consequences of Kuratowski operators

Question

*Let $(X,\tau)$ a topological space and $A\subseteq X$, show that $A\in \mathcal{C}_{\tau} $ , if and only if $ext(A)=X-A$ *

$\Rightarrow)$Suppose that $A\in \mathcal{C}_{\tau}$, by definition $A^{c}\in \tau$, then $int(A^{c})=A^{c}$ , I also have to $ext(A)=int(A^{c})$, hence $ext(A)=X-A$.

$\Leftarrow)$ Conversly, to show that $A$ is closed, we'll prove that $A^{c}$ is open. Since $ext(A)=X-A$ by assumption, and $ext(A)=int(A^{c})$, thus implies $int(A^{c})=A^{c}$, that is to say $A^{c}$ is open.

Hence $A$ is closed.

Note: I have already shown that $ext(A)=int(A^{c})$

I would like to know if the demo is ok, and it is well written.

Answer 1

The proof is correct, although you could take fewer lines by writing it as follows:

Proof. Let $A \subseteq X$. Then \begin{align} A \in \mathcal C_\tau & \quad \textrm{if and only if} \quad X \setminus A \in \tau \\ & \quad \textrm{if and only if} \quad \operatorname{int}(X \setminus A) = X \setminus A \\ & \quad \textrm{if and only if} \quad \operatorname{ext}(A) = X \setminus A. \end{align}