
\label{quotient}

Recall that a family $\mathcal{X}\subseteq[\omega]^\omega$ is said to be \textit{centered}, if $\bigcap F$ is infinite for all $F\in[\mathcal{X}]^{<\omega}$. From this almost natural concept in infinite combinatorics an ideal on $\omega\times\omega$ is defined.

\begin{mydef}
\label{nci}

The Nowhere Centered ideal is defined on $\omega\times\omega$ as follows
$$\mathcal{NC}:=\lbrace A\subseteq\omega\times\omega\mid\forall X\in[\omega]^\omega ~the~set~\lbrace A(n) : n\in X \rbrace~is~not~centered\rbrace.$$
\end{mydef}

Obviously $\omega\times\omega\notin\mathcal{NC}$ and $\mathcal{NC}$ is closed under subsets. Take $A_0 , A_1 \subseteq\omega\times\omega$ and suppose that $A_0 \cup A_1 \notin\mathcal{NC}$. There exists $X\in[\omega]^\omega$ such that $\lbrace A_0 (n)\cup A_1 (n)\mid n\in X\rbrace$ is a centered family. If $\mathcal{U}$ is an ultrafilter extending this family, there exist $i<2$ and $X^\prime \in[X]^\omega$ such that $\lbrace A_i (n)\mid n\in X^\prime \rbrace$ is subset of $\mathcal{U}$, and hence is a centered family. Therefore $A_i \notin\mathcal{NC}$, and we conclude that $\mathcal{NC}$ is closed under finite unions, and hence an ideal.

\begin{obs}
\label{nowhere}
If $X\in[\omega]^\omega$ and $A\in\mathcal{NC}$, then there exist $n<\omega$ and $Y\in[X]^\omega$ such that $\vert A(m)\cap Y\vert<\omega$, for all $m\geq n$.
\end{obs}

Indeed we claim that there exists $n<\omega$ such that $\vert X\setminus\bigcup_{m\in F}A(m)\vert=\omega$, for all $F\in[\omega\setminus n]^{<\omega}$. Otherwise for all $n<\omega$ there exists $F\in[\omega\setminus n]^{<\omega}$ such that $X\subseteq^* \bigcup_{m\in F}A(m)$. But hence any ultrafilter containing $X$ would give us  $B\in[\omega]^\omega$ such that $\lbrace A(m)\mid m\in B\rbrace$ is a centered family, which is a contradiction. Given such $n<\omega$, getting such $Y\in[X]^\omega$ is an easy task.

One could see Observation \ref{nowhere} as a \textit{literary justification} for the naming of the ideal $\mathcal{NC}$, since it has a similar flavour as the observation that whenever we have $\sigma\in\omega^{<\omega}$ and a subtree $T\subseteq\omega^{<\omega}$ whose branches form a nowhere dense set of $\omega^\omega$, then there exists $\tau\in\omega^{<\omega}$, an extension of $\sigma$, such that $\langle\tau\rangle\cap T=\emptyset$. In this case given the rectangle $\omega\times X$ and $A\in\mathcal{NC}$ we get an ``extension" of the rectangle which ``avoids" $A$. A measure of this avoidance as well as a known face to compare $\mathcal{NC}$ to is given by the following ideal.

\begin{mydef}
 $fin\times fin:=\lbrace A\subseteq\omega\times\omega\mid\exists n<\omega~\forall m\geq n~\vert A(m)\vert<\omega\rbrace$.
\end{mydef}

 The ideal $fin\times fin$ is often thought of as the ideal on $\omega\times\omega$ generated by the columns $\lbrace n\rbrace\times\omega$, for $n<\omega$, and the sets below a function $\lbrace(n,m)\in\omega\times\omega\mid m\leq f(n)\rbrace$, for $f\in\omega^\omega$. Since for $A\in fin\times fin$ and $X\in[\omega]^\omega$ there exists $n$ such that $$\bigcap_{i\in X\cap n} A(i)<\omega,$$ it easily follows that $fin\times fin\subseteq\mathcal{NC}$. This inequality is strict. Indeed, if $\lbrace Y_n \mid n<\omega\rbrace\subseteq [\omega]^\omega$ is an almost disjoint family, the set $$Y:=\coprod_{n<\omega}Y_n$$ is an element of $\mathcal{NC}$ which does not lie in $fin\times fin$.

For getting to know the quotient $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ take $A\in\mathcal{NC}^+$. By definition there exists $X\in[\omega]^\omega$ such that $$\lbrace A(n)\mid n\in X\rbrace$$ is a centered family. If $Y\in[\omega]^\omega$ is a pseudointersection of said family, it follows that $$X\times Y\subseteq_{fin\times fin} A,$$ which means that $$\qquad\qquad\qquad X\times Y\subseteq_{\mathcal{NC}} A. \qquad \qquad \qquad (\star)$$ Observe that $Y$ is as much a key in determining $A\in\mathcal{NC}^+$ as $X$ is. Therefore $\mathcal{NC}^+$ is the second coordinate projection of the subset of $[\omega]^\omega \times \mathcal{P}(\omega\times\omega)$ consisting of all $(Y,A)$ satisfying the Borel formula $$\forall n<\omega\exists m\geq n\exists k<\omega\forall l\geq k(l\in Y\Rightarrow l\in A(m)),$$ which means there exist infinite $m<\omega$ such that $Y\subseteq^* A(m)$. It follows that the ideal $\mathcal{NC}$ is coanalytic.\footnote{Since $\mathcal{P}(\omega\times\omega)$ has a natural topology induced by that of the Polish space $2^{\omega\times\omega}$, its ideals can be studied by its topological properties. For elements of descriptive set theory see \cite{kechris}.}

From $(\star)$ it also follows that the set $$\qquad\qquad\lbrace[X\times Y]_{\mathcal{NC}}\mid X,Y\in[\omega]^\omega\rbrace \qquad\quad \circledast$$ is dense in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$. This important fact will be heavily used all through this section. This means that $\mathcal{P}(\omega)\slash fin\oplus \mathcal{P}(\omega)\slash fin$ is (isomorphic to) a dense subalgebra of $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}. $\footnote
$\mathcal{P}(\omega)\slash fin\oplus \mathcal{P}(\omega)\slash fin$ is isomorphic to the clopen sets of $(\beta\omega\setminus\omega)^2$, where $\beta\omega\setminus\omega$ is the space of free ultrafilters on $\omega$. Its non-zero elements are of the form $\bigcup_{i<n}\widetilde{X}_i \times\widetilde{Y}_i$, where each $\widetilde{X}_i$ and $\widetilde{Y}_i$ are the sets of free ultrafilters containing $X_i$ and $Y_i$, some infinite subsets of $\omega$. The very conception of $\mathcal{NC}$ derives from the search for a quotient approximating this free product, and so this relation is highlighted. All cardinal invariants like $\mathfrak{a}(2)$ in this text refer to cardinal invariants of this Boolean algebra.} It is also easy to see that $$\lbrace[\omega\times X]_{\mathcal{NC}}\mid X\in[\omega]^\omega \rbrace$$ is a (regular) subalgebra of $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ isomorphic to $\mathcal{P}(\omega)\slash fin$.

There are several combinatorial ways in which the Boolean algebras $\mathcal{P}(\omega)\slash fin$, $\mathcal{P}(\omega)\slash fin\oplus \mathcal{P}(\omega)\slash fin$ and $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ are similar.

\begin{theorem}
$\mathfrak{s}=\mathfrak{s}(2)=\mathfrak{s}(\mathcal{NC})$.
\end{theorem}

\begin{proof}
The first equality holds because $\mathfrak{s}(A\oplus B)=\min\lbrace\mathfrak{s}(A),\mathfrak{s}(B)\rbrace$ for all non-atomic Boolean algebras $A$ and $B$. This fact can be found in Theorem 10 of \cite{mio}. For proving $\mathfrak{s}\leq\mathfrak{s}(\mathcal{NC})$ first take $\kappa<\mathfrak{s}$ and $\lbrace A_\alpha \mid\alpha<\kappa\rbrace\subseteq\mathcal{NC}^+ $. Since the family $$\lbrace A_\alpha (n)\mid\alpha<\kappa,~n<\omega\rbrace$$ is not splitting in $\mathcal{P}(\omega)\slash fin$, take $Y\in[\omega]^\omega$ witnessing this fact. Define $$K:=\lbrace\alpha<\kappa\mid\exists^{\infty}n<\omega~(Y\subseteq^* A_\alpha (n))\rbrace.$$ For each $\alpha\in K$ consider the set $X_\alpha :=\lbrace n<\omega\mid Y\subseteq^* A_\alpha (n)\rbrace$. The family $\lbrace X_\alpha \mid\alpha\in K\rbrace$ is not splitting in $\mathcal{P}(\omega)\slash fin$. Take $X\in[\omega]^\omega$ witnessing this fact. The set $X\times Y$ witnesses that  $\lbrace A_\alpha \mid\alpha<\kappa\rbrace\subseteq\mathcal{NC}^+ $ is not splitting in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$. Indeed, take $\alpha<\kappa$. If $\alpha\in K$ and $X\subseteq^* X_\alpha$, then $X\times Y\subseteq_{\mathcal{NC}}A_\alpha$. If $\alpha\in K$ and $\vert X\cap X_\alpha \vert<\omega$, then $\vert Y \cap A_\alpha (n)\vert <\omega$ for almost all $n\in X$. On the other hand if $\alpha\notin K$, then $\vert Y \cap A_\alpha (n)\vert <\omega$ for almost all $n<\omega$. From both last cases it follows that $(X\times Y)\cap A_\alpha \in\mathcal{NC}$. We conclude that $\mathfrak{s}\leq\mathfrak{s}(\mathcal{NC})$.

Now take family $\lbrace X_\alpha \mid\alpha<\mathfrak{s}\rbrace\subseteq[\omega]^\omega$ splitting in $\mathcal{P}(\omega)\slash fin$. The family $$\mathcal{S}:=\lbrace X_\alpha \times X_\beta \mid\alpha,~\beta<\mathfrak{s}\rbrace$$ is splitting in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$. Take $A\in\mathcal{NC}^+$, which will be proved to be split by some element of $\mathcal{S}$. Since the set $\circledast$ is dense, w.l.o.g $A=X\times Y$ for some $X,~Y\in[\omega]^\omega$. Take $\alpha<\mathfrak{s}$ and $\beta<\mathfrak{s}$ such that $X_\alpha$ splits $X$ and $X_\beta$ splits $Y$. It follows that both $(X\times Y)\cap(X_\alpha \times X_\beta)\in\mathcal{NC}^+$ and $(X\times Y)\setminus(X_\alpha \times X_\beta)\in\mathcal{NC}^+$, i.e. $X_\alpha \times X_\beta$ splits $A$. We conclude that $\mathfrak{s}(\mathcal{NC})\leq\mathfrak{s}$.

\end{proof}

Concerning reaping families in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ matters are not as straightforward. Nevertheless non surprising bounds were found for $\mathfrak{r}(\mathcal{NC})$. A family $\mathcal{R}\subseteq[\omega]^\omega$ is said to be $\sigma$-\textit{reaping} if for all $\lbrace X_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$ there exists $A\in\mathcal{R}$ such that for all $n<\omega$ either
$A\subseteq^* X_n$ or $\vert A\cap X_n \vert<\omega$. The cardinal $\mathfrak{r}_\sigma$ is defined to be the smallest cardinality of a $\sigma$-reaping family.

\begin{theorem}
$\mathfrak{r}\leq\mathfrak{r}(2)\leq\mathfrak{r}(\mathcal{NC})\leq\mathfrak{r}_\sigma$.
\end{theorem}

\begin{proof}
Take $\kappa<\mathfrak{r}$ and $\lbrace A_{\alpha} ,B_{\alpha}\mid\alpha<\kappa\rbrace\subseteq[\omega]^\omega$. Take $X,Y\in[\omega]^\omega$ such that $X$ splits $A_\alpha$ and $Y$ splits $B_\alpha$, for all $\alpha<\kappa$. Therefore $\widetilde{X}\times\widetilde{Y}$ splits (as clopen subset of $(\beta\omega\setminus\omega)^2$) $\widetilde{A}_\alpha \times\widetilde{B}_\alpha$, for all $\alpha<\kappa$. We conclude that $\kappa<\mathfrak{r}(2)$.

