# Homeomorphisms Between Compact and Hausdorff Spaces

Recall from the Preservation of Compactness under Continuous Maps page that if $X$ and $Y$ are topological spaces, $A \subseteq X$ is compact in $X$, and $f : A \to Y$ is continuous then the image $f(A)$ is compact in $Y$.

On the Closed Sets in Compact Topological Spaces page we saw that if $X$ is a compact topological space and $A \subseteq X$ is closed in $X$ then $A$ is also compact in $X$.

On the Compact Sets in Hausdorff Topological Spaces page that if $X$ is a Hausdorff topological space and $A \subseteq X$ is compact in $X$ then $A$ is also closed in $X$.

We will now use all of these results to prove a very important theorem which says that if $X$ is a compact topological space, $Y$ is a Hausdorff topological space, and $f : X \to Y$ is a continuous bijection from $X$ to $Y$, then $f$ is also a homeomorphism between $X$ and $Y$.

Theorem 1: Let $X$ and $Y$ be topological spaces and let $f : X \to Y$. If $X$ is compact, $Y$ is Hausdorff, and $f$ is a continuous bijection, then $f$ is a homeomorphism between $X$ and $Y$. |

**Proof:**Since $f : X \to Y$ is already given to be bijection and continuous, all that remains to show is that $f$ is open to prove that $f$ is a homeomorphism between $X$ and $Y$.

- Let $U$ be an open set in $X$. Then $U^c$ is closed in $X$. Since $U^c$ is closed in $X$ and $X$ is compact, we have that $U^c$ is therefore compact.

- Since $U^c$ is compact and $f$ is a continuous function we have that $f(U^c)$ is compact in $Y$.

- Since $f(U^c)$ is compact and $Y$ and $Y$ is a Hausdorff space we have that $f(U^c)$ is closed.

- However, $f(U^c) = (f(U))^c$. So $(f(U))^c$ is closed which shows that $f(U)$ is open.

- Hence $f$ is a bijective, continuous, and open map from $X$ to $Y$, so $f$ is a homeomorphism between $X$ and $Y$. $\blacksquare$