0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $U\cap V = \emptyset$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.