Let $(X, \tau)$ be a topological space and $E_1,E_2 \subseteq X$. Prove that $$\overline{\overline{E_1 \setminus E_2} \setminus E_2} = \overline{E_1 \setminus E_2}.$$ Note: $\overline E$ represents the closure of $E$.

I can intuitively understand why the equality stands, but I'm at a loss when I have to write down a formal proof. The usual way of showing equality in these cases does not seem to help much. Usually I would show that $A = B$ iff $A \subseteq B$ and $A \supseteq B$.

EDIT: I have made progress. I'm using the fact that $$x \in \overline E \iff \forall U \in \mathcal U_x, U \cap E \neq \varnothing$$ where $\mathcal U_x$ is the set of all open neighborhoods of $x$. So we have that $$x \in \overline{E_1 \setminus E_2} \iff \forall U \in \mathcal U_x, U \cap (E_1 \setminus E_2) \neq \varnothing \tag 1$$ and similarly $$x \in \overline{\overline{E_1 \setminus E_2} \setminus E_2} \iff \forall U \in \mathcal U_x, U \cap (\overline{E_1 \setminus E_2} \setminus E_2) \neq \varnothing \tag 2.$$

From $(2)$, it follows that $U \cap \overline{E_1 \setminus E_2} \neq \varnothing$, and so (?) $U \cap E_1 \setminus E_2 \neq \varnothing$, which is the same as $(1)$. I'm not really sure about the highlighted passage. It seems intuitively true but I should prove it as well. Does this look right?

  • $\begingroup$ You might start by thinking about what points belong to $\overline{E_1\setminus E_2}$ but not to $\overline{E_1\setminus E_2}\setminus E_2$. $\endgroup$ – almagest Apr 23 '16 at 8:23
  • $\begingroup$ @almagest Thanks for the hint. I've added my attempt in the question. Can you please briefly check it? Thanks! $\endgroup$ – rubik Apr 23 '16 at 16:00

OK, so indeed you can use the criterion as stated.

So pick $x \in \overline{\overline{E_1 \setminus E_2} \setminus E_2}$, and we have to show it is in $\overline{E_1 \setminus E_2}$.

So pick $O$ any open set that contains $x$. Then $O$ must intersect $\overline{E_1 \setminus E_2} \setminus E_2$, say we have $y \in O$ and $y \in \overline{E_1 \setminus E_2}$ and $y \notin E_2$. As $O$ is a neighbourhood of $y$ we apply the criterion again to see that $O \cap (E_1 \setminus E_2) \neq \emptyset$.

As $O$ was an arbitrary open neighbourhood of $x$, $x \in \overline{E_1 \setminus E_2}$.

For the reverse, pick $x \in \overline{E_1 \setminus E_2}$, and let $O$ be any open neighbourhood of $x$. Then we have that there is $y \in O$ with $y \in E_1 \setminus E_2 (\subseteq \overline{E_1 \setminus E_2})$, so $y \notin E_2$ in particular as well. So $y$ is in the intersection of $O$ and $\overline{E_1 \setminus E_2} \setminus E_2$.

As $O$ was arbitrary again, we have that $x \in \overline{\overline{E_1 \setminus E_2} \setminus E_2}$, as required.

  • $\begingroup$ Perfect, your proof flows well and I understand it. I knew I was almost there! Thanks a lot! $\endgroup$ – rubik Apr 23 '16 at 19:10
  • $\begingroup$ @rubik Thanks. I made a minor mistake and made an edit, which should make it a bit clearer. Glad to be of assistance. $\endgroup$ – Henno Brandsma Apr 23 '16 at 21:00
  • $\begingroup$ Oh I see. I don't think it was a mistake before, this is just clearer. $\endgroup$ – rubik Apr 24 '16 at 7:33

More simply: $$\overline {\overline {E_1\backslash E_2}\backslash E_2}\supset \overline {(E_1\backslash E_2)\backslash E_2}=\overline {E_1\backslash E_2}= \overline {\overline {E_1\backslash E_2}}\supset \overline {\overline {E_1\backslash E_2}\backslash E_2}.$$

  • $\begingroup$ Unfortunately I don't understand this very well. I see you used the fact that $\overline{\overline{A} \setminus B} \supseteq \overline{A \setminus B}$. But what about the last inclusion? $\endgroup$ – rubik Apr 23 '16 at 19:15

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