Prove $x_n$ converges IFF $x_n$ is bounded and has at most one limit point

I'm not entirely sure how to go about proving this so hopefully someone can point me in the right direction. The definition I have for a limit point is "$a$ will be a limit point if for a sequence $x_n$ there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} (x_{n_k})=a$".

In the forward direction if $x_n$ converges to a value $a$ then every convergent subsequence of $x_n$ must also converge to $a$. I'm not entirely sure where else to go from here. Intuitively I know that what I'm trying to prove must be true since if $x_n$ converges to $x$ then there exists an $N\in\mathbb{N}$ such that $\forall n\geq N$ we have infinitely many terms satisfying $\forall \varepsilon >0\Rightarrow x-\varepsilon<x_n$ which implies that $x$ is a limit point. I'm not entirely sure how to rule out a second limit point. Would it be sufficient to show that if $a$ and $b$ are limits of a convergent sequence $x_n$ then $a$ must equal $b$?

In the backwards direction if $x_n$ is bounded and has only one limit point then we know that there exists only one point $x$ such that we can find infinitely many terms satisfying $\forall \varepsilon >0$ $\left | x_n-x\right |<\varepsilon$ which implies that $x_n$ converges to $x$.

I'm quite concerned I am proving this incorrectly since I don't think I am using any facts about subsequences although they are mentioned directly in the definition for a limit point.

The arrow from right to left can be proven as follows. Suppose there is subsequence which does not converge to that unique limit point. There is an area around the unique limit point such that our subsequence has infinitely many points outside this area. This constitutes another subsequence which must have a limit point other than the unique one. Contradiction.

First show: a point $x_0$ is a limit point of a sequence $\{x_n\}$ if and only if for every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ centered at $x_0$ with radius of $\varepsilon$ that contains infinite number of $x_n$.

In the leftward direction of the original problem, let $x_0$ be the unique limit point. For every $\varepsilon>0$, there exists an open ball $B(x_0,\varepsilon)$ that contains infinite many of $x_n$. Thus, there are finite many $x_n$ that are outside the open ball. Otherwise, these infinite exceptional points have another limit point, which is not the case since $x_0$ is the unique limit point. Let $N$ be the max index of these exceptional points. Therefore, for all $n>N$, we have $x_n\in B(x_0,\varepsilon)$, which proves $x_n\rightarrow x_0$ as $n\rightarrow \infty$.

• "For every $\varepsilon>0$, there exists an open ball $𝐵(x_0,\varepsilon)$ that contains infinite many of $x_n$. Thus, there are finite many $x_n$ that are outside the open ball." Why? Jul 20, 2022 at 21:13

I'm assuming you're working in $\mathbb R$ here. Since the sequence is bounded it has a convergent subsequence and hence at least one limit point. Therefore by the condition it has exactly one limit point.

I claim that the whole sequence converges to this limit point. This is the same as saying every subsequence converges to this limit point. If this were not the case, meaning there were a subsequence that did not converge to this limit point, then this hypothetical subsequence itself has a limit point since it is bounded. This limit point is different from the unique limit point of the larger sequence by the choice of subsequence. However, being a limit point of the subsequence, this limit point is a limit point of the larger sequence, contradicting uniqueness.

• If a subsequence is not converging to a certain limit point, it can have a subsequence which converges to that limit point. So the reasoning to prove the left arrow is not entirely correct. Did I overlook something? Dec 31, 2016 at 16:29
• @Nadori You're right, it's a bit more subtle than this. Don't have time to fix it right now though. Dec 31, 2016 at 16:38
• I have found a proof inspired by yours. Dec 31, 2016 at 16:44
• @Nadori, can you share the link to the proof you found? Thanks! Oct 6, 2019 at 17:30
• The reasoning of this answer can be saved noting that every bounded (of course, in this case) divergent (otherwise it would be the same unique limit point, so we assume this to be the case by contradiction) subsequence must have two subsubsequences converging to two different limit points. So this triggers the contradiction, because each of them must be the unique limit point of the original sequence.
– xyz
May 14, 2020 at 14:55