The set is closed (resp. open) iff the complement set is open (resp. closed)

Question

There's a theorem in my small danish course book. Let $(M,d)$ be a metric space.

Theorem: The concepts of open and closed are dual: A set $A\subseteq M$ is closed (resp. open) if and only if the complement set $\complement A$ is open (resp. closed).

Proof: The formula $\complement \overline{A}=(\complement A)^{\circ}$ shows that $A=\overline{A}$ only if $(\complement A)^{\circ}=\complement A$, that is $A$ is closed, only if $\complement A$ is open. Using this on $\complement A$ insted of $A$, we get that $A$ is open only if $\complement A$ is closed.

I don't think the proof is useful. Here's what I want to prove; \begin{align*} \overline{A}=A\iff (\complement A)^{\circ}=\complement A\tag{1}\\ A^{\circ}=A\iff \overline{\complement A}=\complement A\tag{2}. \end{align*}

Note this course book has some of few formulas without proofs added.

Case $(1)$. $\implies:$ Assume that $\overline{A}=A$. We'll use the formula

$\complement \overline{A}=(\complement A)^{\circ}$.

Since $\complement \overline{A}= \complement A$, we have $\complement A=(\complement A)^{\circ}$.

$\impliedby:$ Assume that $\complement A=(\complement A)^{\circ}$. We'll use the formula

$\overline{A}=\complement((\complement A)^{\circ})$.

We have $\overline{A}=\complement((\complement A)^{\circ})=\complement(\complement A)=A$.

Case $(2)$. $\implies:$ Assume that $A^{\circ}=A$. We will use the formula

$\overline{A}=A^{\circ}\cup \partial A$.

We have $$\overline{\complement A}=(\complement A)^{\circ}\cup \partial(\complement A)=(\complement A)^{\circ}\cup \partial A=M\setminus A^{\circ}=M\setminus A=\complement A.$$

$\impliedby:$ This one I need help with.

What do you think about my proof so far? I know that there are other proofs available in some websites but I would like to write it differently.

Answer 1

Case 2, $\Longleftarrow$

\begin{align*} A^{\circ}=[\complement(\complement A)]^{\circ}=[\complement(\overline{\complement A})]=[\complement(\complement A)]=A. \end{align*} The first equality uses the simple fact that $A=\complement(\complement A)$. The second equality uses the formula you had already exploited in Case 1, $\Longrightarrow$, with $A$ on the right-hand side of that formula being replaced with $\complement A$. The third equality uses the assumption that $\overline{\complement A}=\complement A$. The fourth equality follows again from $A=\complement(\complement A)$.

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As for Case 2, $\Longrightarrow$, I don't think you even need the formula

$\overline{A}=A^{\circ}\cup \partial A$.

Just use

$\overline{A}=\complement[(\complement A)^{\circ}]$

as before, with $A$ on the left-hand side replaced with $\complement A$:

$$\overline{\complement A}=\complement[(\complement[\complement A])^{\circ}]=\complement(A^{\circ})=\complement A.$$

The rest of your proof seems fine to me.