Finite Topological Space is Compact

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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 by selecting finite subcover

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.

$\blacksquare$


Proof by finiteness of the topology

From Power Set of Finite Set is Finite, the power set of $S$ is finite.

Since the topology $\tau$ is by definition a set of subsets of $S$, it follows that $\tau$ is finite as well.

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

By definition $\VV \subseteq \tau$ and so is also a finite set.

From Set is Subset of Itself, $\VV \subseteq \VV$.

Thus by definition $\VV$ is a finite subcover of $\VV$.

The result follows by definition of compact.

$\blacksquare$


Sources

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