Take $\kappa<\mathfrak{r}(2)$ and $\lbrace A_\alpha \mid\alpha<\kappa\rbrace\subseteq\mathcal{NC}^+$. For $\alpha<\kappa$ take $X_\alpha ,~Y_\alpha \in[\omega]^\omega$ such that $(X_\alpha \times Y_\alpha )\subseteq_{\mathcal{NC}}A_\alpha$. There exist $C_0 ,...,C_{n-1},D_0 ,...,D_{n-1}\in[\omega]^\omega$, for some $n<\omega$, such that $$\widetilde{X}_\alpha \times\widetilde{Y}_\alpha \cap\bigcup_{i<n}\widetilde{C}_i \times\widetilde{D}_i\neq\emptyset\neq\widetilde{X}_\alpha \times\widetilde{Y}_\alpha \setminus\bigcup_{i<n}\widetilde{C}_i \times\widetilde{D}_i,$$ for all $\alpha<\kappa$. Therefore $$X_\alpha \times Y_\alpha \cap\bigcup_{i<n}C_i \times D_i \in\mathcal{NC}^+$$ and $$X_\alpha \times Y_\alpha \setminus\bigcup_{i<n} C_i \times D_i \in\mathcal{NC}^+,$$ for all $\alpha<\kappa$. We conclude that $\kappa<\mathfrak{r}(\mathcal{NC}) $.

Let $\mathcal{R}=\lbrace A_\alpha \mid\alpha<\mathfrak{r}_\sigma \rbrace\subseteq[\omega]^\omega$ be a $\sigma$-reaping family. Consider the family $\lbrace A_\alpha \times A_\beta \mid\alpha,\beta<\mathfrak{r}_\sigma \rbrace$ and take $B\in\mathcal{NC}^+$. Since infinitely many elements of $\lbrace B(n)\mid n<\omega\rbrace$ are infinite, there exists $\alpha<\mathfrak{r}_\sigma$ such that for all $n<\omega
$ either $A_\alpha \subseteq^* B(n)$ or $\vert A_\alpha \cap B(n)\vert<\omega$. Define $X:=\lbrace n<\omega \mid A_\alpha \subseteq^* B(n)\rbrace$. Since $\mathcal{R}$ is a reaping family, there exists $\beta<\mathfrak{r}_\sigma$ such that either $A_\beta \subseteq^* X$ or $\vert A_\beta \cap X\vert<\omega$. In the first case it follows that $A_\beta \times A_\alpha \subseteq_{\mathcal{NC}}B$; in the second one, that $(A_\beta \times A_\alpha)\cap B\in\mathcal{NC}$. In either case we conclude that $\lbrace A_\alpha \times A_\beta \mid\alpha,\beta<\mathfrak{r}_\sigma \rbrace$ is a reaping family in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$. Therefore $\mathfrak{r}(\mathcal{NC})\leq\mathfrak{r}_\sigma$.
\end{proof}

Notwithstanding these results, considering towers and infinite partitions huge differences between the corresponding cardinal invariants of $\mathcal{P}(\omega)\slash fin$ and $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ begin to arise. On one hand any infinite partition of   $\mathcal{P}(\omega)\slash fin$ induces a partition of  $\mathcal{P}(\omega)\slash fin\oplus \mathcal{P}(\omega)\slash fin$, whose infinite partitions induce infinite partitions on $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$. It follows that $$\mathfrak{a}(\mathcal{NC})\leq\mathfrak{a}(2)\leq\mathfrak{a}.$$ Nevertheless the left inequality is strict. This stems from the fact that the all countable subsets of $$\lbrace[\omega\times X]_{\mathcal{NC}}\mid X\in[\omega]^\omega \rbrace$$
have a supremum in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$.

\begin{lemma}
\label{aI=omega}
If $\lbrace Y_n : n<\omega\rbrace\subseteq[\omega]^\omega ,$ then

$$\coprod_{n<\omega}\bigcup_{i\leq n} Y_i = \bigvee_{n<\omega} \omega\times Y_n .$$
\end{lemma}

\begin{proof}
It is immediate that $$\omega\times Y_n \subseteq_{\mathcal{NC}}\coprod_{n<\omega}\bigcup_{i\leq n}Y_i$$ for all $n<\omega$. Take $A\times B\subseteq_{\mathcal{NC}}\coprod_{n<\omega}\bigcup_{i\leq n}Y_i$, with $A,B\in[\omega]^\omega$. If $B\nsubseteq^* \bigcup_{i\leq n}Y_i$ for all $n<\omega$, then there exists $B^\prime \in[B]^\omega$ such that $\vert B^\prime \cap Y_n \vert<\omega$ for all $n<\omega$, but this is clearly a contradiction. Hence there exists $n<\omega$ such that $B\subseteq^* \bigcup_{i\leq n}Y_i$. It follows that $$A\times B\subseteq_{\mathcal{NC}}\bigcup_{i\leq n}\omega\times Y_i .$$ Since every lower bound of $\coprod_{n<\omega}\bigcup_{i\leq n}Y_i$ is met by $\omega\times Y_n$ for some $n<\omega$, the lemma follows.

\end{proof}

It analogously follows that for any countable family $\lbrace X_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$, the set $\coprod_{n<\omega}\bigcap_{i\leq n}X_i$ is the infimum of the family $\lbrace \omega\times X_n \mid n<\omega\rbrace$. Observe that also if $\lbrace X_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$ is an increasing (decreasing) family the set $\coprod_{n<\omega}X_n$ is the supremum (infimum) of the family $\lbrace \omega\times X_n \mid n<\omega\rbrace$.

There are several other instances where a countable family $\lbrace A_n \mid n<\omega\rbrace\subseteq\mathcal{NC}^+$ has supremum in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$, e.g. if there exists $\lbrace X_n \mid n<\omega\rbrace$ a partition of $\omega$ such that $A_n (m)\subseteq^* X_n$ for all $n,m<\omega$. It is also worth noticing, as it will be proved later, that not all countable subfamilies of $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ have supremum. Nevertheless Lemma \ref{aI=omega} proves to be strong enough for  getting interesting results.

\begin{corollary}
$\mathfrak{a}(\mathcal{NC})=\omega$.
\end{corollary}

\begin{proof}
Take $\lbrace A_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$ a partition of $\omega$. Then $$\lbrace\omega\times A_n \mid n<\omega\rbrace\cup\lbrace\omega\times\omega\setminus\bigvee_{n<\omega}\omega\times A_n \rbrace$$ is clearly an $\mathcal{NC}$-mad family witnessing this statement.
\end{proof}

Furthermore we have that $\mathfrak{t}(\mathcal{NC})=\omega$. The equality $\mathfrak{a}(\mathcal{I})=\omega$ not being an uncommon result when $\mathcal{I}$ is a definable ideal on a countable set $X$, as can be seen in Chapter 2 of \cite{farkas}, the cardinal $\overline{\mathfrak{a}}(\mathcal{I})$, defined as the smallest size of an uncountable $\mathcal{I}$-mad family, was introduced. On one hand the study of this cardinal helps deepen the understanding of the combinatorics $\mathcal{P}(X)\slash\mathcal{I}$. On the other hand, be it equal to $\mathfrak{a}(\mathcal{I})$ or not, it often reflects some of the behaviour of $\mathfrak{a}$ (see \cite{farkas}). The next step in this work is to study the cardinal invariant $\overline{\mathfrak{a}}(\mathcal{NC})$. On that matter we have the following result.

\begin{theorem}
\label{t=omega_1}
If $\mathfrak{t}=\omega_1$, then $\overline{\mathfrak{a}}(\mathcal{NC})=\omega_1$.
\end{theorem}

\begin{proof}
Suppose that $\mathfrak{t}=\omega_1$ and that $\lbrace X_\alpha :\alpha <\omega_1 \rbrace\subseteq[\omega]^\omega$ is an increasing tower of $\mathcal{P}(\omega)\slash fin$ (i.e. the complements of its elements form a tower). For $\alpha <\omega_1 $ define $$A_\alpha :=\omega\times X_\alpha \setminus\bigvee_{\beta<\alpha}\omega\times X_\beta .$$

These sets are well defined from lemma \ref{aI=omega} because each $\alpha$ is a countable ordinal. From its very definition it follows that  $\lbrace A_\alpha :\alpha<\omega_1 \rbrace$ is an $\mathcal{NC}$-ad family. Take $A,B\in[\omega]^\omega$ and let $\alpha$ be the least ordinal number less than $\omega_1$ such that $\vert B\cap X_\alpha \vert =\omega$. It follows that $(A\times B)\cap(\omega\times X_\alpha) \in\mathcal{NC}^+$ and that $(A\times B)\cap(\omega\times X_\beta )\in\mathcal{NC}$ for all $\beta <\alpha$. Therefore $(A\times B)\cap A_\alpha \in\mathcal{NC}^+$. Then we have an $\mathcal{NC}$-mad family of size $\omega_1$.
\end{proof}

Recall that $\mathfrak{b}$, is the smallest size of a family $\mathcal{F}\subseteq\omega^\omega$ such that for all $g\in\omega^\omega$ there exists $f\in\mathcal{F}$ such that $f(m)> g(m)$, for infinitely many $m<\omega$ ($f\nleq^* g$, in standard notation).  It is known that adding $\kappa$ many Hechler reals to a GCH model, for an uncountable regular cardinal $\kappa$, we get a model satisfying $\mathfrak{t}=\omega_1$ and $\mathfrak{b}=\kappa=\mathfrak{c}$ (see \cite{baumgartner} or \cite{brendle}). On the other hand one can prove that $\mathfrak{b}\leq\mathfrak{a}(2)$ (see \cite{note_on_products}). From this we get the following consistency result.

\begin{corollary}
\label{small_uncountable_}
If $\kappa$ is an uncountable regular cardinal of a model $V$ of ZFC, there exists a $ccc$ generic extension of $V$ where $\overline{\mathfrak{a}}(\mathcal{NC})=\omega_1 <\mathfrak{b}=\mathfrak{a}(2)=\mathfrak{a}=\kappa$.
\end{corollary}

Since $\mathfrak{b}\leq\mathfrak{a}(fin\times fin)$, $\mathfrak{b}\leq\overline{\mathfrak{a}}(\mathcal{I})$, if $\mathcal{I}$ is an analytic $P$-ideal\footnote{An ideal $\mathcal{I}$ is a $P$-ideal iff $add^* (\mathcal{I})>\omega$.}, and $add(\mathcal{M})\leq\overline{\mathfrak{a}}(nwd)$,\footnote{$add(\mathcal{M})$ is the smallest size of a family of meager sets of the real line whose union is not a meager set. It is also equal to $\kappa$ in this generic extension.} where $nwd$ is the ideal of nowhere dense subsets of the rationals,
these cardinal invariants are also equal to $\kappa$ in the model of Corollary \ref{small_uncountable_}. All this signals a huge contrast between the combinatorial nature of $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ and that of other definable quotients, as well to that of its subalgebras $\mathcal{P}(\omega)\slash fin$ and $\mathcal{P}(\omega)\slash fin\oplus \mathcal{P}(\omega)\slash fin$.

