Compact Subspace of Hausdorff Space is Closed/Proof 2
Theorem
Let $H = \struct {A, \tau}$ be a Hausdorff space.
Let $C$ be a compact subspace of $H$.
Then $C$ is closed in $H$.
Proof
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:
- $\ds \OO = \set {V \in \tau: \exists U \in \tau: x \in U \subseteq V^\complement}$
By Empty Intersection iff Subset of Complement:
- $U \subseteq V^\complement \iff U \cap V = \O$
Hence, by the definition of a Hausdorff space, it follows that $\OO$ is an open cover for $C$.
By the definition of a compact subspace, there exists a finite subcover $\FF$ of $\OO$ for $C$.
By the Principle of Finite Choice, there exists an $\FF$-indexed family $\family {U_V}_{V \mathop \in \FF}$ of elements of $\tau$ such that:
- $\forall V \in \FF: x \in U_V \subseteq V^\complement$
Define:
- $\ds U = \bigcap_{V \mathop \in \FF} U_V$
By General Intersection Property of Topological Space:
- $U \in \tau$
Clearly, $x \in U$.
We have that:
\(\ds U\) | \(\subseteq\) | \(\ds \bigcap_{V \mathop \in \FF} \paren {V^\complement}\) | Set Intersection Preserves Subsets | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\bigcup \FF}^\complement\) | De Morgan's laws | |||||||||||
\(\ds \) | \(\subseteq\) | \(\ds C^\complement\) | 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:
- $C^\complement \in \tau$
That is, $C$ is closed in $H$.
$\blacksquare$