Compact Subspace of Hausdorff Space is Closed

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Let $H = \left({A, \tau}\right)$ be a Hausdorff space.

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

Then $C$ is closed in $H$.


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 $U \left({x}\right)$ and $V \left({x}\right)$ 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 $\displaystyle U_a = \bigcap_{i \mathop = 1}^r U \left({x_i}\right)$ and get $a \in U_a \subseteq A \setminus C$.

Now suppose $C$ is not finite.

The set $\left\{{V \left({x}\right): x \in C}\right\}$ is an open cover for $C$.

As $C$ is compact, it has a finite subcover, say $\left\{{V \left({x_1}\right), V \left({x_2}\right), \ldots, V \left({x_r}\right)}\right\}$.

Let $\displaystyle U_a = \bigcap_{i \mathop = 1}^r U \left({x_i}\right)$.

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

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

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

Hence $b \notin V \left({x_i}\right)$, so $b \notin C$, since $\displaystyle C = \bigcup_{i \mathop = 1}^r V \left({x_i}\right)$.

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 and so $C$ is closed.


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:

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


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

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 $H$.