Other small substructure is given in the following theorem. Since it depends on the existence of \textit{Aronszajn trees}, a note on trees and their notation will be useful for the remaining of this section.

If $\kappa$ and $\lambda$ are infinite ordinals, $T\subseteq\kappa^{<\lambda}$ will be called a (sub)tree if $\sigma\upharpoonright\alpha\in T$, for all $\sigma\in T$ and $\alpha\in dom(\sigma)$. If $\alpha<\lambda$, the $\alpha$-th level of $T$ is the set $$T_\alpha :=\lbrace\sigma\in T\mid dom(\sigma)=\alpha\rbrace.$$ Accordingly the restriction to $\alpha$ is the subtree of $\kappa^{<\alpha}$ $$T_{<\alpha}:=\lbrace\sigma\in T\mid dom(\sigma)<\alpha\rbrace.$$ The (cofinal) \textit{branches} of $T$ are the elements of $$[T]:=\lbrace f\in\kappa^\lambda \mid\forall\alpha<\lambda~f\upharpoonright\alpha\in T\rbrace.$$ We will say that $T$ is \textit{well-pruned} if for all $\sigma\in T$ and all $dom(\sigma)<\alpha<\lambda$ there exists $\tau\in T_{\alpha}$ such that $\sigma\subseteq\tau$.\footnote{For the purposes of this text this definition will be enough. For a more standard definition of well-pruned tree see \cite{kunen}.} A tree $T\subseteq\omega^{<\omega_1}$ is called an \textit{Aronszajn} tree if $0<\vert T_\alpha \vert\leq\omega$, for all $\alpha<\omega_1$,  and $[T]=\emptyset$. The existence of Aronszajn trees is provable in ZFC. Basic and abundant information on this subject can be found in Chapter III of \cite{kunen}.

\begin{theorem}
\label{aronszjan2}
Let $T\subseteq\omega^{<\omega_1}$ be an Aronszajn tree. Take $\lbrace A_\sigma \mid \sigma\in T\rbrace\subseteq[\omega]^\omega$ such that if $\sigma\subseteq\tau\in T$, then $A_\tau \subseteq^* A_\sigma$ and $\lbrace A_\sigma \mid \sigma\in T_\alpha \rbrace$ is a partition of $\omega$, for all $\alpha <\omega_1$. Then the sets $X_\alpha :=\bigvee_{\sigma \in T_\alpha }\omega\times A_\sigma$ for $\alpha<\omega_1$ form an $\mathcal{NC}$-tower.
\end{theorem}

\begin{proof}
Suppose there exist $A,B\in[\omega]^\omega$ such that $A\times B\subseteq_{\mathcal{NC}}X_\alpha$, for all $\alpha<\kappa$. This means that for all $\alpha$ there exists $F_\alpha \in[ T_\alpha ]^{<\omega}$ such that $$B \subseteq^* \bigcup_{\sigma\in F_\alpha} A_\sigma.$$ Extending $\lbrace\bigcup_{\sigma\in F_\alpha}A_\sigma \mid\alpha<\omega_1 \rbrace$ to an ultrafilter we can choose $\sigma_\alpha$, for all $\alpha<\omega_1$, such that $\lbrace A_{\sigma_\alpha}\mid\alpha<\omega_1 \rbrace$ is a centered family. But this means that $\lbrace\sigma_\alpha \mid\alpha<\omega_1 \rbrace$ is a branch of $T$, which is a contradiction.
\end{proof}

While Corollary \ref{small_uncountable_} detached the ideal $\mathcal{NC}$ from many definable ideals, Theorem \ref{aronszjan2} provides it with a companion: it was proved in \cite{szymanski} that $\mathcal{P}(\omega\times\omega)\slash fin\times fin$ has a tower of size $\omega_1$.

Given this suite of results on small substructures of $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$, one could wonder if this tower could be used in a way similar to that of proposition \ref{t=omega_1} to build a partition of size $\omega_1$, whose existence were provable in ZFC. However this is not the case for this idea. To prove this we will first prove some limits that iterating $\bigvee$ and $\bigwedge$, on countable indices, has on this Boolean algebra.

\begin{proposition}
\label{gdelta,fsigma}
Let $T\subseteq\omega^{<\omega}$ be a subtree and take  $\lbrace A_\sigma \mid \sigma\in T\rbrace\subseteq[\omega]^\omega$ such that if $\sigma\subseteq\tau\in T$, then $A_\tau \subseteq^* A_\sigma$ and $\lbrace A_\sigma \mid \sigma\in T_n \rbrace$ is a partition of $\omega$, for all $n<\omega$. Then $$\bigwedge_{n<\omega}\bigvee_{\sigma\in T_n}\omega\times A_\sigma$$ exists in $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ iff there exist $T^n \subseteq T$, for $n<\omega$, finite level subtrees such that
$$[T]=\bigcup_{n<\omega}[T^n].$$
\end{proposition}

\begin{proof}
First suppose that we have such decomposition $\lbrace T^n \mid n<\omega\rbrace$ of $T$. Define $$A:=\coprod_{n<\omega}\bigcup\lbrace A_\sigma \mid\sigma\in\bigcup_{i\leq n}T^{i}_{n}\rbrace.$$ Take a positive set $C\times B\subseteq_{\mathcal{NC}}A$. Without loss of generality, this means that $B\subseteq^* A(n)$ for all $n\in C$. It follows that for all $n\in C$ there exists $F_n \in[T_n ]^{<\omega}$ such that $B\subseteq^* \bigcup_{\sigma\in F_n} A_\sigma$, and hence that $C\times B \subseteq_{\mathcal{NC}}\bigvee_{\sigma\in T_n}\omega\times A_\sigma$ for all $n<\omega$. Therefore $A$ is a lower bound of these sets.

Let $C\times B$ be a lower bound of $\bigvee_{\sigma\in T_n}\omega\times A_\sigma$, for all $n<\omega$. For all $n<\omega$ there exists $F_n \in[T_n ]^{<\omega}$ such that $B\subseteq^* \bigcup_{\sigma\in F_n}A_\sigma$. This means that there exists $f\in[T]$ such that $$\lbrace A_{f\upharpoonright n}\mid n<\omega\rbrace\cup\lbrace B\rbrace $$ is a centered family. If $f\in T^n$, this means that $\lbrace A(m)\mid m\geq n\rbrace\cup\lbrace B\rbrace$ is a centered family. Therefore $A\cap (C\times B)\in\mathcal{NC}^+$. It follows that $A=\bigwedge_{n<\omega}\bigvee_{\sigma\in T_n}\omega\times A_\sigma$.

Now suppose that there exists $A\in\mathcal{NC}^+$ such that
$$A=\bigwedge_{n<\omega}\bigvee_{\sigma\in T_n}\omega\times A_\sigma .$$ For $F\in[\omega]^{<\omega}$ define $$A_F :=\coprod_{n\in\omega\setminus F} \bigcap_{i\in n+1\setminus F}A(i)\setminus\bigcup_{i\in F}A(i). $$

\begin{claim}
$$A= \bigvee_{F\in[\omega]^{<\omega}} A_F .$$

\end{claim}

\begin{proof}
It is clear that $A_F \subseteq_{\mathcal{NC}}A$, for all $F\in[\omega]^{<\omega}$. Take $C\times B\subseteq_{\mathcal{NC}}A$. Without loss of generality this means that $\lbrace A(n)\mid n\in C\rbrace\cup\lbrace B\rbrace$ is a centered family. Extend $C$ to a set $C^\prime$ such that $\lbrace A(n)\mid n\in C^\prime \rbrace\cup\lbrace B\rbrace$ is a centered family, and which is maximal with this property. Let $B^\prime$ be a pseudointersection of said family and suppose that $C^\prime $ is coinfinite.

Since $C^\prime \times B^\prime$ is a lower bound of $\bigvee_{\sigma\in T_n}\omega\times A_\sigma$ for all $n<\omega$, we can take $F_n \in[T_n ]^{<\omega}$ such that $B^\prime \subseteq^* \bigcup_{\sigma\in F_n}A_\sigma $. But it easily follows that $(\omega\setminus C^\prime )\times B^\prime$ is a positive set $\mathcal{NC}$-almost contained in each $\bigvee_{\sigma\in T_n}\omega\times A_\sigma$. Meanwhile, from the maximality of $C^\prime$, it follows that $\vert A(n)\cap B^\prime \vert<\omega$ for all $n\in\omega\setminus C^\prime$. This means that $(\omega\setminus C^\prime )\times B^\prime$ is $\mathcal{NC}$-almost disjoint to $A$, which is a contradiction. Therefore $C^\prime $ is cofinite.

Take $F=\omega\setminus C^\prime$. It is not hard to verify that $C\times B^\prime \subseteq_{\mathcal{NC}} A_F$, whereby the claim is proved.
\end{proof}

From their definition we have that for each $\mathcal{NC}$-positive $A_F$ there exists $\lbrace B_n \mid n<\omega\rbrace\subseteq[\omega]^{\omega}$ a decreasing family such that $A_F=\bigwedge_{n<\omega}\omega\times B_n $. So there exists a family $\lbrace B_{n}^m \mid n,m<\omega\rbrace\subseteq[\omega]^\omega$ such that $ B_{n+1}^m \subseteq B_{n}^m$, for all $m,n<\omega$, and $$A=\bigvee_{m<\omega}\bigwedge_{n<\omega}\omega\times B_{n}^m .$$

\begin{claim}
Fix $m<\omega$. For all $k<\omega$ there exist $F_{k} \in[T_k ]^{<\omega}$ and $n_k <\omega$ such that $B^{m}_{n_k} \subseteq^* \bigcup_{\sigma\in F_k }A_\sigma$.

\end{claim}

\begin{proof}
Suppose on the contrary that there exists $k<\omega$ such that for all $F\in[T_k]^{<\omega}$ and for all $n<\omega$, $B_{n}^{m}\nsubseteq^* \bigcup_{\sigma\in F}A_\sigma$. This means that there exists $B$, a pseudointersection of the family $\lbrace B^{m}_n \mid n<\omega\rbrace$, which is almost disjoint to every element of the family $\lbrace A_\sigma \mid\sigma\in T_k \rbrace$. Therefore $\omega\times B\subseteq_{\mathcal{NC}}\bigwedge_{n<\omega}\omega\times B_{n}^m$ while being $\mathcal{NC}$-almost disjoint to $\bigvee_{\sigma\in T_k}\omega\times A_\sigma$, which is a contradiction.

\end{proof}

Fix $m<\omega$. If $n_k$ is the least integer and $F_k$ the smallest finite subset of $T_k$ such that $B^{m}_{n_k} \subseteq^* \bigcup_{\sigma\in F_k }A_\sigma$, it follows that $$T^{m}:=\bigcup_{k<\omega}F_k$$ is a subtree of $T$.

Unfix $m$. Take $f\in[T]$. Since $A^\prime :=\bigwedge_{n<\omega}\omega\times A_{f\upharpoonright n}\subseteq_{\mathcal{NC}}A$, it follows that there exists $m<\omega$ such that $$ A^\prime \cap \bigwedge_{n<\omega}\omega\times B_{n}^m \in\mathcal{NC}^+ .$$ Take $k<\omega$. If $n_k$ and $F_k$ are as in the previous paragraph, then $f\upharpoonright k\in F_k$, for otherwise we would have that $\vert A_{f\upharpoonright k}\cap B_{n_k}^m \vert<\omega$. Therefore $f\in[T^m ]$ and we conclude that $[T]=\bigcup_{n<\omega}[T_n]$.

