# Contents

## Statement

###### Proposition

(maps from compact spaces to Hausdorff spaces are closed and proper)

Let $f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a continuous function between topological spaces such that

1. $(X,\tau_X)$ is a compact topological space;

2. $(Y,\tau_Y)$ is a Hausdorff topological space.

Then $f$ is

###### Proof

For the first statement, we need to show that if $C \subset X$ is a closed subset of $X$, then also $f(C) \subset Y$ is a closed subset of $Y$.

Now

1. since closed subsets of compact spaces are compact it follows that $C \subset C$ is also compact;

2. since continuous images of compact spaces are compact it then follows that $f(C) \subset Y$ is compact;

3. since compact subspaces of Hausdorff spaces are closed it finally follow that $f(C)$ is also closed in $Y$.

For the second statement we need to show that if $C \subset Y$ is a compact subset, then also its pre-image $f^{-1}(C)$ is compact.

Now

1. since compact subspaces of Hausdorff spaces are closed it follows that $C \subse Y$ is closed;

2. since pre-images under continuous of closed subsets are closed, also $f^{-1}(C) \subset X$ is closed;

3. since closed subsets of compact spaces are compact, it follows that $f^{-1}(C)$ is compact.

## Consequences

###### Corollary

(continuous bijections from compact spaces to Hausdorff spaces are homeomorphisms)

Let $f \colon (X, \tau_X) \longrightarrow (Y, \tau_Y)$ be a continuous function between topological spaces such that

1. $(X,\tau_X)$ is a compact topological space;

2. $(Y,\tau_Y)$ is a Hausdorff topological space.

3. $f \;\colon\; X \longrightarrow Y$ is a bijection of sets.

Then $f$ is a homeomorphism, i. e. its inverse function $Y \to X$ is also a continuous function.

In particular then both $(X,\tau_X)$ and $(Y, \tau_Y)$ are compact Hausdorff spaces.

###### Proof

Write $g \colon Y \to X$ for the inverse function of $f$.

We need to show that $g$ is continuous, hence that for $U \subset X$ an open subset, then also its pre-image $g^{-1}(U) \subset Y$ is open in $Y$. By passage to complements, this is equivalent to the statement that for $C \subset X$ a closed subset then the pre-image $g^{-1}(C) \subset Y$ is also closed in $Y$.

But since $g$ is the inverse function to $f$, its pre-images are the images of $f$. Hence the last statement above equivalently says that $f$ sends closed subsets to closed subsets. This is true by prop. .

###### Remark

The idea captured by corollary is that Hausdorffness is about having “enough” open sets whilst compactness is about having “not too many”. Thus a compact Hausdorff space has both “enough” and “not too many”. This theorem says that both conditions are at their limit: if we try to have more open sets, we lose compactness. If we try to have fewer open sets, we lose Hausdorffness.

Last revised on June 21, 2017 at 04:49:51. See the history of this page for a list of all contributions to it.