Compact Subspace of Hausdorff Space is Closed

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $H = \struct {A, \tau}$ be a Hausdorff space.

Let $C$ be a compact subspace of $H$.


Then $C$ is closed in $H$.


Corollary

A finite subspace of a Hausdorff space is closed.


Proof 1

From Subspace of Hausdorff Space is Hausdorff, a subspace of a Hausdorff space is itself Hausdorff.

Let $a \in A \setminus C$.

We are going to prove that there exists an open set $U_a$ such that $a \in U_a \subseteq A \setminus C$.


For any single point $x \in C$, the Hausdorff condition ensures the existence of disjoint open set $\map U x$ and $\map V x$ containing $a$ and $x$ respectively.

Suppose there were only a finite number of points $x_1, x_2, \ldots, x_r$ in $C$.

Then we could take $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$ and get $a \in U_a \subseteq A \setminus C$.


Now suppose $C$ is not finite.

The set $\set {\map V x: x \in C}$ is an open cover for $C$.

As $C$ is compact, it has a finite subcover, say $\set {\map V {x_1}, \map V {x_2}, \dotsc, \map V {x_r} }$.

Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r \map U {x_i}$.

Then $U_a$ is open because it is a finite intersection of open sets.

Also, $a \in U_a$ because $a \in \map U {x_i}$ for each $i = 1, 2, \ldots, r$.


Finally, if $b \in U_a$ then for any $i = 1, 2, \ldots, r$ we have $b \in \map U {x_i}$.

Because $\displaystyle C \subseteq \bigcup_{i \mathop = 1}^r \map V {x_i}$:

$b \notin \map V {x_i}$, so $b \notin C$

Thus:

$U_a \subseteq A \setminus C$.


Then:

$\displaystyle A \setminus C = \bigcup_{a \mathop \in A \mathop \setminus C} U_a$

So $A \setminus C$ is open.

It follows that $C$ is closed.

$\blacksquare$


Proof 2

For a subset $S \subseteq A$, let $S^{\complement}$ denote the relative complement of $S$ in $A$.


Consider an arbitrary point $x \in C^{\complement}$.

Define the set:

$\displaystyle \mathcal O = \left\{{V \in \tau: \exists U \in \tau: x \in U \subseteq V^{\complement}}\right\}$

By Empty Intersection iff Subset of Complement, we have that:

$U \subseteq V^{\complement} \iff U \cap V = \varnothing$

Hence, by the definition of a Hausdorff space, it follows that $\mathcal O$ is an open cover for $C$.


By the definition of a compact subspace, there exists a finite subcover $\mathcal F$ of $\mathcal O$ for $C$.

By the Principle of Finite Choice, there exists an $\mathcal F$-indexed family $\left\langle{U_V}\right\rangle_{V \in \mathcal F}$ of elements of $\tau$ such that:

$\forall V \in \mathcal F: x \in U_V \subseteq V^{\complement}$

Define:

$\displaystyle U = \bigcap_{V \mathop \in \mathcal F} U_V$

By General Intersection Property of Topological Space, it follows that $U \in \tau$.


Clearly, $x \in U$.

We have that:

\(\displaystyle U\) \(\subseteq\) \(\displaystyle \bigcap_{V \mathop \in \mathcal F} \left({ V^{\complement} }\right)\) Set Intersection Preserves Subsets
\(\displaystyle \) \(=\) \(\displaystyle \left({\bigcup \mathcal F}\right)^{\complement}\) De Morgan's laws
\(\displaystyle \) \(\subseteq\) \(\displaystyle C^{\complement}\) Definition of Cover of Set and Relative Complement inverts Subsets

From Subset Relation is Transitive, we have that $U \subseteq C^{\complement}$.


Hence $C^{\complement}$ is a neighborhood of $x$.

From Set is Open iff Neighborhood of all its Points, we have that $C^{\complement} \in \tau$.

That is, $C$ is closed in $H$.

$\blacksquare$


Sources