\end{proof}

Notice that the existence of said family of finite level subtrees of $T$ gives us the necessary finite steps recipe for approximating the desired substructure.
  Also observe that if the countable amount of columns of $\omega\times\omega$ gives space to finitely approximating some simple sumprema or infima, like those of Lemma \ref{aI=omega} and Proposition \ref{gdelta,fsigma}, said columns lack ``depth" to approximate some countable but barely more complex algebraic operations.

In order to use the tower of theorem \ref{aronszjan2} to get a partition of size $\omega_1$ on $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$, we need, according to the last proposition, an Aronszajn tree $T$ such that for all $\alpha<\omega_1$ there exists $\lbrace S^{\alpha,m}\mid m<\omega\rbrace$ a family of subtrees of finite levels of $T_\alpha$ such that $$[T_\alpha ]=\bigcup_{m<\omega}[S^{\alpha,m}].$$ The last result of this section shows that this is not the case.

\begin{proposition}
\label{limits_of_aronszjan}
Let $T\subseteq\omega^{<\omega_1}$ be an Aronszajn tree. There exists $\alpha\in Lim(\omega_1 )$ such that if we take $\lbrace S^n \mid n<\omega\rbrace$, a family of finite level subtrees of $T_{<\alpha}$, then $$\bigcup_{n<\omega}[S^n ]\neq[T_{<\alpha}].$$
\end{proposition}

\begin{proof}
First, for $t\in T$ set the notation $t\uparrow:=\lbrace s\in T\mid t\subseteq s\rbrace$. Notice that $$T^\prime :=\lbrace t\in T\mid\vert t\uparrow\vert=\omega_1 \rbrace$$ is a well-pruned Aronszajn tree and that any $\alpha\in Lim(\omega_1)$ proving the proposition for $T^\prime$ will prove it for $T$. So without loss of generality suppose that $T$ is well-pruned.

\begin{claim}
There exists $\alpha\in Lim(\omega_1 )$ such that for all $t\in T_{<\alpha}$ there exists $dom(t)\leq\gamma<\alpha$ such that $\vert t\uparrow \cap T_\gamma \vert=\omega$.
\end{claim}
\begin{proof}
Suppose on the contrary that for all $\alpha\in Lim(\omega_1 )$ there exists $t_\alpha \in T_{<\alpha}$ such that $\vert t_\alpha \uparrow\cap T_\gamma \vert<\omega$, for all $dom(t_\alpha )\leq\gamma<\alpha$. Define $g:Lim(\omega_1 )\rightarrow\omega_1$ as $g(\alpha)=dom(t_\alpha )$. This is a regressive function defined on a stationary set. Therefore (see Lemma III.6.14 of \cite{kunen}) there exist $\beta<\omega_1$ and $S\in[\omega_1 ]^{\omega_1 }$ such that $g(\alpha) =\beta$, for all $\alpha\in S$. Furthermore, there exists $t\in T_\beta$ and $S^\prime \in[S]^{\omega_1 }$ such that $t_\alpha =t$ for all $\alpha\in S^\prime$. Since $S^\prime$ is a cofinal subset of $\omega_1$, it follows that $0<\vert t\uparrow\cap T_\gamma \vert<\omega$ for all $dom(t)\leq\gamma<\omega_1$. But a compacity argument gives us a cofinal branch on $T$, which is a contradiction.
\end{proof}

Take the $\alpha$ given by this claim and let $\lbrace S^n \mid n<\omega\rbrace$ be a family of subtrees of $T_{<\alpha}$ with finite levels. Fix $\lbrace\alpha_n \mid n<\omega\rbrace\subseteq\alpha$ an increasing sequence converging to $\alpha$. Begin with $t_0 \in T_{<\alpha}$. The claim gives us $dom(t_0 )\leq\gamma_0 <\alpha$ such that $\vert t_0 \uparrow \cap T_{\gamma_0} \vert=\omega$. Take $t^{\prime}_1 \in(t_0 \uparrow \cap T_{\gamma_0})\setminus S^0$ and choose $t_1 \in t^{\prime}_1 \uparrow\cap T_{\max\lbrace\alpha_0 ,\gamma_0 \rbrace}$. Suppose now that we have $\lbrace t_0 ,...,t_i \rbrace$, a chain in $T_{<\alpha}$, for some $i<\omega$. We know that there exists $dom(t_i)\leq\gamma_i <\alpha$ such that $\vert t_i \uparrow \cap T_{\gamma_i} \vert=\omega$. Therefore we can choose $t_{i+1} \in (t_i \uparrow\cap T_{\max\lbrace\alpha_i ,\gamma_i \rbrace})\setminus S^i$. If $f$ is the cofinal branch of $T_{<\alpha}$ extending the chain $\lbrace t_i \mid i<\omega\rbrace$, by its construction we get that $f\notin\bigcup_{n<\omega}[S^n ]$, whereby proving the proposition.
\end{proof}

We conclude this section giving some observations and stating a couple of questions. All constructions of towers and partitions on $\mathcal{P}(\omega\times\omega)\slash\mathcal{NC}$ given in this section consist of sets of the type $\omega\times X$, $\bigwedge_{n<\omega}\omega\times X_n$ or  $\bigvee_{n<\omega}\omega\times X_n$ where $\lbrace X\rbrace\cup\lbrace X_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$. Thus far these sets have been proved to be very useful. However they are far from being the only kind of sets of this algebra. For example, if $\lbrace X_n \mid n<\omega\rbrace$ is an independent family, i.e. for all disjoint non-empty $F_0 ,F_1 \in[\omega]^{<\omega}$ the set $$\bigcap_{i\in F_0}X_i \setminus\bigcup_{j\in F_1}X_j$$ is infinite, then $\coprod_{n<\omega}X_n$ is an element of $\mathcal{NC}^+$ not as easily pictured as those of the ``nice" types mentioned above. It is also easy to imagine that there are even less nice elements of $\mathcal{NC}^+$.

Although a good idea for getting a $\mathcal{NC}$-mad family of size $\omega_1$ in ZFC was finally rejected by Proposition \ref{limits_of_aronszjan} the next question remains open:

\begin{question}
\label{omega1_partitions}
Is provable in ZFC that there exists a $\mathcal{NC}$-mad family of size $\omega_1$? Is its existence at least consistent with $\mathfrak{p}>\omega_1$?
\end{question}

It is easy to see that any $\mathcal{NC}$-mad family consisting only of elements of the type $\omega\times X$ or $\bigvee_{n<\omega}\omega\times X_n$ must be of size at least $\mathfrak{a}$.
Therefore the only hope for getting a $\mathcal{NC}$-mad family of size $\omega_1$ consisting of ``nice" sets, as those discussed in the last paragraphs, while also having $\mathfrak{p}>\omega_1$, can only be achieved with sets of the form $\bigwedge_{n<\omega}\omega\times X_n$. So an auxiliary question to question \ref{omega1_partitions}, translated to a language more familiar to those who deal with infinite combinatorics and very interesting in itself, is the following one.

\begin{question}
Is it provable in ZFC the existence of a family $\lbrace X_{\alpha}^{n}\mid\alpha<\omega_1 ,~n<\omega\rbrace\subseteq[\omega]^\omega$ such that

\begin{enumerate}
    \item $X_{\alpha}^{n+1}\subseteq X_{\alpha}^{n}$, for every $\alpha<\omega_1$ and every $n<\omega$,

    \item for all $\alpha<\beta<\omega_1$ there exists $n<\omega$ such that $\vert X_{\alpha}^{n}\cap X_{\beta}^{n}\vert<\omega$ and

    \item for every $X\in[\omega]^{\omega}$ there exists $\alpha<\omega_1$ such that $\lbrace X\rbrace\cap\lbrace X_{\alpha}^{n}\mid n<\omega\rbrace$ is a centered family?
\end{enumerate}

Is its existence at least consistent with $\mathfrak{p}>\omega_1$?
\end{question}

Notice that if we define $X_\alpha :=\coprod_{n<\omega} X_{\alpha}^{n}$, for all $\alpha<\omega_1$, point 1 says that $X_\alpha$'s are of the ``infimum" type, point 2 says that they are $\mathcal{NC}$-almost disjoint and point 3 gives the madness of the family.

\section{Higher dimensional relatives of the Nowhere Centered ideal}
\label{third_dimension}

As well as $fin^2 :=fin\times fin$ is a ``two-dimensional" version of the ideal $fin$, we can easily generalize it to ideals on $\omega^k$, for all $0<k<\omega$.

\begin{mydef}
\label{fin^k}
For $0<k<\omega$ we will recursively define an ideal $fin^k$ on $\omega^k$ as follows:

\begin{itemize}
    \item $fin^1 :=fin$

    \item $fin^k :=\lbrace A\subseteq\omega^k \mid\forall^\infty n<\omega ~A(n)\in fin^{k-1}\rbrace$ for all $k>1$, where $A(n):=\lbrace\overline{x}\in\omega^{k-1}\mid(n)\frown\overline{x}\in A\rbrace$.
\end{itemize}
\end{mydef}

It is easy to see that each $fin^k$ is an ideal and a Borel subset of $\omega^k$. Some kinds of sets that are found in these ideals are the following ones:

\begin{lemma}
Take $k\geq 1$. Then the following statements hold:

\begin{enumerate}
    \item $\lbrace\overline{x}\in\omega^k \mid\overline{x}(i)<n\rbrace\in fin^k$, for all $i<k$ and $n<\omega$.

    \item If $A\subseteq\omega^k$ and $\vert\lbrace n<\omega\mid\overline{x}\frown(n)\in A\rbrace\vert<\omega$, for all $\overline{x}\in\omega^{k-1}$, then $A\in fin^k$.

    \item If $1\leq j<k$, $e_0 <...<e_{j-1}<e<k$ and $g:\omega^j \rightarrow\omega$, then $$\lbrace\overline{x}\in\omega^k \mid\overline{x}(e)<g(\overline{x}(e_0 ),...,\overline{x}(e_{j-1}))\rbrace\in fin^k .$$

\end{enumerate}

\end{lemma}

\begin{proof}
Since statements 1 and 2 are simple and even provable from Definition \ref{fin^k} and statement 3, only the last one will be proved. It will be inductively proved for $1\leq j<k<\omega$. For the basic case when $j=1$ and $k=2$ it is easy to see that $\lbrace(x,y)\in\omega^2 \mid y<g(x)\rbrace\in fin^2$, for all $g\in\omega^\omega$.

