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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.

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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!

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    $\begingroup$ 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. $\endgroup$ Commented Dec 19, 2022 at 14:48
  • $\begingroup$ @user103178981 You're very welcome, and it happens to all of us. $\endgroup$
    – FShrike
    Commented Dec 19, 2022 at 15:21

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