Bolzano–Weierstrass theorem for random variables? I am wondering if there is something similar to the Bolzano–Weierstrass theorem for random sequences. Namely, let $\{x_n\}$ be a bounded random sequence. Is it true that, under some reasonable conditions (I cannot be specific), there exists a subsequence of it that converges almost surely to a constant (not random variable)? 
thanks in advance.  
 A: In general, no.  A standard counterexample is to take your probability space to be $\Omega = [0,1]$ with Lebesgue measure, and consider the complex-valued random variables $X_n(\omega) = e^{2 \pi i n \omega}$, which are all bounded in absolute value by 1.  They are orthonormal in $L^2(\Omega)$, so by Bessel's inequality, they converge weakly to 0 in $L^2(\Omega)$.  If a subsequence converges almost surely, then by dominated convergence it converges strongly in $L^2(\Omega)$, and the limit must be the same as the weak limit, i.e. 0.  But all $X_n$ have $L^2$ norm equal to 1, so this is absurd.
You can modify this to get real-valued counterexamples if you prefer.
Also, there is no reason to expect to be able to get a subsequential limit that is constant.  As a very trivial example, let $X$ be any non-constant random variable and consider the sequence $(X,X,X,X,\dots)$; every subsequence converges almost surely to $X$.
If you want to guarantee convergence of a subsequence (which is really a statement about compactness), the usual approach is to choose a weaker mode of convergence.  For instance, it follows from Prohorov's theorem that a uniformly bounded sequence of random  variables has a subsequence converging in distribution.  If your probability space is standard Borel, the Banach-Alaoglu theorem will guarantee that for any $1 < p < \infty$, you have a subsequence converging weakly in $L^p$.
In another direction, the Tychonoff theorem guarantees that your sequence has a subnet converging everywhere.  This subnet is typically not a subsequence.
A: After thinking about it some, here is a result in that direction that uses the methodology of the standard Bolzano-Wierstrass proof. I just made this up, let me know if this is an existing result:
Fix $(\Omega, \mathcal{F}, P)$ as a probability space.
Proposition: Let $\{X_n\}_{n=1}^{\infty}$ be an infinite sequence of mutually independent random variables that take values in a bounded subset of the reals. Then there is a (non-random) constant $c$ and a set $B \subseteq \Omega$ with $P[B]=1$ such that for all $\omega \in B$, the sequence of real numbers $\{X_n(\omega)\}_{n=1}^{\infty}$ contains a subsequence  that converges to $c$.
[The constant $c$ does not depend on $\omega$, but the subsequence that converges to $c$ can depend on $\omega$.]
The proof relies on the second Borel-Cantelli Lemma:
Lemma: (Second Borel-Cantelli)  Let $\{X_1, X_2, \ldots\}$ be mutually independent random variables and let $\{A_1, A_2, \ldots\}$ be events such that $\sum_{n=1}^{\infty} Pr[X_n \in A_n] = \infty$.  Then with probability 1, $X_n \in A_n$ infinitely often. (This is a standard result so I will skip the proof)
Proof of Proposition: Without loss of generality we can assume the random variables take values in the bounded interval $[-M,M]$ for some positive number $M$. Call $[-M,M]$ the interval $I_1$. Chop $[-M,M]$ into two sub-intervals $[-M,0]$ and $[0,M]$.  Then $Pr[X_n\in I_1] = 1$ for all $n$ and so:
\begin{align}
\infty &= \sum_{n=1}^{\infty} Pr[X_n \in I_1] \\
&\leq \sum_{n=1}^{\infty} \left(Pr\left[X_n \in [-M,0]\right] +  Pr\left[X_n \in [0, M]\right]\right)\\
&=\sum_{n=1}^{\infty} Pr\left[X_n \in [-M,0]\right] + \sum_{n=1}^{\infty} Pr\left[X_n \in [0,M]\right]
\end{align}
Thus, either the left or right sub-interval must have an infinite sum. Choose the left-most sub-interval that has an infinite sum and call this interval $I_2$. Then $\sum_{n=1}^{\infty} Pr[X_n \in I_2] = \infty$ and so, by the second Borel-Cantelli lemma, we know (with prob 1) that $X_n \in I_2$ infinitely often. Now chop $I_2$ into two sub-intervals $Left_2$ and $Right_2$. Then:
\begin{align}
\infty &=\sum_{n=1}^{\infty} Pr[X_n \in I_2] \\
&\leq \sum_{n=1}^{\infty} Pr[X_n \in Left_2] + \sum_{n=1}^{\infty} Pr[X_n \in Right_2]
\end{align}
and so again one of the sub-intervals must have an infinite sum.  Choose the left-most sub-interval that has an infinite sum and call this $I_3$.  Then $\sum_{n=1}^{\infty} Pr[X_n \in I_3] = \infty$.  By Borel-Cantelli, with prob 1 $X_n$ must be in $I_3$ infinitely often. Continuing this way, we get a sequence of nested closed sub-intervals $\{I_1, I_2, I_3, \ldots\}$ that have size that vanishes to 0.  So the sub-intervals must converge to a single point $c$. Note that there is nothing random about the sub-intervals, or about the point $c$. So $c$ is a (non-random) constant. With probability 1, for each $k$ the sequence $\{X_n\}$ is in each sub-interval $I_k$ infinitely often. Now form a subsequence as follows:  Choose $n[1] = 1$. For each $k>1$, choose $n[k]$ as the smallest index $m$ such that $m>n[k-1]$ and $X_m \in I_k$.  Such a sub-sequence can be constructed with probability 1.  Then $\{X_{n[k]}\}_{k=1}^{\infty}$ converges to $c$.

The Nate Eldredge example $(X,X,X, \ldots)$ gives a simple counter-example if we try to remove the independence assumption in the above proposition.
