Finite Topological Space is Compact

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space where $S$ is a finite set.


Then $T$ is compact.


Proof by selecting finite subcover

Let $\mathcal V$ be an open cover of $T$.

For each $x \in S$, define $\mathcal V_x$ to be $\left\{{V \in \mathcal V : x \in V}\right\}$


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

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


Since there were only finitely many $\mathcal V_x$, this provides finitely many open sets $V_x \in \mathcal V$ such that $\displaystyle S \subseteq \bigcup_{x \in S} V_x$.


Thus $\left\{{\mathcal V_x : x \in S}\right\}$ is a finite subcover of $\mathcal V$.

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 $\mathcal V$ be an open cover of $S$.

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

From Set is Subset of Itself, $\mathcal V \subseteq \mathcal V$.

Thus by definition $\mathcal V$ is a finite subcover of $\mathcal V$.

The result follows by definition of compact.

$\blacksquare$


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