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. 
 A: 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.  
A: 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.
