Finite Topological Space is Compact/Proof 1

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Theorem

Let $T = \struct {S, \tau}$ be a topological space where $S$ is a finite set.

Then $T$ is compact.

Proof

From Cardinality of Empty Set, the empty set is finite.

Suppose $S$ is empty.

Then by Empty Set is Compact Space, $T$ is compact.

Suppose $S$ is non-empty.

Let $\VV$ be an open cover of $T$.

For each $x \in S$, define $\VV_x$ to be $\set {V \in \VV : x \in V}$

Since $S$ is finite, and since by definition of a cover, each $x\in S$ is contained in at least one $V$ in $\VV$, we have that $\set {\VV_x : x \in S}$ is a finite collection of nonempty sets.

From Principle of Finite Choice, there is a choice function which selects one $V_x$ from $\VV_x$ for each $x \in S$. By definition of $\VV_x$, such a $V_x$ contains $x$.

Since there were only finitely many $\VV_x$, this provides finitely many open sets $V_x \in \VV$ such that $\ds S \subseteq \bigcup_{x \mathop \in S} V_x$.

Thus $\set {\VV_x : x \in S}$ is a finite subcover of $\VV$.

The result follows by definition of compact.

$\blacksquare$