Prove that $\partial A = \mathrm{Cl}(A) \cap \mathrm{Cl}(X − A)$

Question

My textbook doesn't give me the fact that $\partial A = \mathrm{Cl}(A) \cap \mathrm{Cl}(X − A)$. We're asked to prove it. I'm given $\partial A = \mathrm{Cl}(A) − \mathrm{Int}(A)$. In a previous question we proved (with the help of math.stackexchange.com folks) that this also equals $\mathrm{Cl}(A) \cap (X − \mathrm{Int}(A))$.

I have very little knowledge of set theory and proofs, so from there I'm not sure how to prove this. As always, I appreciate any help.

Edit: Let A be a subset of a topological space X. The interior of A, denoted Å or Int(A), is the union of all open sets contained in A. The closure of A, denoted Ā or Cl(A), is the intersection of all closed sets containing A. $\partial A = \mathrm{Cl}(A) − \mathrm{Int}(A)$

The question comes from “Introduction to Topology: Pure and Applied” by Colin Adams and Robert Franzosa.

"Let $A$ be a subset of a topological space $X$. Prove that $\partial A = \mathrm{Cl}(A) \cap \mathrm{Cl}(X − A)$."

Answer 1

Basic fact relating closures and interiors via complements, for all $A \subseteq X$: $$X\setminus \operatorname{Int}(A) = \operatorname{Cl}(X\setminus A)\tag{1}$$

Suppose $O$ is open and $O \subseteq A$. Then $X\setminus A \subseteq X\setminus O$ and the latter set is closed. The closure of a set is the smallest closed set containing it so by this minimality: $\operatorname{Cl}(X\setminus A) \subseteq X\setminus O$. Taking $O=\operatorname{Int}(A)$ then shows one inclusion of $(1)$.

OTOH let $C$ be a closed set such that $X\setminus A \subseteq C$. Then $X\setminus C$ is open and $X\setminus C \subseteq A$ and as the interior of a set is the largest open subset of that set, we get $X\setminus C \subseteq \operatorname{Int}(A)$ and $X \setminus \operatorname{Int}(A) \subseteq C$. Now taking $C=\operatorname{Cl}(X\setminus A)$ gives the other inclusion and hence equality.

Now $\partial A = \operatorname{Cl}(A)\setminus \operatorname{Int}(A) = \operatorname{Cl}(A) \cap (X\setminus \operatorname{Int}(A))$ as you already knew, and now applying (1) this equals $\operatorname{Cl}(A) \cap \operatorname{Cl}(X\setminus A)$ as required.

Answer 2

By your previous work it suffices to show that $X-\mbox{int}(A)=\mbox{Cl}(X-A)$.

Let $x\in \mbox{Cl(X-A)}$, then any open subset $U$ containing $x$ must satisfy $U\cap (X-A)\neq \emptyset$. Thus $x$ cannot be an interior point of $A$, otherwise there would exist an open subset $V$ with $x\in V$ and $V$ properly contained in $A$, i.e. $V\cap X-A=\emptyset$. So, $x$ is not in the interior of $A$ but $x\in X$, thus $x\in X-\mbox{int}(A)$. This shows that $\mbox{Cl}(X-A)\subset X-\mbox{int}(A)$.

Can you show the other inclusion, $X-\mbox{int}(A)\subset \mbox{Cl}(X-A)$? Once we have shown inclusion in both directions, we have equality and we are done.