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Let $x \in X$ and $x_n$ be a sequence in $X$. If every subsequence of $x_n$ has a subsequence that converges to $x$. Show that $x_n$ must also converge to $x$.

I'm trying to get a contradiction here by supposing the opposite. So suppose that $x_n$ does not converge to $a$. This means that there exists $\varepsilon >0$ such that $d(x,x_n)\ge\varepsilon$ for some $n \ge K \in \mathbb{N}$. I'm not sure how I should continue from here? What definitions can I use to proceed?

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  • $\begingroup$ In what sense is it meant for a sequence to converge, here? Are we assuming that $X$ is a metric space? $\endgroup$ – Math1000 Mar 30 at 13:13
  • $\begingroup$ It's essential you have infinitely many $n$ with that property... See my answer.. $\endgroup$ – Henno Brandsma Mar 30 at 13:49
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The negation of convergence of $x_n \to x$ is:

$$\exists \varepsilon>0: \forall N \exists n > N: d(x_n, x) \ge \varepsilon\tag{1}$$

Fix this $\varepsilon>0$ for the remainder. Now proceed by recursion: Let $n_1$ be the $n$ given (by $(1)$) for the choice $N=1$, so that we know $d(x_{n_1}, x) \ge \varepsilon$.

Now having defined $n_1 < \ldots < n_k$ in $\Bbb N$ so that $x_{n_k}$ satisfies $d(x_{n_i}, x) \ge \varepsilon$, for all $i \le k$, pick $N=n_k$ in $(1)$ and we have some $n$, which we call $n_{k+1} > n_k$ and which also obeys $d(x_{n_{k+1}}, x) \ge \varepsilon$.

So from the negation of $x_n \to x$ we have constructed a subsequence $(x_{n_k})_k$ of $(x_n)_n$ with all terms at least $\varepsilon$ away from $x$. It follows that no subsequence of $(x_{n_k})_k$ can converge to $x$, as the open ball $B(x,\varepsilon)$ contains no points at all of that subsequence. This contradicts the assumption on $(x_n)_n$ so $x_n \to x$ must hold.

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You are almost done. For any $N$ there is an $n>N$ such that $|x_n-a| > \epsilon.$ Now, take the subsequence $(x_n)$ with $n$ as in the previous sentene.

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  • $\begingroup$ $|x_n - a| > \varepsilon$ is the same as $d(x_n, a) > \varepsilon$ since $X$ is a metric space right? $\endgroup$ – user854387 Mar 30 at 13:06
  • $\begingroup$ Yes, correct, just shorthand. $\endgroup$ – Igor Rivin Mar 30 at 13:07
  • $\begingroup$ So I have a subsequence $y_n$ of $x_n$ with $d(y_n, a) \ge \varepsilon$, but this is a contradiction since $y_n \to a$? How do I know that $d(y_n, a) \ge \varepsilon$? $\endgroup$ – user854387 Mar 30 at 13:09
  • $\begingroup$ @DanielLi You assukmed that in the first sentence. $\endgroup$ – Igor Rivin Mar 30 at 14:08

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