If $A$ and $X/A$ are compact and $A$ closed, then $X$ is compact.

Question

If $(X,\mathscr{T})$ is a compact topological space and $A\subset X$ is closed then we know that $A$ and $X/A$ are both compact. My question is, does the converse also hold? If $A$ and $X/A$ are compact and $A$ closed, does it imply that $X$ is compact? I believe the answer is yes. If $\mathscr{C}$ is an open cover of $X$, then we only have to show that there is a finite subset of $\mathscr{C}$ whose elements form a union which contains $X/A$. But I am not sure how to do this

Answer 1

Yes, $X$ is compact.

Let $\mathscr{U}$ be an open cover of $X$. $A$ is compact,so there is a finite $\mathscr{U}_0\subseteq\mathscr{U}$ that covers $A$. Let $U_0=\bigcup\mathscr{U}_0$, and let

$$\mathscr{V}=\{U_0\}\cup\{U\setminus A:U\in\mathscr{U}\setminus\mathscr{U}_0\}\,.$$

The image of $\mathscr{V}$ under the quotient map is an open cover of $X/A$, so it has a finite subcover, which pulls back to yield a finite $\mathscr{V}_0\subseteq\mathscr{V}$ covering $X$. $\mathscr{U}_0\cup(\mathscr{V}_0\setminus\{U_0\})$ is then a finite open refinement of $\mathscr{U}$, which therefore has a finite subcover.