If every subsequence contains a convergent subsequence, then $(a_n)$ converges to the same limit

Question

If every subsequence $(a_{n_k})$ of $(a_n)$ contains a subsequence $(a_{n_{k_l}})$ that converges to $L$, Prove that $\lim_{n \to \infty} a_n = L$

try:

We may argue by contradiction. If $\lim a_n \neq L$ then $\exists \epsilon >0$ such that $\forall N >0$ $\exists n > N$ such that $|a_n - a | \geq \epsilon $.

Now,$a$ is a limit point of subsequence $(a_{n_k})$. If I can prove that $a$ is also a limit point of $a_n$ then we will have reached a contradiction.

But, since every subsequence of $(a_{n_k})$ converges to $a$, then $a_{n_k} \to a$ as a sequence is a subsequence of itself. Therefore we have found a subsequence of $a_n$ that converges to $a$. In fact, $a$ is a limit point of $(a_n)$ and we have reached a contradiction.

Is this correct?

Answer 1

I assume that your $a$ is the $L$ of the problem statement; I’ll use $L$ here.

You cannot assert that every subsequence of $\langle a_{n_k}:k\in\Bbb N\rangle$ converges to $L$: all that is guaranteed is that some subsequence of $\langle a_{n_k}:k\in\Bbb N\rangle$ converges to $L$. In particular, you cannot conclude that $\langle a_{n_k}:k\in\Bbb N\rangle$ converges to $L$. However, you still have a contradiction, because the subsequence $\langle a_{n_k}:k\in\Bbb N\rangle$ was chosen so that $|a_{n_k}-L|\ge\epsilon$ for each $k\in\Bbb N$: no subsequence of $\langle a_{n_k}:k\in\Bbb N\rangle$ can possibly converge to $L$, because that would require that there be some $k_0\in\Bbb N$ such that $|a_{n_k}-L|<\epsilon$ whenever $k\ge k_0$, and there isn’t even one $k\in\Bbb N$ such that $|a_{n_k}-L|<\epsilon$.

Answer 2

In your attempt, you twice mention a certain sub-sequence $(a_{n_k})_k$ without saying which sub-sequence it is.

From your first line, with $a=L,$ suppose $\neg (a_n\to L).$ Then there exists $\epsilon >0$ such that the set $S=\{n\in\Bbb N:|a_n-L|\ge \epsilon\}$ is not empty and has no largest member. But then if $\Sigma$ is any sub-sub-sequence of the sub-sequence $(a_n)_{n\in S}$ then $\Sigma$ cannot converge to $L$ because $|t-L|\ge \epsilon$ for $every$ term $t$ of $\Sigma.$