Now suppose that we have proved it for all $j^\prime <k^\prime $ such that either $j^\prime <j$ or $k^\prime <k$, for some $1\leq j<k<\omega$. Take $e_0 <...<e_{j-1}<e<k$ as well as $g:\omega^j \rightarrow\omega$. If $A_g :=\lbrace\overline{x}\in\omega^k \mid\overline{x}(e)<g(\overline{x}(e_0 ),...,\overline{x}(e_{j-1}))\rbrace$ and $e_0 =0$, then $A_g (n)\in fin^{k-1}$, for all $n<\omega$, by hypothesis induction for $(j-1,k-1)$. If $0<e_0$, equally $A_g (n)\in fin^{k-1}$, for all $n<\omega$, by hypothesis induction for $(j,k-1)$. Either way $A\in fin^k$ and the lemma is proved.
\end{proof}

In order to relate these paragraphs with the main subject of Section \ref{quotient}, notice that the relation $\subseteq^*$, though not explicit in Definition \ref{nci}, is fundamental in defining the Nowhere Centered ideal. Its relation with the ideal $fin\times fin$ has also been highlighted. Therefore it is not surprising that the ideals $fin^k$ help get higher dimensional relatives of $\mathcal{NC}$.

\begin{mydef}
 If $k\geq 2$, define $$\mathcal{NC}^k :=\lbrace A\subseteq\omega^k \mid\forall(A_0 ,...,A_{k-2})\in([\omega]^\omega)^{k-1}~ \forall^\infty n<\omega ~\prod_{i<k-1}A_i \nsubseteq_{fin^{k-1}}A(n)\rbrace.$$
\end{mydef}

Observe that $\mathcal{NC}^2 =\mathcal{NC}$ and that $fin^k \subseteq\mathcal{NC}^k$, for all $k<\omega$. Also that $\mathcal{NC}^k$ is a downward closed coanalytic subset of $\mathcal{P}(\omega^k )$, for all $k\geq 2$.
In order to prove that they are ideals we will need a lemma.

\begin{lemma}
\label{cubes}
Take $k\geq 1$ and a family $\lbrace A_{n}^0 ,A_{n}^1 \mid n<\omega\rbrace\subseteq (fin^k )^+$. If there exists $(X_0 ,..., X_{k-1})\in([\omega]^\omega )^k$ such that $$\prod_{l<k}X_l \subseteq_{fin^k} A_{n}^0 \cup A_{n}^1 ,$$ for all $n<\omega$, then there exist $(Y_0 ,...,Y_{k-1})\in\prod_{l<k}[X_l]^\omega$, $Z\in[\omega]^\omega$ and $i<2$ such that $$\prod_{l<k}Y_l \subseteq_{fin^k} A_{n}^i ,$$ for all $n\in Z$.
\end{lemma}

\begin{proof}
It will be inductively proved for $k\geq 1$. Suppose that $k=1$ and that we have $\lbrace A_{n}^0 ,A_{n}^1 \mid n<\omega\rbrace\subseteq[\omega]^\omega$ and $X\in[\omega]^\omega$ such that $X\subseteq^* A_{n}^0 \cup A_{n}^1$ for all $n<\omega$. Then any ultrafilter $\mathcal{U}$ containing $X$ will choose $i_n <2$ such that $A_{n}^{i_n}\in\mathcal{U}$, for all $n<\omega$. There exist $Z\in[\omega]^\omega$ and $i<2$ such that $i=i_n$, for all $n\in Z$. Take $Y\in[\omega]^\omega$, a pseudointersection of $\lbrace X\rbrace\cup\lbrace A_{n}^i \mid n\in Z\rbrace$ and the basic case is proved.

Suppose now that the lemma is proved for some $k\geq 1$ and that we have $\lbrace A_{n}^0 ,A_{n}^1 \mid n<\omega\rbrace\subseteq (fin^{k+1})^+$ and $(X_0 ,..., X_{k})\in([\omega]^\omega )^{k+1}$ such that $$\prod_{l\leq k}X_l \subseteq_{fin^{k+1}} A_{n}^0 \cup A_{n}^1 ,$$ for all $n<\omega$. Observe that for all $n<\omega$
$$\prod_{1\leq l\leq k}X_l \subseteq_{fin^k} A_{n}^0 (m)\cup A_{n}^1 (m),$$ for almost all $m\in X_0$.

Using the inductive hypothesis, we can recursively construct, for $n<\omega$, sequences $\lbrace (Y_{0}^n ,...,Y_{k}^n )\mid n<\omega\rbrace$ and $\langle i_n \mid n<\omega\rangle\in 2^\omega$ such that

\begin{itemize}
    \item $Y_{l}^0 \in[X_l ]^\omega$,

    \item $Y_{l}^{n+1}\in[Y_{l}^n ]^\omega$, for all $l\leq k$ and $n<\omega$, and

    \item
    $$\prod_{1\leq l\leq k}Y_{l}^n \subseteq_{fin^k}A_{n}^{i_n}(m),$$ for all $m\in Y_{0}^n$, for all $n<\omega$.
\end{itemize}

For $l\leq k$, take $Y_l \in[\omega]^\omega$, a pseudointersection of the family $\lbrace Y_{l} \mid n<\omega\rbrace$. There exist $Z\in[\omega]^\omega$ and $i<2$ such that $i_n =i$, for all $n\in Z$. If $n\in Z$, then $$\prod_{1\leq l\leq k}Y_{l}^n \subseteq_{fin^k}A_{n}^i (m),$$ for almost all $m\in Y_{0}$. Therefore $$\prod_{l\leq k}Y_{l} \subseteq_{fin^{k+1}}A_{n}^i ,$$ for all $n\in Z$.
\end{proof}

\begin{proposition}
\label{next_level}
For all $k\geq 2$ the family $\mathcal{NC}^k$ is a proper ideal of $\omega^k$.

\end{proposition}

\begin{proof}
Since $\omega^k \notin\mathcal{NC}^k$ and it is downward closed, it only remains to be proved that $A_0 \cup A_1 \in\mathcal{NC}^k$, for all $A_0 ,A_1 \in\mathcal{NC}^k$.  Take $A_0$ and $A_1$ subsets of $\omega^k$ such that $A_0 \cup A_1 \in \mathcal{P}(\omega^k )\setminus\mathcal{NC}^k$. There exists $(A_0 ,..., A_{k-2})\in([\omega]^\omega )^{k-1}$ such that $$\prod_{l<k-1}A_l \subseteq_{fin^k} A_{0}(n)\cup A_{1}(n),$$ for infinitely many $n<\omega$. Applying Lemma \ref{cubes} we get $(A^{\prime}_0 ,...A^{\prime}_{k-2})\in([\omega]^\omega )^{k-1}$ and $i<2$ such that $$\prod_{l<k-1}A^{\prime}_l \subseteq_{fin^k} A_{i}(n),$$ for infinitely many $n<\omega$, which means that $A_i \notin\mathcal{NC}^k$. Therefore, if $A_0 ,A_1 \in\mathcal{NC}^k$, then $A_0 \cup A_1 \in\mathcal{NC}^k$.

\end{proof}

Observe that for all $A\in(\mathcal{NC}^k )^+$ there exists $(X_0 ,...,X_{k-1})\in([\omega]^\omega )^k$ such that $\prod_{i<k}X_i \subseteq_{fin^k}A$. Obviously the set $$\lbrace\omega^{k-1}\times X\mid X\in[\omega]^\omega \rbrace$$ form a subalgebra of $\mathcal{P}(\omega^k)\slash\mathcal{NC}^k$ isomorphic to $\mathcal{P}(\omega)\slash fin$.  All results from Lemma \ref{aI=omega} to Theorem \ref{aronszjan2} have a version on these quotients. But of more importance to this section is how the increase in dimension upgrades the reach of Lemma \ref{aI=omega}.

\begin{nota}
Take $1\leq j<k<\omega$ and $E\in[k-1]^j$. If $\lbrace A_{\overline{n}}\mid\overline{n}\in\omega^E \rbrace\subseteq \mathcal{P}(\omega)$, define $$\coprod_{\overline{n}\in\omega^E}A_{\overline{n}} :=\lbrace\overline{x}\in\omega^k\mi
\overline{x}(k-1)\in A_{\overline{x}\upharpoonright E}\rbrace.$$
\end{nota}

\begin{theorem}
\label{countable_iteration}
Take $1\leq j<k<\omega$ and $l_0 <...<l_{j-1}<k-1$, and define $E:=\lbrace l_i \mid i<j\rbrace$. Then for all family $\lbrace A_{\overline{n}}\mid\overline{n}\in\omega^j \rbrace\subseteq[\omega]^\omega$ the following equality holds in $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$:

$$\coprod_{\overline{n}\in\omega^F}(\bigcap_{i_0 \leq\overline{n}(l_0 )}\bigcup_{i_1 \leq\overline{n}(l_1 )}...~ A_{(i_0 ,...,i_{j-1})})=$$

$$=\bigwedge_{n_0 <\omega}\bigvee_{n_1 <\omega}...~\omega^{k-1}\times A_{(n_0 ,...,n_{j-1})}.$$

\end{theorem}

\begin{proof}
It will be inductively proved for $j\leq 1$. Take $k\leq 2$, $l<k-1$ and a family $\lbrace A_n \mid n <\omega\rbrace\subseteq[\omega]^\omega$. Define $$A:=\coprod_{\overline{n}\in\omega^{\lbrace l\rbrace}}\bigcap_{i\leq\overline{n}(l)}A_i.$$ If $n<\omega$, then $A\setminus(\omega^{k-1}\times A_n )\subseteq\lbrace\overline{x}\in\omega^k \mid\overline{x}(l)<n\rbrace$ which is an element of $\mathcal{NC}^k$. Therefore $A\subseteq_{\mathcal{NC}^k}\omega^{k-1}\times A_n$, for all $n<\omega$. Take $B_0 ,...,B_{k-1}\in[\omega]^\omega$ such that $$\prod_{i<k}B_i\subseteq_{fin^k}\omega^{k-1}\times A_n ,$$ for all $n<\omega$. Therefore $B_{k-1}\subseteq^* A_n $, for all $n<\omega$, and hence it follows that $$\prod_{i<k}B_i \setminus A\in fin^k .$$ We conclude that $A=\bigvee_{n<\omega}A_{(n)}$.

Suppose we have proved the theorem for some $j^\prime \geq 1$. For $j=j^\prime +1$, take $k\geq j+1$, $l_0 <...<l_{j-1}<k-1$, and a family $\lbrace A_{\overline{n}}\mid\overline{n}\in\omega^j \rbrace\subseteq[\omega]^\omega$. With $E:=\lbrace l_i \mid i<j\rbrace$. Define $$A:=\coprod_{\overline{n}\in\omega^F}\bigcap_{i_0 \leq\overline{n}(l_0)}\bigcup_{i_1 \leq\overline{n}(l_1)}... ~A_{(i_0 ,...,i_{j-1})}$$ and $$A_n :=\coprod_{\overline{n}\in\omega^{F\setminus\lbrace l_0 \rbrace}}\bigcup_{i_1 \leq\overline{n}(l_1)}\bigcap_{i_2 \leq\overline{n}(l_2)}... ~A_{(n,i_1 ,...,i_{j-1})},$$ for all $n<\omega$. By hypothesis induction $$A_n =\bigvee_{n_1 <\omega}\bigwedge_{n_2 <\omega}...~\omega^{k-1}\times A_{(n,n_1,...,n_{j-1})},$$ for all $n<\omega$. We need to prove that $A=\bigwedge_{n<\omega}A_n$. Take $n<\omega$ and $\overline{x}\in A$ such that $\overline{x}(l_0 )\geq n$. It follows that $$\overline{x}(k-1)\in\bigcup_{i_1 \leq\overline{x}(l_1 )}\bigcap_{i_2 \leq\overline{x}(l_2 )}...~A_{(n,i_1 ,...,i_{j-1})}$$ and hence that $\overline{x}\in A_n$.  We conclude that $A\subseteq_{\mathcal{NC}^k}A_n$, for all $n<\omega$.

