Proof of $\textrm{int}(S)=X-\overline{X-S}$

Question

I'm looking to prove this and I think I have it done one way, but I don't understand the reverse case for $\textrm{int}(S)\supseteq X-\overline{X-S}$. Also, I'm not sure if this proof is valid so far.

Consider a point $x$ in the interior of $S$. There must then exist some ball about $x$ of radius $r>0$ that is completely contained within $\textrm{int}(S)$. This is not true of any points on the boundary of $S$ as balls about those points by definition contain points both inside and outside the set. Therefore, $x\not\in\partial S=\bar S\cap \overline{X-S}$. It is obvious that $x\not\in X-S$, so $x\not\in\overline{X-S}$. Therefore, because this is the only restriction on $x$, $x$ must be in the complement of this set, so $x\in X-\overline{X-S}$. Therefore, $\textrm{int}(S)\subseteq X-\overline{X-S}$.

Answer 1

If $\tau$ denotes the topology then: $$\mathsf{int}(S)=\cup\{U\in\tau\mid U\subseteq S\}\tag1$$

Now note that $X-S\subseteq\overline{X-S}$ so that $X-\overline{X-S}\subseteq X-(X-S)=S$.

Moreover as a complement of a closed set $X-\overline{X-S}$ is open so that:$$X-\overline{X-S}\in\{U\in\tau\mid U\subseteq S\}$$ Then based on $(1)$ we conclude: $$X-\overline{X-S}\subseteq\mathsf{int}(S)$$

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Edit (to show that $(1)$ is also valid in metric spaces.)

In a metric space $x\in\mathsf{int}(S)$ if and only some $r>0$ exists such that the open ball $B(x,r)$ is a subset of $S$. Based on that it can be proved that $(1)$ is valid:

From this definition it follows directly that $\mathsf{int}(S)\subseteq\cup\{U\in\tau\mid U\subseteq S\}$.

If conversely $x\in U\subseteq S$ for some open set $U$ then some $r>0$ exists such that $x\in B(x,r)\subseteq U\subseteq S$ so that $x\in\mathsf{int}(S)$.

Answer 2

Your proof attempt is not convincing to me. It can be done directly:

$x$ is an interior point of $A$ iff there is a neighbourhood $O$ of $x$ such that $O \subseteq A$. (in a metric space you can read ball $B(x,r)$ for some $r>0$ instead of "neighbourhood of $x$" if you prefer).

$x$ is a point of $\overline{A}$ iff for all neighbourhoods $O$ of $x$ (or balls around $x$ etc.) we have that $O \cap A \neq \emptyset$.

So $x \in \operatorname{int}(S)$ means that we have at least one neighbourhood $O$ of $x$ inside of $S$. This means that $O$ does not intersect $X-S$, and this witnesses that by definition $x \notin \overline{X-S}$ (because to be in that set every neighbourhood of $x$ would have to intersect $X-S$ and we have one that does not). So $x \in X-\overline{X-S}$.

Note that the reverse inlcusion is proved exactly the same way using again that $x \notin \overline{X-S}$ or $x \in X - \overline{X-S}$ means that (where $\mathcal{N}_x$ is the set of neighbourhoods of $x$)

$$\lnot(\forall O \in \mathcal{N}_x: O \cap (X-S) \neq \emptyset)$$

which logically is equivalent to

$$\exists O \in \mathcal{N}_x: O \cap (X-S) = \emptyset$$

which means

$$\exists O \in \mathcal{N}_x: O \subseteq S$$

which means $x \in \operatorname{int}(S)$. Hence $\operatorname{int}(S) = X-\overline{X-S}$ as required.