Convergence in probability with subsequences after hours of trying I wonder whether someone of you could may help me by proving this claim :-)
Assume $X_n$ is a sequence of random variables which converges in probability to $X$. Now, let $X_{nk}$ be a subsequence of $X_n$. Then, there is a subsequence $X_{\overline{nk}}$ of $X_{nk}$ such that 
$$ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} \quad \forall k \geq 1$$
Unfortunately, I do not how to prove this statement... The proof given was "It holds if one waits long enough". But obviously, this is not very mathematically...
Thanks in advance!
 A: The definition of convergence in probability implies that for any $\epsilon > 0$ and $\delta > 0$, we can always find a $N_{\epsilon}$ such for all $n \geq N_{\epsilon}$, we have $P(|X_n - X| > \epsilon) < \delta$. Essentially, we are looking at convergence of the sequence of probabilities $P(|X_n - X| > \epsilon)$ to zero.
Now, every subsequence of the convergent sequence also converges to the same limit. The sequence of $P(|X_{nk} - X| > \epsilon)$ also converges to zero - which implies that we can pick an index from the sequence $X_{nk}$ which satisfies $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k}) \leq \frac{1}{k^2} $ for each $k \geq 1$ and construct a separate sequence $X_{\overline{nk}}$ from these picked indices. Note that for $k_2 > k_1$, $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_2}) \leq \frac{1}{k_2^2} $ implies that $ P( |X_{\overline{nk}} - X | \geq \frac{1}{k_1}) \leq \frac{1}{k_1^2} $. Therefore, we can pick an increasing set of indices. 
Naturally, a sequence  $X_{\overline{nk}}$ constructed as above satisfies the property discussed in the question. 
A: Let $Y_k=X_{n_k}$. Then $Y_k \to X$ in probability. Hence $P(|Y_k-Y| \geq\frac 1 j) \to 0$ for each $j$. So there exists $k_j$ such that $P(|Y_{k_j}-Y| \geq \frac 1 j) < \frac  1{j^{2}}$. We can choose the $k_j$ to be increasing. Now $P(|X_{n_{k_j}} -X| \geq \frac  1j) \leq \frac 1 {j^{2}}$ for all $j$. 