Now take $B_0 ,...,B_{k-1}\in[\omega]^\omega$ such that $$\prod_{i<k}B_i\subseteq_{fin^k}A_n,$$ for all $n<\omega$. Then for all $n_0 <\omega$ we have that $$\prod_{i<k}B_i \subseteq_{fin^k}\bigvee_{n_1 <\omega}\bigwedge_{n_2 <\omega}...~\omega^{k-1}\times A_{(n,n_1,...,n_{j-1})}.$$ Notice that we can construct $g_1 :\omega\rightarrow\omega$ and get $B_{i}^1 \in[B_i ]^\omega$, for $i<k$, such that for all $n_0 <\omega$ we have that $$\prod_{i<k}B_{i}^1 \subseteq_{fin^k}\bigwedge_{n_2 <\omega}\bigvee_{n_3 <\omega}...~\omega^k \times A_{(n_0 ,g_1 (n_0 ),n_2 ,...,n_{j-1})}.$$ Similarly we can get $B_{i}^3$, for $i<k$, and $g_3 :\omega^2 \rightarrow\omega$ such that for all $n_0 <\omega$ and for all $n_2 <\omega$ $$\prod_{i<k}B_{i}^3 \subseteq_{fin^k}\bigwedge_{n_4 <\omega}\bigvee_{n_5 <\omega}...~\omega^k \times A_{(n_0 ,g_1 (n_0 ),n_2 ,g_3 (n_0 ,n_1 ),...,n_{j-1})}.$$  Following this path we finally get $B_{i}^{\prime}\in [B_i ]^\omega$, for $i<k$, and $g_1 :\omega\rightarrow\omega$, $g_3 :\omega^2 \rightarrow\omega$, $g_5 :\omega^3 \rightarrow\omega$,..., for impair numbers $<j$, such that for all $n_0 ,n_2, n_4 ,... <\omega$ (as many variables as pair numbers $<j$), we have that $$\prod_{i<k}B_{i}^\prime \subseteq_{fin^k}\omega^{k-1}\times A_{(n_0 , g_1 (n_0),n_2 ,g_3 (n_0 ,n_2),n_4 ,g_5(n_0 ,n_2 ,n_4),....)}$$ and hence that $$B_{k-1}^\prime \subseteq^* A_{(n_0 , g_1 (n_0)),n_2 ,g_3 (n_0 ,n_2),n_4 ,g_5(n_0 ,n_2 ,n_4),....)}.$$

If $C:=\prod_{i<k}B_{i}^\prime \setminus A$, we need to prove that $C\in\mathcal{NC}^k$ to finish the proof. For impair $i<j$ define $h_i :\omega^{\frac{i+1}{2}}\rightarrow\omega$ as follows: $$h_i (n_0, n_1 ,...)=\max\lbrace g_i (i_0 ,i_1 ,...)\mid i_0 \leq n_0 , i_1 \leq n_1 ...\rbrace.$$ Since $$\lbrace\overline{x}\in\omega^k \mid\overline{x}(l_1 )<h_1 (\overline{x}(l_0 ))\vee\overline{x}(l_3 )<h_3 (\overline{x}(l_0 ),\overline{x}(l_2 ))\vee...\rbrace$$ is an element of $\mathcal{NC}^k$, to prove that $C\in\mathcal{NC}^k$ we just need to prove that $$C^\prime :=C\cap\lbrace\overline{x}\in\omega^k \mid\overline{x}(l_1 )\geq h_1 (\overline{x}(l_0 ))\wedge\overline{x}(l_3 )\geq h_3 (\overline{x}(l_0 ),\overline{x}(l_2 ))\wedge...\rbrace$$ lies in $\mathcal{NC}^k$. If $\overline{y}\in\omega^{k-1}$ and $\overline{y}(l_1 )\geq h_1 (\overline{y}(l_0 ))$, $\overline{y}(l_3 )\geq h_3 (\overline{y}(l_0 ),\overline{y}(l_2 ))$ and so forth, then $$\lbrace n<\omega\mid\overline{y}\frown(n)\in C^\prime \rbrace\subseteq B_{k-1}^\prime \setminus\bigcap_{i_0 \leq\overline{y}(l_0 )}\bigcup_{i_1 \leq\overline{y}(l_1)}...~A_{(i_0 ,...,i_{j-1})}$$ which is a finite set. Therefore $C^\prime \in\mathcal{NC}^k$ and we conclude the proof.

\end{proof}

Theorem \ref{countable_iteration} is powerful enough to settle Question \ref{omega1_partitions} on $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$, for $k\geq 2$, while it is left open on $\mathcal{P}(\omega^2 )\slash\mathcal{NC}$. Whether these results give some light or cast shadows on the questions of Section \ref{quotient} is not known.

\begin{theorem}
Let $T\subseteq\omega^{<\omega_1}$ be an Aronszajn tree. Take $\lbrace A_\sigma \mid \sigma\in T\rbrace\subseteq[\omega]^\omega$ such that if $\sigma\subseteq\tau\in T$, then $A_\tau \subseteq^* A_\sigma$ and that the set $\lbrace A_\sigma \mid \sigma\in T_\alpha \rbrace$ is a partition of $\omega$ for all $\alpha <\omega_1$. If $3\leq k<\omega$, then
\begin{enumerate}
    \item the sets $X_\alpha :=\bigvee_{\sigma \in T_\alpha }\omega^{k-1}\times A_\sigma$, for $\alpha<\omega_1$, form a tower in $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$ and
    \item the sets $Y_\alpha := (\bigwedge_{\beta <\alpha }X_\beta )\setminus X_\alpha$, for $0 <\alpha <\omega_1$, form a partition in $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$.
\end{enumerate}
\end{theorem}
\begin{corollary}
$\overline{\mathfrak{a}}(\mathcal{NC}^k )=\omega_1$, for all $3\leq k<\omega$.
\end{corollary}

From the proof of Theorem \ref{countable_iteration} it is clear how the first $k-1$ dimensions of $\omega^k$ give us room enough to finitely approximate in the last coordinate the desired algebraic operation. It is also clear that the same idea cannot be equally applied to a family indexed by $\omega^k$. If anything, Proposition \ref{gdelta,fsigma} hints to the possibility that there is a limit on the times that $\bigvee$ and $\bigwedge$, with countable indexes, can be iterated on $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$.

\begin{conj}
\label{limits_of_dimension}
For all $2\leq k<\omega$, there exists $\lbrace A_{\overline{x}}\mid\overline{x}\in\omega^k \rbrace\subseteq[\omega]^\omega$ such that $$\bigwedge_{n_0 <\omega}\bigvee_{n_1 <\omega}...~\omega^{k-1}\times A_{(n_0 ,n_1 ,...,n_{k-1})}$$ does not exist in $\mathcal{P}(\omega^k )\slash\mathcal{NC}^k$.
\end{conj}

We finish this section with a question about similar results for $\omega\leq\alpha<\omega_1$. Take $X$ a countable set and an ideal $\mathcal{I}$ such that there exists $\mathcal{A}\leq \mathcal{P}(X)\slash\mathcal{I}$ isomorphic to $\mathcal{P}(\omega)\slash fin$. For $\alpha<\omega_1$, recursively define \begin{itemize}
    \item $\Sigma_{0}^0 (\mathcal{A}):=\mathcal{A}$

    \item $\Pi_{\alpha}^0 (\mathcal{A}):=\lbrace A\in \mathcal{P}(X)\slash\mathcal{I}\mid A\in\Sigma_{\alpha}^0 (\mathcal{A})\rbrace$

    \item $\Sigma_{\alpha}^0 (\mathcal{A}):=\lbrace A\in \mathcal{P}(X)\slash\mathcal{I}\mid\exists\lbrace A_n \mid n<\omega\rbrace\subseteq\bigcup_{\beta<\alpha}\Pi_{\beta}^0 (\mathcal{A}) ~A=\bigvee_{n<\omega}A_n \rbrace.$
\end{itemize} For $\alpha<\omega_1$, we will say that $\mathcal{A}$ is $\alpha$-$\sigma$-\textit{closed}, if it is $\beta$-$\sigma$-\textit{closed}, for all $\beta<\alpha$, and for all $\lbrace A_n \mid n<\omega\rbrace\subseteq\bigcup_{\beta<\alpha}\Pi_{\beta}^0 (\mathcal{A})$ there exists $A\in \mathcal{P}(X)\slash\mathcal{I}$ such that $A=\bigvee_{n<\omega}A_n$. With this notation, Theorem \ref{countable_iteration} says that $\lbrace\omega^{k-1}\times X\mid X\in[\omega]^\omega \rbrace$ is $k-1$-$\sigma$-closed while Conjecture \ref{limits_of_dimension} says that it is not $k$-$\sigma$-closed, for all $2\leq k<\omega$.

\begin{question}
Does there exist a sequence $\lbrace\mathcal{I}_{\alpha}\mid\alpha<\omega_1 \rbrace$, of definable ideals on countable sets $X_\alpha$, and a sequence $\lbrace\mathcal{A}_\alpha \mid\alpha<\omega_1 \rbrace$, of subalgebras of $\mathcal{P}(X_\alpha )\slash\mathcal{I}_\alpha$ isomorphic to $\mathcal{P}(\omega)\slash fin$, such that $\mathcal{A}_\alpha$ is $\alpha$-$\sigma$-closed, but not $\alpha+1$-$\sigma$-closed, for all $\alpha<\omega_1$?
\end{question}

\section{The ideal $\mathcal{NC}$ as a substructure of $\mathcal{P}(\omega\times\omega)$ modulo $[\omega\times\omega]^{<\omega}$}
\label{add_cov_non_cof}

In this section we will briefly study $\mathcal{NC}$ as such. Some close ideals and Kat\v{e}tov relations between them will be helpful.

\begin{mydef}
 The Kat\v{e}tov-Blass order $\leq_{KB}$ on ideals is defined as follows: if $\mathcal{I}$ and $\mathcal{J}$ be two ideals on $\omega$,  then $\mathcal{I}\leq_{KB}\mathcal{J}$ if there exists $f:\omega\rightarrow\omega$ finite-to-one such that $f^{-1}[I]\in\mathcal{J}$ for all $I\in\mathcal{I}$.

 If the function $f$ is not required to be finite-to-one, we get the Kat\v{e}tov relation $\mathcal{I}\leq_K \mathcal{J}$.
\end{mydef}

\begin{mydef}
 \begin{itemize}

         \item $\Delta :=\lbrace(n,m)\in\omega\times\omega\mid m\leq n\rbrace$

         \item $\mathcal{ED}_{fin}:=\lbrace X\subseteq\Delta\mid\exists n<\omega~\forall m<\omega~\vert X(m)\vert\leq n\rbrace$

        \item A graph is a pair $(X,E)$, where $E\subseteq[X]^2$. We say that $Y\subseteq X$ induces a complete subgraph of $(X,E)$ if $[Y]^2 \subseteq E$.

         \item $\mathcal{G}_c :=\lbrace E\subseteq[\omega]^2 \mid\forall Y\in[\omega]^\omega Y~does~not~induce~a~complete~subgraph\rbrace.$
     \end{itemize}
\end{mydef}

The ideal $\mathcal{ED}_{fin}$ can also be defined as the ideal on $\Delta$ generated by the functions from $\omega$ to $\omega$ bounded by the identity function.

\begin{proposition}
\label{katetov relations}
The following relations hold:

\begin{itemize}

\item $fin\times fin\leq_{KB} \mathcal{NC}$

\item $\mathcal{ED}_{fin} \leq_{KB} \mathcal{NC}$

\item $\mathcal{NC}\leq_{KB}\mathcal{G}_c$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $fin\times fin\subseteq\mathcal{NC}$, the Kat\v{e}tov-Blass inequality follows.

For the second relation take the function $f:\omega\times\omega\rightarrow\Delta$ defined by the rule $$f(n,m)=(\max\lbrace n,m\rbrace,\min\lbrace n,m\rbrace),$$ for all $n,m<\omega$. Let $A\subseteq\Delta$ be a function. The set $f^{-1}[A]$ is clearly the union $$A\cup\lbrace (A(n),n)\mid n<\omega\rbrace.$$ The set $A$ is an element of $fin\times fin$. Notice that the second term of the union is equal to the set $$B:=\coprod_{m<\omega} A^{-1}(m).$$ Since $\lbrace A^{-1}(m)\mid m<\omega\rbrace$ is a pairwise disjoint family, it follows that $B\in\mathcal{NC}$. We conclude that $f$ is a finite-to-one Kat\v{e}tov function.

