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For the purposes of this post, the closure $\text{cl}(A)$ of $A$ is defined as $A\cup A'$, where $A'$ is the derived set of $A$. I wish to show that $\text{cl}(A)$ is closed. Proofs of the statement are found all over the internet, for example, take Asinomás' proof here. The standard proof goes as follows:

Let $x$ be a limit point of $\text{cl}(A)$. It is sufficient to show that any neighborhood $N$ of $x$ contains a point in $A$, so that $x$ is also a limit point of $A$ and thus $x\in A'\subseteq \text{cl}(A)$. Since $x$ is a limit point of $\text{cl}(A)=A\cup A'$, we must have either

  1. $N$ contains a point $a\in A$, in which case we are done, or

  2. $N$ contains a point $a'\in A'$, in which case it also contains a point $a\in A$, and we are done.

In metric spaces, the statement in bold is easy to prove. In general topological spaces, I fail to see why it holds, and I was wondering if someone could explain it to me.

Sam
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  • Well, it all revolves around the definition of $A'.$ But the proof is pretty much the same, for some definition of neighborhood. – Thomas Andrews Aug 09 '22 at 00:39
  • @ThomasAndrews I still do not see it. In case 2), there is a neighborhood $N'$ of $a'$ that contains a point $a\in A$, but what guarantees that $a$ is contained in $N$? Why is it assumed we can force $N'$ to be a subset of $N$? – Sam Aug 09 '22 at 00:41
  • Since $a'\in N,$ $N$ is an open set (aka neighborhood) containing $a'.$ So... – Thomas Andrews Aug 09 '22 at 00:51
  • How do you define "neighborhood" in a general topological space? – Thomas Andrews Aug 09 '22 at 00:52
  • @ThomasAndrews As in https://en.wikipedia.org/wiki/Topological_space#Definition_via_neighbourhoods. Once they are defined as in the link (and open sets are defined) we may say $N$ is a neighborhood of $x$ if and only if there is an open set $U$ such that $x\in U\subseteq N$. In a nutshell, neighborhoods need not be open. – Sam Aug 09 '22 at 00:55
  • But I may use $U$, since $U$ is a neighborhood of all its points (including $a'$), and thus contains a point $a\in A$ (due to $a'$ being a limit point of $A$). I see it now, thanks. – Sam Aug 09 '22 at 01:12

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