How to prove a quotient space is again compact and Hausdorff

Question

I'm struggling with a topology question and I'm having a hard time proving the following: If I know that $X$ is a compact Hausdorff space, $A$ is a closed subspace and $X/A$ denotes the quotient space of $X$ which identifies $A$ to a single point.

Then I'd like to show, that $X/A$ is a compact Hausdorff space.

So far I have that since $A$ is a closed subspace of a compact space $X$ then $A$ is compact as well.

Since $A$ is a subspace of a Hausdorff space then $A$ is a Hausdorff space as well.

Since $X/A$ is the quotient space of $X$, which identifies $A$ to a single point, then by the definition of a quotient space (Munkres p. 137) then $X/A$ is a partition of $X$ into disjoint subsets whose union is $X$. Also there exist a surjective map $p:X \to X/A$ that carries each point of $X$ to the the element of $X/A$ containing it. Since $p$ is a quotient map this is equivalent to saying that $p$ is continuous.

And now I can't seem to get any further. How do I prove that $X/A$ is compact and Hausdorff?

Answer 1

Let the quotient map be $q:X \rightarrow X/A$. $X/A$ is compact because if $\{U_\alpha\}$ is any open cover of $X/A$ then $\{q^{-1}(U_\alpha)\}$ is an open cover of $X$. Since $X$ is compact then there is a finite subcover $\{q^{-1}(U_{\alpha_i})| i = 1,...,n\}$. But then $\{U_{\alpha_i}|i=1,..,n\}$ is a finite subcover of $X/A$ because $q(q^{-1}(U_\alpha)) = U_\alpha$. Therefore, $X/A$ is compact.

To show that $X/A$ is Hausdorff we pick two points $x,y\in X/A$. If $x,y\neq q(A)$ then pick open subsets $U, V \subset X$ such that $q^{-1}(x) \in U$, $q^{-1}(y) \in V, U\cap V = \emptyset, U\cap A = \emptyset, V\cap A = \emptyset$. Then $x\in q(U), y\in q(V)$ and both $q(U)$ and $q(V)$ are open in $X/A$ such that $q(U) \cap q(V) = \emptyset$. If $x \neq y = q(A)$ then you can use the fact that $A$ is compact and Hausdorff to show that there is are open subsets $U, V \subset X$ such that $q^{-1}(x)\in U, q^{-1}(y)\in A\subset V$ and $U\cap V = \emptyset$.

Answer 2

Notice that $X$ is normal.
Assume distinct $x',y'$ in the quotient space.

• Case $x',y' \neq a$, the point to which $A$ identifies:
  $x,$y the points in $X$ that quotient map maps to $x',y'$ can be separated by open sets. Use those sets to separated $x',y'$ by open sets. Note $x,y$ are not in $A$.

• Case $x' \neq a, y' = a$:
  the quotient map maps $x$ to $x'$, $A$ to $y'$. Since $x\notin A$, $x$ and $A$ can be separated by open sets. As before use those sets to separate $x',y'$ by open sets.