For the last inequality consider the function $f:[\omega]^2 \rightarrow\omega\times\omega$ defined by the rule $$f(\lbrace n,m\rbrace)=(\min\lbrace n,m\rbrace,\max\lbrace n,m\rbrace).$$ Take $A\in\mathcal{NC}$ and $X\in[\omega]^\omega$. Since $A\in\mathcal{NC}$, it follows that there exists non-empty $F\in[X]^{<\omega}$ such that $\vert\bigcap_{i\in F}A(i)\vert<\omega$. In particular $\vert\bigcap_{i\in F}A(i)\cap X\vert<\omega$. Take $k\in X\setminus\bigcap_{i\in F}A(i)$ bigger than every element of $F$. Then there exists $i\in F$ such that $k\notin A(i)$. Since $f(\lbrace i,k\rbrace)=(i,k)\notin A$, it follows that $\lbrace i,k\rbrace\notin f^{-1}[A]$. Therefore $X$ does not induce a complete subgraph of $(\omega,f^{-1}[A])$, and we conclude that $f^{-1}[A]\in\mathcal{G}_c$.
\end{proof}

Both the additivity and the cofinality of the Nowhere Centered ideal are well-known cardinals. In order to prove that its cofinality is equal to $\mathfrak{c}$ we will use the fact that if $\kappa<\mathfrak{c}$ and $\lbrace X_\alpha \mid\alpha<\kappa\rbrace$ is a family of infinite coinfinite subsets of $\omega$, then there is coinfinite $X\in[\omega]^\omega$ such that $X\nsubseteq^* X_\alpha$ for every $\alpha<\kappa.$ Indeed, consider any family $\lbrace A_\beta \mid\beta<\mathfrak{c}\rbrace$ such that $\lbrace\omega\setminus A_\beta \mid\beta<\mathfrak{c}\rbrace$ is some mad family. Suppose that for all $\beta<\mathfrak{c}$ there exists $\alpha<\kappa$ such that $ A_\beta \subseteq^* X_\alpha$. Since $\kappa<\mathfrak{c}$, there exist $\beta_0 <\beta_1 <\mathfrak{c}$ and $\alpha<\kappa$ such that $A_{\beta_0}\cup A_{\beta_1}\subseteq^* X_\alpha$, but this means that $\omega\subseteq^* X_\alpha$, which is a contradiction.

\begin{theorem}
$add^* (\mathcal{NC})=\omega$ and $cof^* (\mathcal{NC})=\mathfrak{c}$.
\end{theorem}
\begin{proof}
Take the family $\mathcal{A}:=\lbrace\lbrace n\rbrace\times\omega\mid n<\omega\rbrace$ which is a subset of $\mathcal{NC}$. Clearly if $A\subseteq\omega\times\omega$ and $\lbrace n\rbrace\times\omega\subseteq^* A$, for all $n<\omega$, then $A\in\mathcal{NC}^+$. Therefore $\mathcal{A}$ witnesses that $add^* (\mathcal{NC})=\omega$.

Now take $\kappa <\mathfrak{c}$ and $\mathcal{C}:=\lbrace A_\alpha \mid\alpha <\kappa\rbrace$ a subfamily of $\mathcal{NC}$. Consider the family $K_0 :=\lbrace\alpha <\kappa\mid A_\alpha (0)\nsupseteq^* \omega\rbrace$ and define $C_0 :=\omega$. Since $\kappa<\mathfrak{c}$, there exists $B_0 \subseteq\omega$ such that $\vert B_0 \vert=\vert\omega\setminus B_0 \vert=\omega$ and that $B_0 \nsubseteq^* A_\alpha (0)$ for all $\alpha\in K_0$. Suppose that for some $0<m<\omega$ we have defined a disjoint family $\lbrace B_i \mid i<m\rbrace\subseteq[\omega]^\omega$ such that $$C_m :=\omega\setminus\bigcup_{i<m}B_i$$ is infinite. Take $K_m :=\lbrace\alpha <\kappa\mid A_\alpha (m)\nsupseteq^* C_m \rbrace$. Since $\kappa <\mathfrak{c}$, there exists $B_m \subseteq C_m $ such that $\vert B_m \vert=\vert C_m \setminus B_m \vert=\omega$ and that $B_m \nsubseteq^* A_\alpha (m)$ for all $\alpha\in K_m$. Thus we can recursively construct a centered family $\lbrace C_m \mid m<\omega\rbrace$ and a disjoint family $\lbrace B_m \mid m<\omega\rbrace$ such that $B_m \subseteq C_m$, for all $m<\omega$, and such that $B_m \nsubseteq^* A_\alpha (m)$, for all $\alpha\in K_m$, for all $m<\omega$.

Define $B:=\coprod_{m<\omega}B_m $, which is clearly an element of $\mathcal{NC}$. Take $\alpha <\kappa$. Since $\lbrace C_m \mid m<\omega\rbrace$ is a centered family and $A_\alpha$ is an element of $\mathcal{NC}$, there exists $n<\omega$ such that $\alpha\in K_n$. Then $B_n \nsubseteq^* A_\alpha (n)$, and hence it follows that $B\nsubseteq^* A_\alpha$. Therefore $B$ witnesses that $\mathcal{C}$ is not a cofinal subset of $\mathcal{NC}$ and we conclude that $cof^* (\mathcal{NC})=\mathfrak{c}$.
\end{proof}

Since the covering and the uniformity of $\mathcal{NC}$ are not as straightforward as the other two cardinal invariants we will use the following result for helping us to bound them.

\begin{lemma}
\label{cov and non inequalities}
 Let $\mathcal{I}$ and $\mathcal{J}$ be ideals on a countable set. If $\mathcal{I}\leq_K \mathcal{J}$, then $cov^* (\mathcal{J})\leq cov^* (\mathcal{I})$ and $non^* (\mathcal{I})\leq non^* (\mathcal{J})$.
\end{lemma}

\begin{theorem}
\label{covering}
$\min\lbrace\mathfrak{b},\mathfrak{s}\rbrace\leq cov^*(\mathcal{NC})\leq\mathfrak{b}$.
\end{theorem}

\begin{proof}

Since $\min\lbrace\mathfrak{b,s}\rbrace\leq cov^* (\mathcal{G}_c)$ and $cov^* (fin\times fin)=\mathfrak{b}$ (see Chapter 1 of \cite{davidmeza}), the theorem follows from Proposition \ref{katetov relations} and Lemma \ref{cov and non inequalities}.
\end{proof}

Recall that a family $\mathcal{F}\subseteq[\omega]^\omega$ is said to be $\omega$-\textit{hitting} if for all countable family $\lbrace X_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$ there exists $X\in\mathcal{F}$ such that $\vert X\cap X_n \vert=\omega$, for all $n<\omega$.

\begin{theorem}
\label{cov(NC)<b}
It is consistent that $\omega_1 =\mathfrak{s}=cov^* (\mathcal{NC})<\mathfrak{b}=\omega_2$.
\end{theorem}

\begin{proof}
We begin with $V$ a model of ZFC, and the model that will do is $V[G]$ where $G$ is a generic filter on $\mathbb{H}_{\omega_2}$, i.e. the finite support iteration of length $\omega_2$ of Hechler forcing. It is known that $V[G]\models\mathfrak{b}=\omega_2$ and that finite support iterations of Hechler forcing preserve $\omega$-hitting families (see \cite{baumgartner} or \cite{brendle}). It will be enough to prove that  $\mathbb{H}_{\omega_1}$ adds an $\omega$-hitting subfamily of $\mathcal{NC}$ of size $\omega_1$.

We know that in each step of this iteration a Cohen real is added. A version of Cohen forcing (due to Hechler \cite{hechler}) is the following one: $$\mathbb{P}:=\lbrace p;\omega\times\omega\rightarrow 2\mid dom(p)=n_p \times m_p ~for~some~n_p ,~m_p <\omega\rbrace$$ where $p\leq q$ iff \begin{itemize}

\item $p\supseteq q$ and

\item $\vert\lbrace j\in n_q \mid \mathcal{P}(j,i)=1\rbrace\vert\leq 1$ for all $i\in m_p \setminus m_q$.
1
\end{itemize} Since $\mathbb{P}$ is an atomless countable forcing notion, it is equivalent to Cohen forcing. If $H$ is a $(V,\mathbb{P})$-generic filter for some model $M$ of $ZFC$, it follows that $NC_H :=\lbrace (j,i)\mid\exists p\in H \mathcal{P}(j,i)=1\rbrace$ is an element of $\mathcal{NC}\cap V[H]$ which splits all elements of $[\omega\times\omega]^\omega \cap V$. Therefore, after adding $\omega_1$ many Cohen reals, we get in the generic extension an $\omega$-hitting subfamily of $\mathcal{NC}$ of size $\omega_1$.
\end{proof}

Recall that Mathias forcing is the set $$\mathbb{M}:=\lbrace(s,X)\mid s\in[\omega]^{<\omega}~X\in[\omega]^\omega \rbrace,$$ with $(t,Y)\leq (s,X)$ if $t\supseteq s$, $Y\subseteq X$ and $t\setminus s\subseteq X$. If $G\subseteq\mathbb{M}$ is a generic filter, the Mathias real added by it is $m_G =\bigcup\lbrace s\mid\exists X\in[\omega]^\omega ~(s,X)\in G\rbrace. $

While, as noticed in the last proof, Cohen forcing adds a new element of $\mathcal{NC}$ which hits all reals in the base model, observe that Mathias forcing adds an infinite subset of $\omega\times\omega$ which has finite intersection with all elements of $\mathcal{NC}$ in the base model. Let $ \dot f$ be the name of the enumeration of the Mathias real and take $(s,X)\in\mathbb{M}$ and $A\in\mathcal{NC}$. We know there exist $Y\in[X]^\omega$, $n\geq\vert s\vert$ and $g\in\omega^\omega$ such that $A(m)\cap Y\subseteq g(m)$, for all $m\geq n$. If necessary, extend $s$ for having $\vert s\vert =n$. Now take $Z\in[Y]^\omega$ such that for all $i<\omega$ the $i$-th element of $Z$ is strictly  greater than $g(n+i)$. Then $$(s,Z)\Vdash\forall i<\omega ~\dot f(n+i)\in Y\setminus A(n+i)$$ and therefore $(s,Z)$ forces that the intersection of $A$ and $\dot f$ is finite.

From the inequalities of Theorem \ref{covering} a question remains open:

\begin{question}
Is is consistent that $\min\lbrace\mathfrak{b},\mathfrak{s}\rbrace<cov^* (\mathcal{NC})$?
\end{question}

We conclude the section talking about the uniformity. A family $\mathcal{R}\subseteq[\omega]^\omega$ is called a \textit{hereditarily reaping} if $\mathcal{R}\cap[X]^\omega$ is reaping in $X$, for every $X\in\mathcal{R}$. It is not hard to see that there exists a hereditarily reaping family of size $\mathfrak{r}$.

\begin{theorem}
\label{cov(M),b<non(NC)<r}
$\max\lbrace cov(\mathcal{M}),\mathfrak{b}\rbrace\leq non^* (\mathcal{NC})\leq\mathfrak{r}$.\footnote{$cov(\mathcal{M})$ is the smallest size of a family of meager sets of the real line whose union is the whole line. Also, if $\kappa<cov(\mathcal{M})$ and $\mathcal{D}$ is a family of dense subsets of (countable) Cohen forcing, then there exists a $\mathcal{D}$-generic filter. For details see \cite{blass}.}
\end{theorem}

\begin{proof}
First we will prove that $cov(\mathcal{M})\leq non^* (\mathcal{NC})$. Take $\mathcal{F}\subseteq[\omega\times\omega]^\omega $ of size $\kappa<cov(\mathcal{M})$. Consider the version of Cohen forcing $\mathbb{P}$ defined in the proof of Theorem \ref{cov(NC)<b}. Clearly there exists $\mathcal{D}:=\lbrace\mathcal{D}_\alpha \mid\alpha<\kappa\rbrace$, a family of dense subsets of $\mathbb{P}$, such that if $H$ is a $\mathcal{D}$-generic filter, then $NC_H$ infinitely intersects any element of $\mathcal{F}$. Since $\kappa<cov(\mathcal{M})$, there exists such a filter.

