I'm just starting to learn some basic topology, and I've mostly encountered definitions. For instance, I read that for a topological space $X$, for any $E\subseteq X$, define its closure $\overline{E}$ as the set of points $p\in X$ such that each neighborhood $N$ of $p$ has a nonempty intersection with $E$.

I wanted to verify that this is indeed a closure operation, and here's what I have:

(1) Suppose $p\in E$, then for any $N$, $p\in N$, so $N\cap E\neq\emptyset$, so $p\in\overline{E}$, and thus $E\subseteq\overline{E}$.

(2) Suppose $p\in\overline{E\cup F}$. Then for any $N$, $N\cap(E\cup F)\neq\emptyset$, so $(N\cap E)\cup(N\cap F)\neq\emptyset$, and thus either $N\cap E\neq\emptyset$ or $N\cap F\neq\emptyset$, so $p\in\overline{E}$ or $p\in\overline{F}$, and thus $\overline{E\cup F}\subseteq\overline{E}\cup\overline{F}$.

(3) For any $p\in X$, the intersection of any $N$ and $\emptyset$ is empty, so there are no points such that every neighborhood has nonempty intersection with $\emptyset$. Hence $\overline{\emptyset}=\emptyset$.

(4) I'm stuck showing that $\overline{\overline{E}}=\overline{E}$. From (1), I know that $\overline{E}\subseteq\overline{\overline{E}}$, but I can't show the other containment. I took $p\in\overline{\overline{E}}$, and so every $N\cap\overline{E}\neq\emptyset$. I want to show $p\in N\cap\overline{E}$, and I think this would be easy if $X$ contains isolated points. But then I found out there are such things as perfect sets or dense-in-itself sets, so that can't work. Is there some trick to show this containment?

Thanks, I know this is probably very simple, but I'm going a little mad staring at it.


From $N\cap\overline{E}\neq\emptyset$ you know that $N$ is a neighborhood of some element of $\overline{E}$, and thus $N\cap E\neq\emptyset$ by your definition of $\overline{E}$.

  • $\begingroup$ Thank you, I knew it had to be staring me in the face. $\endgroup$ – yunone Oct 27 '10 at 5:55
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    $\begingroup$ You're welcome. Thanks for the well thought out question. $\endgroup$ – Jonas Meyer Oct 27 '10 at 5:57
  • $\begingroup$ Sorry to bring this up again, but I was reviewing this, and can't recall why from $N\cap\bar{E}\neq\emptyset$ it follows that $N$ is a neighborhood of some element of $\bar{E}$. Is there some reason why for all the points in $N\cap\bar{E}$, at least one will have a neighborhood that coincides with this neighborhood $N$ of $p$? $\endgroup$ – yunone Nov 13 '10 at 6:28
  • $\begingroup$ Not a problem. Note that an open set is a neighborhood of each of its points (and I'm assuming $N$ is taken to be open, which I mention because there are other definitions of neighborhood). Thus if $N\cap\overline{E}$ contains an element $x$, then because $x$ is in $N$ it has $N$ as a neighborhood. But $x$ is also in $\overline{E}$. (So you can replace "at least one" with "all".) Does that clear it up? $\endgroup$ – Jonas Meyer Nov 13 '10 at 8:19
  • $\begingroup$ Yes, thanks for jogging my memory! $\endgroup$ – yunone Nov 15 '10 at 9:09

Your proof of (2) is incorrect. The quantifiers are wrong. You have shown that for any $N$ (neighbourhood of $x$, in the remainder): $N$ intersects $E$ or $N$ intersects $F$. But you have to show (for any $N$: $N$ intersects $E$) or (for any $N$, $N$ intersects $F$), i.e. that you intersect the same set for any $N$. So you have to argue by contradiction: suppose $x \in \overline{E \cup F}$, but $x \notin \overline{E}$ and $x \notin \overline{F}$. The latter 2 imply that there is a neighbourhood $N_1$ of $x$ that misses $E$ and a neighbourhood $N_2$ of $x$ that misses $F$. But then $N_1 \cap N_2$ is a neighbourhood of $x$ as well and this misses $E \cup F$, contradicting that $x \in \overline{E \cup F}$. Done.

  • $\begingroup$ Whoops, thanks for pointing out that error! I'll keep that in mind from now on. $\endgroup$ – yunone Dec 6 '10 at 3:42

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