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

Question

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.

Answer 1

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.

Answer 2

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 $n\rightarrow \infty$.

Answer 3

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.