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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?

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Theoneandonly
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  • I'm confused by the definition of a subsequence here. In particular, a sequence is manifestly a subsequence of itself. If this is not allowed, you can remove finitely many terms in the lead, and keep the tail intact, from either of which the result follows. – failedentertainment May 15 '20 at 02:32
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    @failedentertainment: It’s not that simple: the condition ensures that since the original sequence is a subsequence of itself, it has some subsequence that converges to $L$, but that could conceivably be some very sparse subsequence. – Brian M. Scott May 15 '20 at 03:00
  • you can see https://math.stackexchange.com/questions/397978/every-subsequence-of-x-n-has-a-further-subsequence-which-converges-to-x-the – user159888 May 15 '20 at 03:22

2 Answers2

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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 $|a_{n_k}-L|<\epsilon$.

Brian M. Scott
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  • Well, take some subsequence $(a_{n_{k_l}})$ of $(a_{n_k})$ that converges to $L$: means that $\forall \epsilon > 0$ we can find some $N>0$ such that

    $$ l > N \implies |a_{n_{k_l}} - L | < \epsilon $$

    Now, using this same $\epsilon > 0$ in my first stament and putting $N = l$, then we can find $n > l$ such that $|a_{n_{k_l}} - L | \geq \epsilon $ this is contradiction

     – Theoneandonly May 15 '20 at 03:35
  • Is this also another way to do it sir? – Theoneandonly May 15 '20 at 03:36
  • @Theoneandonly: That’s essentially the argument that I gave in my answer: the differences are more cosmetic than substantial. – Brian M. Scott May 15 '20 at 03:37
  • I am just still trying to visualize this result. Its a little hard to imagine it. Can you illuminate a little? – Theoneandonly May 15 '20 at 03:38
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    @Theoneandonly: I think that what makes it hard to visualize is that it’s actually true: saying that every subsequence of $\langle a_n:n\in\Bbb N\rangle$ has a subsequence converging to $L$ makes $\langle a_n:n\in\Bbb N\rangle$ sound rather complicated, and when we try to prove something about it we don’t want to assume anything more than we’re actually given, but in fact it turns out that the sequence just converges to $L$ itself, so that most of the complexity that we were trying to imagine turns out not to be there after all. – Brian M. Scott May 15 '20 at 03:42
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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 $\Sigma.$

DanielWainfleet
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