Compactness

Definition: open covers

An open cover of a space $X$ is a collection $U$ of open subsets of $X$ whose union is $X$. A subcover is a subcollection of elements of an open cover $U$ that still covers $X$.

Definition: compact

A space $X$ is compact if every open cover of $X$ has a finite subcover And a subset $A\subseteq X$ is compact if it is compact as a subspace

A subset of an euclidean space is compact if and only if it is bounded and closed

Example

• Every finite space is compact. (since it only has finitely many open sets )
• Every space with the trivial topology is compact (Simple Examples)
• A subset of a discrete space is compact $⟺$ it is finite (Simple Examples)
  • Suppose $A\subseteq X$ discrete ($|A|=\infty$), then we can find $U:=\{\{a\}{\}}_{a\in A}$ as the union of all singletons in the discrete subset, and we cannot find a finite subcover of $U$ (removing any singleton of $U$ breaks the “coverness”)

Lemma

NOTE

A subset $A\subseteq X$ is compact if and only if every collection $\{U_{\alpha }{\}}_{\alpha }$ of open subsets of $X$ with ${\bigcup }_{\alpha }{U}_{\alpha }\supseteq A$ has a finite subcollection $\{U_{{\alpha }_k}{\}}_{k=1}^n$ satisfying ${\bigcup }_{k=1}^n{U}_{{\alpha }_k}\supseteq A$

This is basically saying the same thing as the definition.

• “every collection $\{U_{\alpha }{\}}_{\alpha }$ of open subsets of $X$ with ${\bigcup }_{\alpha }{U}_{\alpha }\supseteq A$ ” means every open cover of the subspace $A$.
• “has a finite subcollection $\{U_{{\alpha }_k}{\}}_{k=1}^n$ satisfying ${\bigcup }_{k=1}^n{U}_{{\alpha }_k}\supseteq A$” means has a finite subcover that still covers $A$.

NOTE

let $(x_i)$ be a sequence in a space $X$ that converge to a limit point $y$, then the subspace $A=\{x_i:i\in \mathbb{N}\}\cup \{y\}$ is compact