The set of all subsequential limits of a sequence is closed.

Question

Let $X$ be a metric space and $\{x_n\}$ be a sequence in $X$. Let $E$ be the set of all subsequential limits of $\{x_n\}$. Then $E$ is closed.

I have a idea of this proof. Assume $E$ is a infinite set. Let $p\in X-E$. Then $\exists r>0$ such that $B(p,r)\cap E=\emptyset$. The reason is if $p\in \bar E$, then there is some subsequential limit $y$ close to $p$, and there exists a subsequence $x_{n_k}$ converges to $y$. So, $d(x_{n_k},p)<d(x_{n_k},y)+d(y,p)<\varepsilon$ which is a contradiction.

So, $B(p,r)\subset X-E$ and $E$ is closed.

Answer 1

Your argument is not precise . You want to make sure that $n_k$ is strictly increasing.

Suppose there is no $r>0$ such that $B(p,r)\cap E =\emptyset$. For each $i$ there exists $y_i \in B(p,\frac 1 i)$ such that $y_i \in E$. Since $y_i$ is a subsequential limit of $(x_n)$ there exist infinitely many $n_i $ such that $d(y_i,x_{n_i}) <\frac 1 i$. Inductively we can choose the $n_i$'s such that $n_i >n_{i-1}$. Now $(x_{n_i})$ is a subsequence of $(x_n)$ convergin g to $p$, a contradiction.

Answer 2

A more direct but conceptual way is the following. Suppose $(x_n)\in X^{\mathbb{N}}$ is a sequence in the metric space $X$. Denote $L$ the set of all limits of the subsequences of $(x_n)$. Then \begin{align} L = \bigcap_{k\geqslant 0}\overline{\{x_n ~|~ n \geqslant k\}} \end{align} It is an intersection of closed subset, thus it is a closed subset.

This is because a point $l$ is in $L$ if and only if for every $\varepsilon >0$ and every $n \geqslant 0$, there exists $k \geqslant n$ such that $d(l,x_k) \leqslant \varepsilon$.

Answer 3

Allow me to give you an indication of a relation that holds in very general settings. Consider a first countable topological space $(X, \mathscr{T})$, in other words a space in which every point $x \in X$ admits a countable neighbourhood base.

Given an arbitrary family $x$ of objects (sets, ultimately) indexed by set $I$ and an arbitrary subset $J \subseteq I$, let us write $x_J\colon=\{x_i\}_{i \in J}$ for the set of components of indices $i \in J$ of the family $x$. For arbitrary natural number $m \in \mathbb{N}$, let us denote the unbounded natural interval of all natural numbers $n \geqslant m$ by $[m, \rightarrow)$.

Given an arbitrary sequence of points $x \in X^{\mathbb{N}}$, it is not difficult to show that the set of limits of subsequences of $x$ is equal to the set of points adherent to the sequence $x$, set which is by definition expressed as $\displaystyle\bigcap_{n \in \mathbb{N}}\overline{x_{[n, \rightarrow)}}$. As an intersection of closed subsets, this subset is itself closed.

Metric spaces are in particular first countable (for any point $x \in X$ the collection $\left\{\mathrm{B}_q(x)\right\}_{q \in \mathbb{Q_{+}^{\times}}}$ of open balls centered at $x$ and of rational radius is a neighbourhood base), so the above statement particularly applies.

Answer 4

I'm sorry your proof is not correct as we could have $X\subset E$: think about $(\cos(n))_{n\in\mathbb{N}}$.(see this link)

The most elementary way I know to prove this is by diagonal subsequence extraction.

Answer 5

Your idea is very nice, but perhaps your proof could be a little more precise. Note, however, that there is no reason to assume that $E$ is an infinite set - the proof also works for finite $E$.

Here is a suggestion.

It should be clear that a point $\xi \in X$ is the limit of some subsequence of $(x_n)$ if and only if each open neigbhorhood of $\xi$ contains infinitely many $x_n$.

Consider $p \in \overline E$ and let $U$ be an open neigbhorhood of $p$. Then $U$ contains some $\xi \in E$. But $U$ is an open neigbhorhood of $\xi$ and thus contains infinitely many $x_n$.