Advanced Probability Theory: Why does convergence in prob not imply convergence a.s. I totally understand why convergence a.s. implies convergence in probability, and could picture why convergence in probability does not imply convergence a.s.
I got confused when we learned that $X_n$ converges to X iff for every sub-sequence there exists a further sub-sequence which converges a.s. to X.  
I remember learning a property that: every sub-sequence that converges to some value C iff the sequence converges to C.  
So wouldn't that imply that a sequences convergences in prob iff it convergenes a.s.?  
 A: Given random process $\{X_n\}_{n=1}^{\infty}$, the key issue is whether or not we are allowed to choose our subsequence indices $\{n[k]\}_{k=1}^{\infty}$ based on knowledge of the $\{X_n\}$ values, or not. Let’s call a pre-determined infinite subsequence $\{X_{n[k]}\}_{k=1}^{\infty}$ an infinite subsequence where the indices $n[k]$ are chosen in advance, without any knowledge of the actual $\{X_n\}$ values.  Specifically, using notation $\{X_n(\omega)\}_{n=1}^{\infty}$, the subsequence must be chosen without knowledge of the outcome $\omega \in \Omega$ (where $\Omega$ is the probability space).  
Fact 1: If $\{X_n\}$ converges in probability to $X$, then there exists a pre-determined subsequence $\{X_{n[k]}\}_{k=1}^{\infty}$ over which $X_{n[k]}$ converges to $X$ almost surely. 
[Note that this is a stronger statement than usual since, unfortunately,  most statements of this fact do no use the term “pre-determined,” even though the standard proof indeed shows the subsequence can be pre-determined.] 

Here is an example that illustrates the difference between “pre-determined” and “not-pre-determined.” 
Let $\{X_n\}_{n=1}^{\infty}$ be independent Bernoulli variables with 
$$ P[X_n=1] = 1/n, P[X_n=0] = 1-1/n$$
This is a standard example where convergence in probability to 0 holds, but convergence with probability 1 does not hold. 
For this sequence it is true that for any pre-determined infinite subsequence $\{X_{n[k]}\}_{k=1}^{\infty}$  we can find a (pre-determined) sub-subsequence over which the variables converge to 0 almost surely. 
However, with probability 1, we can find an infinite subsequence $\{X_{n[k]}\}_{k=1}^{\infty}$ where indices $n[k]$ depend on the actual realizations of the $\{X_n\}$ process and such that $X_{n[k]}=1$ for all $k \in \{1, 2, 3, …\}$.  Clearly there is no sub-subsequence over which the variables converge to 0 almost surely.  This does not contradict the previous paragraph since the infinite subsequence we are choosing in this current paragraph is not pre-determined. 
Now, considering sample-path convergence of $\{X_n(\omega)\}_{n=1}^{\infty}$ for a particular realization $\omega$, you are correct to apply the real-analysis fact that this converges to some value $X(\omega)$ if and only if for every subsequence there is a sub-subsequence that converges to $X(\omega)$.  The requirement of every subsequence includes those that are not pre-determined (i.e., those that may depend on $\omega$).  So this condition fails in the above example of independent Bernoulli variables, which is why you cannot apply this real-analysis fact to conclude almost-sure convergence from in-probability convergence.

In summary, here is where the argument breaks: 
\begin{align}
&\mbox{$\{X_n\}$ converges to 0 in probability} \\
&\implies \mbox{For every pre-determined infinite subsequence $\{X_{n[k]}\}$}\\
& \quad \quad \mbox{ there is a sub-subsequence that converges to 0 almost surely} \\
&\overset{(a)}{\implies} \mbox{For every sample path and every   infinite subsequence $\{X_{n[k]}\}$}\\
& \quad \quad \mbox{ there is a sub-subsequence that converges to 0 } \\
&\implies \mbox{$\{X_n\}$ converges to 0 on every sample path}\\
&\implies \mbox{$\{X_n\}$ converges to 0 almost surely} 
\end{align} 
The incorrect step is implication (a).
