Contents

# Contents

## Statement

###### Lemma

(compact subspaces in Hausdorff spaces are separated by neighbourhoods from points)

Let

1. $(X,\tau)$ be a Hausdorff topological space;

2. $Y \subset X$ a compact subspace.

Then for every $x \in X \backslash Y$ there exists

1. an open neighbourhood $U_x \supset \{x\}$;

2. an open neighbourhood $U_Y \supset Y$

such that

• they are still disjoint: $U_x \cap U_Y = \emptyset$.

As a direct consequence compact subspaces of Hausdorff spaces are closed.

###### Proof

By the assumption that $(X,\tau)$ is Hausdorff, we find for every point $y \in Y$ disjoint open neighbourhoods $U_{x,y} \supset \{x\}$ and $U_y \supset \{y\}$. By the nature of the subspace topology of $Y$, the restriction of all the $U_y$ to $Y$ is an open cover of $Y$:

$\left\{ (U_y \cap Y) \subset Y \right\}_{y \in Y} \,.$

Now by the assumption that $Y$ is compact, there exists a finite subcover, hence a finite set $S \subset Y$ such that

$\left\{ (U_y \cap Y) \subset Y \right\}_{y \in S \subset Y}$

is still a cover.

But the finite intersection

$U_x \coloneqq \underset{s \in S \subset Y}{\cap} U_{x,s}$

of the corresponding open neighbourhoods of $x$ is still open, and by construction it is disjoint from all the $U_{s}$, hence also from their union

$U_Y \coloneqq \underset{s \in S \subset Y}{\cup} U_s \,.$

Therefore $U_x$ and $U_Y$ are two open subsets as required.

Created on May 19, 2017 at 10:10:29. See the history of this page for a list of all contributions to it.