4

I had a question on what exactly I need to show in proving that the interior of the interior of a set is equal to the interior of a set.

I'm given a set $A\subset X$, where $X$ is a topological space, and I want to show $int(A) = int(int(A))$, where $int(A)$ is defined as the union of all open sets contained in $A$. Is the following proof valid?

Let $\mathcal{O}$ be the set of all open sets $O\subset A$ and $\mathcal{G}$ be the set of all open sets $G\subset int(A)$. We will show that if $S\in \mathcal{O} \implies S\in \mathcal{G}$. Suppose $S$ is open, $S\subset A$. Then $\forall x\in S, \exists \epsilon > 0$ s/t $N_{\epsilon}(x)\subset S$. By definition of $int(A)$, $int(A) = \cup_{O\in \mathcal{O}} O$, so in particular $S\subset int(A)$. But then $N_{\epsilon}(x)\subset S$ still holds so $S$ is open in $int(A)$, and therefore if $O\in \mathcal{O} \implies O\in \mathcal{G}$. Then the union of all $O$ in $\mathcal{O}$ is in the union of all $G\in \mathcal{G}$, so that $int(A) \subset int(int(A))$

Going the other way, if $S\subset int(A)$ and $S$ is open, then $\forall x\in S, \exists \epsilon > 0$ s/t $N_{\epsilon}(x)\subset S$. Since $int(A) \subset A$, $S\subset A$ and so $N_{\epsilon}(x)\subset S$ still holds, so $S$ is open in $A$, then $G\in \mathcal{G} \implies G\in \mathcal{O}$, so the union of all $G\in \mathcal{G}$ is in the union of all $O\in \mathcal{O}$, so $int(int(A))\subset int(A)$, and each is a subset of the other so they are equal.

Any help would be great.

Thanks!

rohan
  • 41

3 Answers3

3

We know $int(int(A)) \subseteq int(A)$, and both are open.
$int(int(A))$ contains every open set $O \subseteq int(A)$, and obviously $int(A) \subseteq int(A)$, so: $$int(A) \subseteq int(int(A))$$ Conclusion: $$int(int(A)) = int(A)$$

user76568
  • 4,542
3
  • For every open (${}^1$) set $O$, $\mathrm{int}(O)=O$
  • By definition, $\mathrm{int}(S)$ is open
  • Apply the first line to $O=\mathrm{int}(S)$

Note (${}^1$): This property is even a characterization of being open.

Did
  • 279,727
  • Your "by definition" clause could, instead, be a theorem (depending on how int is defined). – GEdgar Aug 23 '14 at 20:06
  • 1
    @GEdgar The OP provides a definition of the interior as the union of every open subset, which implies trivially that the interior is indeed open. Hence, no, this is not a theorem in the framework delimited by the OP. – Did Aug 23 '14 at 20:34
1

The definition of interior in general implies that it is the unique largest open set (in terms of set inclusion) included in the set. Now, $T\equiv\operatorname{int} S$ is open and $\operatorname{int} T$ is the largest open set contained in $T$. But $T$ itself is open, so we cannot have a strictly larger open set still included in $T$! It follows that $\operatorname{int} T=T$.

triple_sec
  • 23,377