# Basic properties of interior, closure and boundary in Bredon

I am self reading Bredons topology and geometry and want to prove the following exercises from the book. However, I am not sure whether what I have done is correct and I also don't know how to show the disjointness in $$(3)$$. Any comment is greatly appreciated!

Let $$X$$ be a space and $$A$$ be a subset of $$X$$. Then the closure of $$A$$ in $$X$$ is defined as the smallest closed set that contains $$A$$, that is, the intersection of all closed sets that contain $$A$$. The interior is defined analogously. Furthermore the boundary of $$A$$ is defined to be $$\overline{A} \cap \overline {X \setminus A}$$.

$$(1)$$ Let $$X$$ be a space and $$A,B \subseteq X$$. Then $$\overline{A}=\{x \in X \ | \ \text{for all open } U \ \text{with} \ x \in U \ \text{we have} \ U \cap A \neq \emptyset\}.$$

$$(2)$$ $$X \setminus \mathrm{int}(A)=\overline {X\setminus A}$$ and $$X \setminus \overline A = \mathrm{int}(X \setminus A).$$

$$(3)$$ Prove that $$X=\mathrm{int}(A) \cup \partial A \cup X \setminus \overline A$$ where the union is disjoint.

Proof. $$(1)$$ Let $$x \in \overline{A}$$ and suppose there is some open $$U$$ with $$x \in U$$ and $$U \cap A= \emptyset$$. Then $$X \setminus U$$ is closed and $$A \subseteq X \setminus U$$, since otherwise there would be some $$y \in A$$ with $$y \not\in X \setminus U$$. By definition of the closure and since $$A \subseteq X \setminus U$$ we get $$x \in X \setminus U$$ which is a contradiction.

Conversly, let $$x$$ be in the set on the RHS and $$C \supseteq A$$ be closed. If $$C$$ does not contain $$x$$, then $$x \in X \setminus C$$ which is open and thus gives $$(X \setminus C) \cap A \neq \emptyset$$ which is a contradiction since $$A \subseteq C$$.

$$(2)$$ We have $$X \setminus \mathrm{int}(A) = X \setminus \bigcup_{U \in \mathcal{O}, U \subseteq A} U= \bigcap_{U \in \mathcal{O}, U \subseteq A} X \setminus U = \bigcap_{X \setminus A \subseteq C \ \mathrm{closed}} C$$ which proves the first claim.

For the second claim we have $$X \setminus \overline{A}= X \setminus \bigcap_{A \subseteq C \text{closed}} C = \bigcup_{U \subseteq X \setminus A \ \mathrm{open}} U.$$

$$(3)$$ If we show that $$\overline A= \mathrm{int}(A) \cup \partial A$$ is a disjoint union, we have that this is the disjoint from $$X \setminus \overline A$$ and thus we are done. It holds that $$\mathrm{int}(A) \cup \partial A= \mathrm{int}(A) \cup (\overline A \cap \overline{X \setminus A})= (\mathrm{int}(A) \cup \overline A) \cap ( \mathrm{int}(A) \cup \overline {X \setminus A})=\overline A \cap X = \overline A.$$ However, I don't know how to show that this is indeed disjoint.

Your proofs for $$(1)$$ and $$(2)$$ look good to me, except you should have written: $$X\setminus\overline{A}=X\setminus\bigcap_{A\subseteq C\text{ closed }}C=\bigcup_{U\subseteq X\setminus A\text{ open }}U$$
For $$(3)$$, I agree with your proof that $$\mathrm{int}(A)\cup\partial A=\overline{A}$$. To see that this is a disjoint union, consider that $$\mathrm{int}(A)\cap\partial A=\emptyset$$ iff. $$\partial A\subseteq X\setminus\mathrm{int}(A)=\overline{X\setminus A}$$. But, $$\partial A=\overline{A}\cap\overline{X\setminus A}\subseteq\overline{X\setminus A}$$ by definition!
• Thanks. No clue what I did in $(2)$ there, but you're right, I will edit the question. Of course your proof of the disjointness makes sense, no clue how I did not see that! Again, thanks a lot. Commented Dec 19, 2022 at 14:48