On $\mathcal{A}$ with $\forall C\in\mathcal{P}_{\infty}(\mathbb{N})\exists A\in\mathcal{A}:A\cap C\in\mathcal{P}_{\infty}(\mathbb{N})$

Question

Edit: I came up with an answer myself, see below.

Denote $\mathcal{P}_{\infty}(\mathbb{N})=\{A\subseteq\mathbb{N}\ |\ A\text{ infinite}\}$ and let $\mathcal{A}\subseteq \mathcal{P}_{\infty}(\mathbb{N})$ be such that for all $C\in\mathcal{P}_{\infty}(\mathbb{N})$ there exists $A\in\mathcal{A}$ with $A\cap C\in\mathcal{P}_{\infty}(\mathbb{N})$. Furthermore, suppose that $\mathcal{A}$ is stable under finite union, i.e. $A\cup B\in\mathcal{A}$ for all $A,B\in\mathcal{A}$. Is it true that there exists $N\in\mathbb{N}$ and $A\in\mathcal{A}$ such that $\{N,N+1,...\}\subseteq A$?

This question came up when I tried to characterize the sets $\mathcal{B}\subseteq \mathcal{P}_{\infty}(\mathbb{N})$ such that a sequence $a\in\mathbb{R}^\mathbb{N}$ converges to some $a_\infty\in\mathbb{R}$ if and only if $a|_{B}$ converges to $a_\infty$ for every $B\in\mathcal{B}$. If the statement in question would be true, then it would imply that there must be $B_1,...,B_n\in\mathcal{B}$ and $N\in\mathbb{N}$ with $\{N,N+1,...\}\subseteq B_1\cup...\cup B_n$.

I tried to generalize the concepts that arise here and when doing so, I realized that it seems to be linked to a notion that I never learned about in class yet but already heard of, namely that of a filter. This is because if we define the quasi-order $\lesssim$ on $\mathcal{P}(\mathbb{N})$ by $$ \forall A,B\in\mathcal{P}(\mathbb{N}):\quad A\lesssim B\iff \exists N\in\mathbb{N}:A\cap[N,\infty[\supseteq B\cap[N,\infty[, $$ then $\mathcal{A}$ can be extended to a filter to of $(\mathcal{P}(\mathbb{N}),\lesssim)$ by also including all subsets of elements of $\mathcal{A}$ into $\mathcal{A}$. The problem is that apart from being able to understand the definition, I know nothing about filters. Is there a theorem or concept which simplifies matters here?

Answer 1

After doing some more research, I saw that the most natural notion to use here is that of an ideal in the set-theoretical sense. In fact, if we define $\mathcal{A}':=\{A'\subseteq\mathbb{N}\ |\ \exists A\in\mathcal{A},F\subseteq\mathbb{N}\text{ finite}:\ A'\subseteq A\cup F\}$, then $\mathcal{A}'$ is a set-theoretical ideal on $\mathbb{N}$. Furthermore, my question is then equivalent to whether $\mathcal{A}'=\mathcal{P}(\mathbb{N})$ (or simply whether $\mathbb{N}\in\mathcal{A}'$).

And in this context, we can come up with the following counter-example: Take $\mathcal{A}$ to be the set formed by the infinite sets of an asymptotical density of $0$, then $\mathcal{A}'=\mathcal{A}\cup\{F\subseteq\mathbb{N}\text{ finite}\}$ is the ideal formed by all the sets of an asymptotical density of $0$. Then if $C\subseteq\mathbb{N}$ is some infinite set, we clearly can find $A\in\mathcal{A}$ with $A\subseteq C$ and in particular $A\cap C=A\in\mathcal{P}_{\infty}(\mathbb{N})$. But no $A\in\mathcal{A}$ contains $\{N,N+1,...\}$, as that would mean that $A$ would have an asymptotical density of $1$.

Answer 2

I don't know the answer to your first question, but consider the following for the characterisation you are looking for.

Let $\mathcal{A}\subset\mathcal{P}_{\infty}(\mathbb{N})$ be such that for all $C\in\mathcal{P}_{\infty}(\mathbb{N})$ there exists an $A\in\mathcal{A}$ with $A\cap C\in\mathcal{P}_{\infty}(\mathbb{N})$.

We will show that if $a=(a_{n})$ is a sequence then it converges to some $a_{\infty}\in\mathbb{R}$ if and only if $a|_{A}$ converges to $a_{\infty}$ for every $A\in\mathcal{A}$.

First suppose that $a$ converges to some $a_{\infty}\in\mathbb{R}$. Evidently every subsequence of $a$ converges to $a_{\infty}$ and thus $a|_{A}$ converges to $a_{\infty}$ for every $A\in\mathcal{A}$.

Now suppose that $a$ does not converge to some $a_{\infty}\in\mathbb{R}$. Then we can find a $C,D\in\mathcal{P}_{\infty}(\mathbb{N})$ such that either $a|_{C}$ diverges to $\pm\infty$ or $a|_{C}$ converges to some $c_{\infty}\in\mathbb{R}$ and $a|_{D}$ converges to some different $d_{\infty}\in\mathbb{R}$. By definition of $\mathcal{A}$ we can find $A,B\in\mathcal{A}$ such that $A\cap C,B\cap D\in\mathcal{P}_{\infty}(\mathbb{N})$.

If $a|_{C}$ diverges to $\pm\infty$, then $a|_{A\cap C}$ diverges to $\pm\infty$ and thus $a|_{A}$ does not converge.

If $a|_{C}$ converges to $c_{\infty}$ and $a|_{D}$ converges to $d_{\infty}$, then $a|_{A\cap C}$ converges to $c_{\infty}$ and $a|_{B\cap D}$ converges to $d_{\infty}$. It follows that either $a|_{A}$ or $a|_{b}$ does not converge, or $a|_{A}$ converges to $c_{\infty}$ and $a|_{B}$ converges to $d_{\infty}$. As $c_{\infty}\neq d_{\infty}$ we find that in all cases there exist $A,B\in\mathcal{A}$ such that either $a|_{A}$ or $a|_{B}$ does not converge, or $a|_{A}$ and $a|_{B}$ converge to different values which proves our statement.

As a final note, you have the wrong definition of filters. Filters are closed under finite intersection and taking supset.