Compact Sets of Double Pointed Topology

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S, \vartheta}\right)$ be a topological space.

Let $D$ be a doubleton endowed with the indiscrete topology.

Let $\left({S \times D, \tau}\right)$ be the double pointed topology on $S$.


Then $X \subseteq S \times D$ is compact in $\tau$ iff for some compact set $C$ of $\vartheta$:

$\operatorname{pr}_1 \left({X}\right) = C$

where $\operatorname{pr}_1$ denotes the first projection on $S \times D$.


Corollary

$\left({S \times D, \tau}\right)$ is compact if and only if $\left({S, \vartheta}\right)$ is compact.


Proof

Necessary Condition

Suppose that $X \subseteq S \times D$ is a compact set in $\vartheta$.

It is to be shown that $C = \operatorname{pr}_1 \left({X}\right)$ is compact in $\tau$.

This follows from Compactness Properties Preserved under Projection Mapping.

$\Box$


Sufficient Condition

Suppose $\operatorname{pr}_1 \left({X}\right) = C$ for some compact $C \subseteq S$.

Let $\mathcal U$ be an open cover for $X$, i.e. $\displaystyle \bigcup \mathcal U \supseteq X$.

By Open Sets of Double Pointed Topology, each $U \in \mathcal U$ is of the form:

$U' \times D$

with $U'$ open in $\vartheta$; note that $U' = \operatorname{pr}_1 \left({U}\right)$.

By Cartesian Product Distributes over Union, it follows that:

$\displaystyle \bigcup \mathcal U = \left({\bigcup_{U \mathop \in \mathcal U} \operatorname{pr}_1 \left({U}\right)}\right) \times D$

Since $X \subseteq \displaystyle \bigcup \mathcal U$, it follows by Image Preserves Subsets and Image of Union under Relation that:

$C = \operatorname{pr}_1 \left({X}\right) \subseteq \bigcup_{U \mathop \in \mathcal U} \operatorname{pr}_1 \left({U}\right)$

showing that:

$\mathcal V := \left\{{\operatorname{pr}_1 \left({U}\right): U \in \mathcal U}\right\}$

is an open cover for $C$.

Since $C$ is compact, we find a finite subcover $\mathcal V'$ of $\mathcal V$ such that:

$C \subseteq \displaystyle \bigcup \mathcal V'$


Finally, we will show that:

$\mathcal U' := \left\{{V \times D: V \in \mathcal V'}\right\} \subseteq \mathcal U$

forms a finite subcover of $\mathcal U$.

That is, that:

$X \subseteq \displaystyle \bigcup \mathcal U'$

So let $\left({s, d}\right) \in X$.

By assumption on $X$, we have that $s \in C$, and since $\mathcal V'$ is a open cover for $C$:

$\exists V \in \mathcal V': s \in V$

It is immediate that $\left({s, d}\right) \in V \times D$ by definition of Cartesian product.

Since $V \times D \in \mathcal U'$, it follows by definition of set union that:

$\left({s, d}\right) \in \displaystyle \bigcup \mathcal U'$

That is, $\mathcal U'$ is a finite subcover for $X$, and hence $X$ is compact.

$\blacksquare$