Proof that the set of subsequential limits is closed

Question

I have tried to prove the following result, but it looks a bit messy in my opinion. How can it be improved?

Proposition. Let $a_n$ be a bounded sequence with values in $\mathbf{R}$. Let $A=\{l\in\mathbf{R} \mid \text{ there exists a subsequence of $a_n$ >converging to }l\}$. Then $A$ is closed.

Lemma: For each subsequence of $a_n$ which converges to $l$ it is true that $\forall {\varepsilon}>0$ $a_n\in B(l,{\varepsilon})$ for arbitrarily large $n$ -- that is, for each $m\in\mathbf{N} \; \exists M\geq m \mid a_M\in B(l,{\varepsilon})$.

Let $a_{k_n}$\,be such a subsequence. By definition, for each $\varepsilon>0 \; \exists N$ such that $\forall n\ge N \; a_{k_n}\in B(l,\varepsilon)$. Let $m\in\mathbf{N}$. Then $k_m\ge m$, since $a_{k_n}$ is a subsequence of $a_n$.

• if $m\ge N$, take $M:=k_m$;
• if $m<N$, take $M:=k_N$.

Proof. Let $x$ be a closure point of $A$. By definition, $\forall \varepsilon>0 \; B(x,\varepsilon)\cap A\neq\varnothing$, which means that $\forall \varepsilon>0$ there exists a subsequence $a_{k_n}$ which converges to ${l}_\varepsilon\in B(x,\varepsilon)$.

This means in particular that for all $m\in\mathbf{N}$ there exists a subsequence $a_{k^{(m)}_n}$ that converges to $l_m\in B(x,1/2^{m+1})$ .

Define the sequence $x_n$ inductively:

• $x_0=0$;
• $x_n=\min\{m>x_{n-1} \mid a_m \in B(l_m,1/2^{m+1}) \}$.

The sequence is well-defined, because there exists a subsequence converging to $l_m$, and by the lemma there are infinite indices for which the sequence lies in the neighborhood. So the set is not empty and by well-ordering there exists a minimum element.

It follows that for each $n\in\mathbf{N}$ $$x-1/2^{m}<l_m-1/2^{m+1}<a_{x_n}<l_m+1/2^{m+1}<x+1/2^{m},$$

so by squeeze theorem (strict inequalities imply non-strict inequalities) $a_{x_n}\to x$.

Answer 1

I suppose it is easier to show that $\Bbb R\setminus A$ is open:

If $b\in\Bbb R\setminus A$, then there is no subsequence converging to $b$, hence for some $r>0$, there are only finitely many $n$ with $|a_n-b|<r$. Then for $b'$ with $|b-b'|<\frac r2$, there are only finitely many $n$ with $|a_n-b'|<\frac r2$, hence there is no subsequence converging to $b'$.

Answer 2

Notice that $A$ is precisely the set of all limit points of the sequence $(a_n)_n$, meaning that $x \in A$ if and only if every open neighbourhood of $x$ contains infinitely many terms of the sequence $(a_n)_n$.

To show $A$ is closed, it suffices to prove it contains its limit points. Let $y \in \mathbb{R}$ be a limit point of $A$.

Let $U$ be an open neighbourhood of $y$. Since $y$ is a limit point of $A$, there exists $x \in A \cap U, x \ne y$. Then $x$ is a limit point of the sequence $(a_n)_n$ and $U$ is an open neighbourhood of $x$ so $U$ contains infinitely many terms of the sequence.

Since $U$ was arbitrary, we conclude that $y$ is a limit point of the sequence $(a_n)_n$, i.e. $y \in A$.

Therefore $A$ is closed.