limit of limit of sets is a limit of sets Let $\mathcal A$ be a collecction of subsets of some set $A$. Whenever $\{ X_n\}_{n\in\mathbb N}$ is a sequence of sets such that $\liminf_n X_n \triangleq \bigcup_{n\geq 1}\bigcap_{m\geq n} X_m$ and $\limsup_n X_n \triangleq \bigcap_{n\geq 1}\bigcup_{m\geq n} X_m$ are both equal to some set $X$ then we say that $\lim_n X_n$ exists and equal $X$.
Suppose we have a sequence of sets $\{ X_{n,k} \}_{n,k\in\mathbb N}$ in $\mathcal A$ such that for all $n$, $\lim_n X_{n,k}=X_n$ for some $X_n$ (that may not be in $\mathcal A$) and suppose that $\lim_n X_n = X$ for some $X$. I want to prove (or disprove) the following

There is a sequence of sets $\{ Y_n \}_{n\in\mathbb N}$ in $\mathcal A$ such that $\lim_n Y_n=X$.


My attempt uses a bijective mapping $\phi:\mathbb N\to\mathbb N^2$ and the sequence of sets $Y_n=X_{\phi(n)}$, it feels that this might do the trick but I can't quite finish the argument. It would be enough to show that $\liminf_n X_n\subseteq \liminf_n Y_n$ and $\limsup_n Y_n\subseteq \limsup_n X_n$, proving those two should be very similar so let's focus on the first one.
\begin{align*}
\bigcup_{n\geq1}\bigcap_{m\geq n} \bigcup_{k\geq1}\bigcap_{\ell\geq k} X_{m,\ell}\subseteq \bigcup_{n\geq1}\bigcap_{m\geq n} X_{\phi(m)}
\end{align*}
here it feels like we would like to swap the two middle intersection and union but this would result in the other inclusion. I am not sure what kind of condition ensures that we can do the swapping or if this idea is even a good one in the first place. Any idea is most welcome.

An equivalent formulation using characteristic functions : it is known that $\lim_n X_n=X$ if and only if $\psi_{X_n}$ converges to $\psi_X$ point-wise. From $\mathcal A$ we can get $\mathcal F=\{ \psi_X : X \in\mathcal A\}$ which is a set of $\{0,1\}$ valued functions. Consider $\overline{\mathcal F}=\{ \psi : \psi = \lim_n \psi_n, \psi_n\in\mathcal F \}$ the sequential closure of $\mathcal F$. What I am trying to prove is essentially that $\overline{\overline{\mathcal F}}=\overline{\mathcal F}$.
 A: Here is a counterexample.  Let $A=\mathbb{N}^\mathbb{N}$ and let $p_i:A\to\mathbb{N}$ denote the $i$th projection.  Let $X_n=p_0^{-1}(\{0,\dots,n\})$ and let $X_{n,k}=X_n\cap p_{n+1}^{-1}(\{0,\dots,k\})$.  Then $(X_n)$ converges to $A$ and $(X_{n,k})_k$ converges to $X_n$ for each $n$ (and these are even increasing sequences, so the limit is just the union).  However, I claim there is no sequence in $\mathcal{A}=\{X_{n,k}:n,k\in\mathbb{N}\}$ that converges to $A$.  To prove this, consider any sequence $Y_m=X_{n(m),k(m)}$ in $\mathcal{A}$.  If $n(m)$ does not go to $\infty$, then there is a subsequence on which $n(m)$ is constant and this subsequence cannot converge to $A$ since all of its terms are contained in $X_n$.  On the other hand, if $n(m)$ goes to $\infty$ then passing to a subsequence we may assume the values of $n(m)$ are distinct for distinct values of $k$.  Now take an element $a\in A$ with $p_{n(m)+1}(a)=k(m)+1$ for each $m$.  Then $a\not\in Y_m$ for all $m$, so $(Y_m)$ cannot converge to $A$.

Here is a more abstract perspective on the question.  As you say, you are asking whether the set of sequential limits of a subset of $\{0,1\}^A$ must itself be closed under sequential limits.  You could ask this more generally for any topological space in place of $\{0,1\}^A$, and there are well-known counterexamples, such as the Arens space, which is the free topological space with a sequence $(x_n)$ converging to a point $x$ and a sequence $(x_{n,k})$ converging to $x_n$ for each $n$ (see this answer for an explicit construction of it).  Then your question is just whether such a counterexample can be embedded in $\{0,1\}^A$ for some set $A$.  To embed a $T_0$ topological space in a space of the form $\{0,1\}^A$, you just need to give a family of clopen subsets (which correspond to maps to $\{0,1\}$) that generate the topology.  It is easy to see that, for instance, the Arens space has a basis of clopen sets, so it embeds in $\{0,1\}^A$ for some $A$.

Here is an example where $\mathcal{A}$ is an algebra of sets.  Let $A$ be the Cantor set and let $\mathcal{A}$ be its algebra of clopen subsets.  Note that any limit of a sequence of elements of $\mathcal{A}$ is a $G_\delta$ set, since it is the intersection of a sequence of unions of clopen sets.  Also, it is easy to see that any finite subset of $A$ is a limit of a sequence of clopen sets.  Now let $Q$ be a countable dense subset of $A$.  By the Baire category theorem, $Q$ is not $G_\delta$, so it is not a limit of a sequence of elements of $\mathcal{A}$.  However, each finite subset of $Q$ is such a limit, and $Q$ is the limit of a sequence of finite subsets.
