# Showing that if a sub-subsequence converges then the sequence also converges

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?

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

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.

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.

• $|x_n - a| > \varepsilon$ is the same as $d(x_n, a) > \varepsilon$ since $X$ is a metric space right? – user854387 Mar 30 at 13:06
• Yes, correct, just shorthand. – Igor Rivin Mar 30 at 13:07
• 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 $d(y_n, a) \ge \varepsilon$? – user854387 Mar 30 at 13:09
• @DanielLi You assukmed that in the first sentence. – Igor Rivin Mar 30 at 14:08