Bolzano-Weierstrass for sequences of sets

Let $\mathcal{A}_n,\,n\in\mathbb{N}$ be a sequence of subsets of, say, $\mathbb{R}$. Let $\limsup_{n\rightarrow\infty} \mathcal{A}_n = \{x:x\in\mathcal{A}_n\mbox{ for infinitely many } n\}$, and $\liminf_{n\rightarrow\infty} \mathcal{A}_n = \{x:x\in\mathcal{A}_n\mbox{ for all but finitely many } n\}$ as usual. We say the sequence $\mathcal{A}_n,\,n\in\mathbb{N}$ has a limit if $\limsup_{n\rightarrow\infty} \mathcal{A}_n = \liminf_{n\rightarrow\infty} \mathcal{A}_n$ and write $\lim_{n\rightarrow\infty} \mathcal{A}_n$ for the limit. My question is, given an arbitrary sequence $\mathcal{A}_n,\,n\in\mathbb{N}$, is there a subsequence $\mathcal{A}_{n_k},\,k\in\mathbb{N}$ such that $\lim_{k\rightarrow\infty} \mathcal{A}_{n_k}$ exists? This should be well-known, but I could not find a reference.

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I think this is false. Set up a correspondence between all infinite subsequences of $\mathbb N$ and $\mathbb R$, say $\Phi: \mathbb R \to 2^{\mathbb N}$. Now, define $\mathcal A_n := \{ r \in \mathbb R: n \in \Phi(r) \text{ and$n$has odd index in } \Phi(r)\}$. Here by "index" I mean the position of $n$ when $\Phi(r)$ is listed in increasing order, or in other words the cardinality of $[1,n] \cap \Phi(r)$.

No matter what subsequence of $\{\mathcal A_n\}$ one chooses, it corresponds to some $\Phi(r)$, and within that subsequence $r$ alternates between included and excluded, so the subsequence fails to converge at $r$.

On the other hand, it should be true for subsets of $\mathbb N$ by an argument similar to the infinite Ramsey theorem. Either infinitely many $\mathcal A_n$ contain $1$ or infinitely many exclude $1$, so one can inductively pass to a subsequence where $\{\mathcal A_n\}$ converges at $1$, then a subsubsequence that converges at $2$, etc., and finally taking a diagonal of this sequence of sequences.

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which spaces of sets would posess the Bolzano Weierstrass Property? Possibly related: math.stackexchange.com/questions/202012/… –  Alex R. Jan 4 '13 at 0:10
@Alex Good question: between the argument for $\mathbb R$ and the converse for $\mathbb N$, it would seem that the only remaining question is for sets whose cardinality falls strictly between the two (in particular we'd need CH to be false for there to be anything left to ask). –  Erick Wong Jan 4 '13 at 0:17
Erick, your comment gives some sort of Martin's Axiom feel to this idea. Although this is just a late-night raving and might have no mathematical content to it... –  Asaf Karagila Jan 4 '13 at 0:21
Hi Erick, thanks for your reply. I was wondering what kind of correspondence $\Phi$ you had in mind. For example, I picked $\Phi$ to be the decimal expansion of a given $r\in\mathbb{R}$. For example, $\Phi(\pi)$ is the sequence 3,1,4,1,5,... In this case, the sequence $\mathcal{A}_n$ itself has a limit (which is empty). –  Anon Jan 4 '13 at 0:25
@AsafKaragila Martin's axiom sounds on the money (thanks for pointing this out, I was unaware of it). Wikipedia gives as a consequence that every compact Hausdorff space with smallish cardinality is sequentially compact, which seems to apply to the space $\{0,1\}^S$. –  Erick Wong Jan 4 '13 at 0:45
The smallest cardinality of a set $X$ such that some sequence of subsets of $X$ has no convergent subsequence is called the splitting number and usually denoted by $\mathfrak s$. It is one of the cardinal characteristics of the continuum that set theorists like me study. As indicated in Erick Wong's answer and its comments, we always have $\aleph_0<\mathfrak s\leq 2^{\aleph_0}$. If the continuum hypothesis holds, the only cardinal in that range is $\aleph_1$, so $\mathfrak s=\aleph_1$. But if the continuum hypothesis fails, so there are two or more cardinals in that range, then the usual axioms of set theory (Zermelo-Fraenkel axioms, including choice) do not determine the value of $\mathfrak s$. In particular, Martin's Axiom implies that $\mathfrak s=2^{\aleph_0}$, so this is consistent with $2^{\aleph_0}$ being as large as you want. On the other hand, it is consistent that $\mathfrak s=\aleph_1$ even if the continuum hypothesis fails.
A good deal is known about the relationship between $\mathfrak s$ and other cardinal characteristics of the continuum. For example, any set of reals of cardinality $<\mathfrak s$ is of first Baire category and has Lebesgue measure zero.