If $ (a_n) $ is a sequence s.t. for every integer $ p > 1 $ . $ ( a_{pk} )_{k=1}^{\infty} \rightarrow L $ , then $ (a_n) \rightarrow L $ Prove/Disprove: If $ ( a_n) $ is a sequence s.t. for every integer $ p > 1 $ the subsequence $ ( a_{pk} )_{k=1}^{\infty} $ converges to the same limit, then $ (a_n) $ converges to the same limit.
Answer: The statement's false. Take for example the sequence
$ \begin{aligned}
a_n=\begin{cases}1 & \text{if }n  \text{ is prime } \\
0 & \text{else }n \text{ is not prime } \end{cases}
\end{aligned}$
My question: The sequence above is one such counter-example, but I still have quite difficulty understanding it. Maybe you can please elaborate more on why this subsequence provides a counter-example? does there exists an easier counter-example than above?

By definition a sequence $a_{n}$ converges to $x\in \mathbb{R}$ if for every $\epsilon >0$ there exists a natural $N$ such that for all naturals $n$ greater than $N$ $|a_{n}-x|<\epsilon$. If you want to prove that the sequence does not converge to any $x$ in $\mathbb{R}$ , you have to prove the above statements negation that is ; for every $x\in \mathbb{R}$ there exists an epsilon such that for all natural $N$ there exists a natural $n$ greater than $N$ such that $|a_{n}-x|>\epsilon$. Since the primes are  infinite you can easily conclude form here that your sequence does not converge but $a_{kp}$ does converge since $kp$ is never prime for all $k,p > 1$ .
