# nLab map from compact space to Hausdorff space

### Context

#### Topology

topology

algebraic topology

# Contents

## Properties

There is a classic theorem about maps from compact spaces to Hausdorff spaces which states:

###### Theorem

Let $C$ be a compact space and $H$ be a Hausdorff space. Let $f\colon C \to H$ be a continuous bijection. Then $f$ is a homeomorphism.

An immediate corollary is that the two spaces involved are both compact Hausdorff spaces.

The idea that this captures 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.

The above theorem is a special case of a slightly more general result.

###### Theorem

Let $C$ be a compact space and $H$ be a Hausdorff space. Let $f\colon C \to H$ be a continuous map. Then $f$ is closed and proper.

###### Proof

To prove this, we need to show that a closed subset of $C$ is taken to a closed subset of $H$ and that the preimage of a compact subset of $H$ is compact in $C$. Both follow from the fact that closed subsets of a compact set are compact and that compact subsets of a Hausdorff space are closed.

For the first, let $D \subseteq C$ be closed. As it is a closed subset of a compact space, it is compact. Since $f$ is continuous, $f(D)$ is a compact subset of $H$. Thus as $H$ is a Hausdorff space, $f(D)$ is closed.

For the second, let $G \subseteq H$ be compact. As it is a compact subset of a Hausdorff space, it is closed. Thus as $f$ is continuous, $f^{-1}(G)$ is closed in $C$. Hence as $C$ is compact, $f^{-1}(G)$ is compact.

Revised on April 26, 2016 03:19:01 by Urs Schreiber (79.204.25.229)