The limit points of a set form a closed set

Question

For any set $A$, let $L$ be the set of limit points of $A$. Then $L$ is closed.

My proof is below. Please verify or critique. Note that proofs are readily available; my question is for verification and critique of my proof and its exposition.

Let $(\ell_n)$ be a sequence in $L$ converging to $\ell$. Then there exists a sequence $(a_n)$ in $A$ converging to $\ell$. Since each $\ell_n$ is a member of $L$, it is a limit point of $A$, and so for every $n \in \mathbb{N}$, there exists an $a_n$ in A such that $|a_n - \ell_n| < \frac{1}{n}$. Then $|a_n - \ell| \leq |a_n - \ell_n| + |\ell_n - \ell| = 1/n + |\ell_n - \ell| \to 0.$ QED.

Is my proof correct, rigorous, clear, and well written?

Answer 1

In order to determine if the proof is correct or not let us contextualize a little by restating the proposition.

Let $X$ be a space. For any set $A\subseteq X$, let $L$ be the set of limit points of $A$. Then $L$ is closed.

To adhere to standard notation, we will use the notation $L=A'$ and call it the derived set of $A$. We will check for which $X$ the proof presented in the question stands. We will start with well-behaved spaces and move towards more general ones as far as we can go.

The proof, as it is written, holds true for $X=\mathbb{R}^n$, in fact, it holds for any normed vector space (assuming $|\cdot|$ is the norm on $X$). It's well written, clear and rigorous. (Maybe, I'd change the "Then there exists a sequence $(a_n)$ [...]" to "We'll show that there exists a sequence $(a_n)$ [...]" but that's me being picky).

As you asked in a comment, if $X$ is any metric space, then we have a problem: $a_n-\ell_n$ isn't defined (since there may be no algebraic operations). Furthermore, $|\cdot|$ isn't defined since $X$ may not have a norm. To fix both issues, in your proof write $d(x,y)$ whenever you wrote $|x-y|$ (where $d$ is the metric of $X$). Since every normed vector space is a metric space, this proof would be more general and still work for the cases previously mentioned.

If we wanna get more abstract, letting $X$ be any first countable space, makes the proof no longer work. There is no distance and therefore you can't impose that $a_n$ be at a distance of at most $1/n$ of $\ell_n$ (I imagine the proposition is till true though, haven't tried).

In regards of $X$ being complete or not, it doesn't matter. Doesn't play a role in your proof. Your proof stands true for the spaces already mentioned (metric, if you fix the detail) irregardless of $X$ being complete or not.

Answer 2

This is true in any metric space.

One case you didn't cover was the empty set. Since the empty set has no limit points, its set of limit points is the empty set and the empty set is closed.

Suppose $A\ne\emptyset$. Let $L$ be the set of limit points of $A$. Then $A\subset L$ so $L$ is not empty. To show that $L$ is closed we show that it contains its limit points.

Let $p$ be a limit point of $L$. Then for any $\epsilon\gt 0$ we know that the open ball around $p$, $B_{\epsilon}(p)$ contains an element $l$ of $L$. Since the open ball is open, we can find an open ball $B_{\delta}(l)$ such that $B_{\delta}(l)\subset B_{\epsilon}(p)$.

Since $l$ is a limit point of $A$, there exists $a\in A$ such that $a\in B_{\delta}(l)\subset B_{\epsilon}(p)$ so $a\in B_{\epsilon}(p)$. Since this is true for all $\epsilon$, $p$ is also a limit point of $A$ so $p\in L$.

Since this is true for all limit points $p$ of $L$, $L$ contains its limit points and is therefore closed.