Show that a set $S$ is closed if and only if$S=\operatorname{int}(S)\cup \operatorname{Boundary}(S)$.
I´m using the following definitions:
- A set is closed if it is the complement of an open set.
- $O$ is open if every point of $O$ has a neighborhood contained in $O$.
- $\operatorname{int}(S)$ is the set of all interior points of $S$. An interior point $x$ of $S$ is such that $x$ has a neighborhood that is contained in $S$.
- $\operatorname{Boundary}(S)$ is the set of all boundary points of $S$. A boundary point $x$ of $S$ is such that every neighborhood of $x$ is both in $S$ and its complement.
I already proved that if $S=\operatorname{int}(S)\cup \operatorname{Boundary}(S)$ then $S$ is closed.
How would you prove the other direction?
The containments $int(S)⊂S⊂int(S)∪Boundary(S)$ is easy, but how do you prove the other one?
Best Answer
What you have left to show is that, if $S$ is closed, then $\mathrm{bd}(S) \subset S$. We assume $S$ is closed, so $\forall y \in S^C$, there exists open set $O$ containing $y$ such that $O \subset S^C$.
Now, consider arbitrary $x \in \mathrm{bd}(S)$. By the definition of the boundary, this means that for any open set $N$ containing $x$, $N$ must contain points in $S$. It follows that no neighborhood $N$ of $x$ satisfies $N \subset S^C$. Because $S^C$ is open, this means that $x \not\in S^C$, and therefore $x \in S$ (otherwise, if $x \in S^C$ and has no neighborhood contained in $S^C$, we contradict the assumption that $S$ is closed).
Therefore we have that, if $S$ is closed, then $x \in \mathrm{bd}(S)$ implies that $x \in S$ (i.e. $\mathrm{bd}(S) \subset S$).
Now knowing that, if $S$ is closed, then both $\mathrm{int}(S)$ and $\mathrm{bd}(S)$ are subsets of $S$, we can deduct that $S$ closed implies that $\mathrm{int}(S) \cup \mathrm{bd}(S) \subset S$.