Now suppose that $\kappa<\mathfrak{b}$ and that $\mathcal{F}:=\lbrace A_\alpha \mid\alpha<\kappa\rbrace$ is a family of infinite subsets of $\omega\times\omega$. For all $A\in[\omega\times\omega]^\omega$ there exists either $n<\omega$ such that $\vert(\lbrace n\rbrace\times\omega)\cap A\vert=\omega$ or $f\in\omega^\omega$ such that $\vert A\cap f\vert=\omega$. Therefore w.l.o.g. there exists a partition $K_0 \cup K_1$ of $\kappa$, such that \begin{itemize} \item for all $\alpha\in K_0$ there exists $n_\alpha <\omega$ such that $A_\alpha\subseteq\lbrace n_\alpha \rbrace\times\omega$ and
\item for all $\alpha\in K_1$ the set $A_\alpha$ is a partial function from $\omega$ to $\omega$.
\end{itemize}  Since $\kappa<\mathfrak{b}$, there exists $f\in\omega^\omega$ such that $A_\alpha \leq^* f$, for all $\alpha\in K_1$. Therefore $B_1 :=\lbrace (n,m)\mid m\leq f(n)\rbrace$ is an element of $\mathcal{NC}$ such that $B_1 \cap A_\alpha$ is infinite, for all $\alpha\in K_1$. Recall that $\mathfrak{b}\leq\mathfrak{r}$. The set $\lbrace A_\alpha(n_\alpha )\mid\alpha\in K_0\rbrace$ is split by some real $C_0 \in[\omega]^\omega$. Suppose that for some $n<\omega$ we have defined a disjoint family $C_0 , ..., C_n \in[\omega]^\omega$ such that $D_n :=\bigcup_{i\leq n}C_i $ splits $A_\alpha (n_\alpha )$, for all $\alpha\in K_0$.  The family $$\lbrace A_\alpha (n_\alpha )\setminus D_n \mid\alpha\in K_0 \rbrace$$ is split by some $C_{n+1}\in[\omega\setminus D_n]^\omega$. It follows that both $\bigcup_{i\leq n+1} C_i$ and $C_{n+1}$ split $A_\alpha (n_\alpha )$, for all $\alpha\in K_0$. Thus we can recursively construct a disjoint family $\lbrace C_n \mid n<\omega\rbrace\subseteq[\omega]^\omega$ such that $C_n$ splits $A_\alpha (n_\alpha )$, for all $n<\omega$ and all $\alpha\in K_0$. Take $$B_0 :=\coprod_{n<\omega}C_n ,$$ which is an element of $\mathcal{NC}$, and $\alpha\in K_0$.  Since $C_{n_\alpha}$ splits $A_{\alpha}(n_\alpha )$, it follows that $B_0 \cap A_\alpha$ is infinite. Therefore $B:=B_0 \cup B_1$ is an element of $\mathcal{NC}$ such that $B\cap A_\alpha$ is infinite, for all $\alpha<\kappa$. We conclude that $\mathfrak{b}\leq non^* (\mathcal{NC})$.

Now we prove that $non^* (\mathcal{NC})\leq\mathfrak{r}$. Let $\mathcal{R}$ be a hereditarily reaping family of size $\mathfrak{r}$. We claim that $$\lbrace\lbrace n\rbrace\times X\mid X\in\mathcal{R},~n<\omega\rbrace$$ witnesses that $non^* (\mathcal{NC})\leq\mathfrak{r}$. Suppose that $A\in[\omega\times\omega]^\omega$ is such that $A\cap(\lbrace n\rbrace\times X)$ is infinite, for all $n<\omega$ and all $X\in\mathcal{R}$. Take $X \in\mathcal{R}$. Since $A\cap(\lbrace 0\rbrace\times X)$ is infinite there exists $X_0 \in[X]^\omega \cap\mathcal{R}$ such that $X_0 \subseteq^* A(0)$ or $X_0 \cap A(0) $ is finite. Since this last case cannot happen, we conclude that $X_0 \subseteq^* A(0)$. Suppose that for some $n<\omega$ we have defined a decreasing family $X_0 ,..., X_n$ of elements of $\mathcal{R}$ such that $X_{i}\subseteq^* A(i) $, for all $i\leq n$. Since $A\cap(\lbrace n+1\rbrace\times X_n )$ is infinite, there exists $X_{n+1} \in[X_n ]^\omega \cap\mathcal{R}$ such that $X_{n+1} \subseteq^* A(n+1)$ (for the case when $X_{n+1}\cap A(n+1)$ is finite cannot happen). So we can recursively construct a decreasing subfamily of $\mathcal{R}$ which witnesses that $\lbrace A(n)\mid n<\omega\rbrace$ is a centered family and that $A\notin\mathcal{NC}$. We conclude that $non^* (\mathcal{NC})\leq\mathfrak{r}$.

\end{proof}

In the realm of consistency we have the following result.

\begin{theorem}
It is consistent that $\max\lbrace cov(\mathcal{M}),\mathfrak{b}\rbrace<non^* (\mathcal{NC}).$
\end{theorem}

\begin{proof}
In Theorem 1.6.12 of \cite{davidmeza} we have a model of $cof(\mathcal{M})=\omega_1$\footnote{$cof(\mathcal{M})$ is the smallest size of a family $\mathcal{F}$ of meager subsets of the real line such that for all meager set $X$ there exists $Y\in\mathcal{F}$ such that $X\subseteq Y$.} and $non^* (\mathcal{ED}_{fin})>\omega_1$. From Proposition \ref{katetov relations} it follows that in this model also $non^* (\mathcal{NC})>\omega_1$ holds. Since $cov(\mathcal{M}),\mathfrak{b}\leq cof(\mathcal{M})$, the inequality holds in this model.
\end{proof}

Concerning the uniformity of the Nowhere Centered ideal a question remains open:

\begin{question}
Is it consistent that $non^* (\mathcal{NC})<\mathfrak{r}$?
\end{question}

redpajama_set_name: RedPajamaArXiv


If for any reason you are not completely satisfied with your purchase or simply change your mind we offer a full 30 Day money back guarantee. No questions asked!
At Najo we stand by the quality of our jewellery and offer a 12 month warranty on all our pieces for any manufacturing faults.
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Sterling Silver can be easily cleaned and polished using a simple cleaning cloth to keep your piece looking as new as the first day you take it home! To keep it safe from harm also make sure you store your Najo jewellery in a soft pouch or jewellery box.
The nature of sterling silver is that it will tarnish and discolour over time. Exposure to moisture, humidity, acidity from skin, and exposure to such things as soaps and perfumes can all cause the silver to discolour.
Regular cleaning though with a silver polishing cloth, and occasionally a metal cleaner or silver foam will keep your piece looking beautiful for years to come.
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text:
LUCIE ATAYA
Four Star Rating
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Review: Anti-Virus, by Freya Mcmillan
'I loved Anti-Virus — it's brilliant and eerie and beautifully spooky, a must read for any fan of dystopian fiction with a highly realistic feel.'
I found Freya McMillan's debut Anti-Virus to be intriguing and highly disturbing — in all the right ways.
We're catapulted in the world of post-pandemic Great Britain (… sounds familiar?) where, following a near-cataclysmic virus spread, the nation lives tightly controlled, with every aspect of people's existence strictly regimented by the government. For Callie, former field agent for the secret services, it's all for the greater good. But the day she stumbles upon something that shakes her to her core and starts pulling at loose threads, her blindly idealistic vision of the system she serves all comes crumbling around her.
Anti-Virus is suspenseful and dark and uncomfortably familiar in so many ways. I loved following Callie and her kick-ass girlfriend Jak through a maze of secrets and conspiracies, and was amazed at the level of detail the author put into building such a convincing plot. All the way through, it had me thinking that some of these scenarios were not so far from the realms of possibilities — especially after the last couple of years we've all endured.
I'd have to say my favourite part of the book was watching Callie's character evolve and grow. From a terrified recluse dealing with incredible trauma to a woman ready to face her own demons to do what's right. Freya McMillan's handling of that delicate transition was masterful, and I enjoyed every bit of it.
I loved Anti-Virus — it's brilliant and eerie and beautifully spooky, a must read for any fan of dystopian fiction with a highly realistic feel.
I was given a free ARC reader copy of this book in exchange for an honest review.
In Body I Trust, by Lauren Dow
Review: Counter Ops, by Jessica Scurlock
Review: Heart of a Runaway Girl, by Trevor Wiltzen
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redpajama_set_name: RedPajamaCommonCrawl


Just a quick note about the "Gonstead" technique.
I confess I'm wary of Chiropractors, so ill have deeper look at this in future posts. This post is a statement of fact not a recommendation, or for that matter a condemnation!


It should come as no shock that over 80% of businesses are active on one or more social media platforms. With the number of active users on social media growing daily, no company can afford to bypass the benefits social platforms provide (brand awareness, lead generation, website visits – just to name a few.) The financial services industry is no exception. In two of our most recent studies (found here and here), we quantified that more than 9 out of every 10 of the world's top asset managers and global private equity firms are active on at least one social media platform.
with such a large number of businesses active on social media it's not enough to simply exist – you must build a presence.
Consumer attention spans are shrinking – if you want prospects to pay attention you have to be concise. This is especially important in financial services, where industry jargon often drives people to either consciously or subconsciously tune out. Because users on social media tend to consume content on-the-go, long-form social media content is often indigestible and, as a result, ignored. Organizations that can boil down convoluted topics into easy-to-read, compelling copy will find the most success.
How short is short enough? We recommend sharing no more than three sentences, or around 150 characters, in a single post, regardless of platform. This isn't always possible, but is a good goal to keep in mind.
Social media posts that include an image, infographic, or video receive up to 200% more engagement than text-only posts.1 Posts that include a visual have also been found to have better brand recall by consumers as well.
Financial services and technology providers should make use of eye-catching infographics and videos to help distill long messages into more engaging formats. In fact, video is one of the most popular mediums used by the top 200 asset managers globally, with 93% producing video content regularly, according to our 2017 study.
Not every post, however, lends itself to a video or infographic. In these instances, we've found many clients benefit from a subscription to a stock image website to easily source pictures or artwork that can accompany social media posts. There are a number of options, including several that offer free imagery under creative commons licenses, but we've found Shutterstock to be a low-cost, high-quality alternative that meets the needs of most clients.
To be sure, there can be tremendous value in simply engaging directly with the larger universe of users who are accessible and active on social media. It's important to keep in mind, however, the larger objectives of the social media strategy. Whenever possible, you will want to include a call to action or even a link to your website in your social copy. After all, if you're not trying to get users to engage with your business, why put the time into social media? It goes both ways, too - 73.4% of users follow a brand on social media because they're interested in the product or service 1. Give the fans what they want!
For financial service companies, using social media as a distribution channel for original content often pays the most dividends. Going back to our previously mentioned studies, we've found that content marketing has been adopted by the vast majority of the world's top firms in both asset management and private equity (88% and 66%, respectively.) Original content that offers users valuable information is a highly effective way to promote your brand, grow your following, and acquire new prospects. For that reason, if you're producing content, be sure to share your firm's insights, blogs, whitepapers, and other pieces on social media for optimal engagement.
Now that you've got the basics down, time to play ball!