Accumulation points and closed sets

Denote by $F$ the set of all accumulation points of $(x_{n})$. We define an accumulation point $x \in \mathbb{R}$ if there exists a subsequence $(x_{n_{k}})$ of $(x_{n})$ (being the latter a bounded sequence) such that $\lim(x_{n_{k}})=x$. Prove $F$ is a bounded, closed set. As well, show that $\limsup(x_{n})=\sup(L)$.

I'm not sure on where to start with this problem. From most definitions I've seen in other books, they define an cluster point (I don't know if this is equivalent with an accumulation point) as being a point of a set $A$ if $\forall \delta > 0$, there exists at least one point $x \in A, x \neq c$ so that $|x-c|<\delta$; but I'm not sure on how to relate it with the definition given.

If this is the case, then I was thinking of the following proof for the first part of the exercise (I'm not sure if it is relevant to this exercise): if $t$ is not a cluster point of $F$, then it belongs in $F^{c}$. This implies that $\exists \epsilon > 0$ such that $(t-\epsilon,t+\epsilon)$ does not contain any points in $F$. As $t \in F^{c}$, then we would have that $(t-\epsilon, t + \epsilon) \subseteq F^{c}$. As $t$ is an arbitrary point of $F^{c}$, then $\forall t \in F^{c}$ we can find an open set contained in $F^{c}$ which implies that it will be open, so it follows $F$ is closed.

The main problem here would be how to relate it to the concept of sequences. As well, I'm not sure on how to start the second part of the exercise. Thank you for all your help and patience.

If $x$ is not in $F$, it means that there is no subsequence of $(x_n)$ that converges to $x$. This means that there is some $\epsilon$ such that there are at most finitely many $k$ such that $x_k \in (x-\epsilon, x+\epsilon)$.

(for suppose not: every $\frac{1}{n}$-neighbourhood of $x$ would contain infinitely many $x_k$ and this would allow us to construct a subsequence of $(x_n)$ converging to $x$.

For every $y \in (x-\epsilon, x + \epsilon)$ we have some $\delta > 0$ such that $(y-\delta, y+\delta) \subset (x-\epsilon, x+ \epsilon)$, and so every such $y$ also has a neighbourhood with only finitely many $x_k$, making it not in $F$. So $\mathbb{R} \setminus F$ is open.

If all $x_n$ are $\le M$, so will any subsequential limit be, as $(-\infty, M]$ is closed (so closed under sequence limits). Same for lower limits. So $F$ is bounded whenever the sequence $(x_n)$ is.

As to the last point: by closedness $\sup(F) \in F$. So there is a subsequential limit converging to $\sup(F)$. Can you show using the definition of $\limsup x_n$ that it equals this number?

• Thanks for the answer. However, I don't understand how we can say that there will be a $\delta$ such that $(y-\delta, y+\delta) \subset (x-\epsilon, x+\epsilon)$. And as well, from this how can we say that for every $x \in F^{c}$ there will be an open set contained in this set? Oct 19, 2014 at 21:28
• This just says that $(x-\epsilon, x+\epsilon)$ is an open set. You can use $\delta = \min(y - (x-\epsilon), (x+\epsilon)-y)$ if you want an explicit $\delta$. The $(x- \epsilon, x+\epsilon)$ is the required open set around $x$ that is contained in $F^c$. Oct 20, 2014 at 7:10
• Your proof technique here is contrapositive right?
– user960654
Dec 23, 2021 at 13:41
• @HomerSimpson what do you mean? I just show the complement is open. That’s applying a definition not a contrapositive. The second paragraph is a small supporting subproof from contradiction to support the claim in the first. Dec 23, 2021 at 13:48
• You’re right . Didn’t see rest of the proof
– user960654
Dec 23, 2021 at 14:06

Here's some help for the first part at least.

You can show that $F$ is closed by showing $F$ contains all of $its$ accumulation points. Let $(y_m)\in F$ be a sequence converging to $y$. This implies for each $m$ there is a subseuqence $(x_{m,k})_{k=1}^\infty$ of $(x_n)$ converging to $y_m$. For each $m$, choose a number $k(m)$ large enough so that $$|x_{m,k(m)}-y_m|\le \frac1m.$$ Then the sequence $x_{m,k(m)}$ will converge to $y$, since $$|x_{m,k(m)}-y|\le |x_{m,k(m)}-y_m|+|y_m-y|\le \frac1m+|y_m-y|\to 0$$ as $m\to\infty$. Since $x_{m,k(m)}\to y$, this shows $y\in F$, so $F$ is closed.