I was asked the following question about a set $X$ being Hausdorff.

If $X=A\cup B$ where $A,B$ are closed and Hausdorff (as subspaces). Show that $X$ is Hausdorff.

I was able to prove it directly using sets but my first thought was using adjunction spaces. I had the following problem earlier which I solved which said:

If $X=C\cup D$ where $C,D$ are closed and cover $X$, we form the adjunction space along their intersection (the inclusion map), then we get a homeomorphism between $(C\amalg D/\sim)=C\cup_{C\cap D}D\cong X$.

So for my original problem to show that $X=A\cup B$. We know that $A\amalg B$ is Hausdorff so I want to show that $(A\amalg B)/\sim$ is Hausdorff. However, I have read that it is not always true that a quotient space is Hausdorff. One question I have is when will a quotient space be Hausdorff and another question is other neat ways to solve my original question.


Let $x, y \in A \cup B$. If both $x,y \in A$, then they have disjoint neighborhoods because $A$ is Hausdorff. Similarly if $x,y \in B$. If neither of these cases apply, then $A \setminus B$ and $B \setminus A$ are disjoint neighborhoods of the points.

Your more general question, when will a quotient space $X/\sim$ be Hausdorff, is trickier. A necessary condition is for the set $\sim\, \subset X^2$ defining the relation to be closed, which in particular implies that each equivalence class is closed. If the quotient map is open, then that is also sufficient (see here). It is also sufficient if $X$ is compact Hausdorff (see here). Quotients of Hausdorff spaces by actions of compact groups are always Hausdorff (as quotients by group actions are always open).


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.