$Y$ is compact if and only if every open cover of $Y$ has a finite subcover

Theorem.
Let $X$ be a topological space and $Y$ a subset of $X$. Then the following statements are equivalent.

1. $Y$ is compact as a subset of $X$.

2. Every open cover of $Y$ (with open sets in $X$) has a finite subcover.

Proof. Suppose $Y$ is compact, and ${\{U_i\}}_{i\in I}$ is an arbitrary open cover of $Y$, where $U_i$ are open sets in $X$. Then ${\{U_i\cap Y\}}_{i\in I}$ is a collection of open sets in $Y$ with union $Y$. Since $Y$ is compact, there is a finite subset $J\subset I$ such that $Y={\cup }_{i\in J}(U_i\cap Y)$. Now $Y=({\cup }_{i\in J}{U}_i)\cap Y\subset {\cup }_{i\in J}{U}_i$, so ${\{U_i\}}_{i\in J}$ is finite open cover of $Y$.

Conversely, suppose every open cover of $Y$ has a finite subcover, and ${\{U_i\}}_{i\in I}$ is an arbitrary collection of open sets (in $Y$) with union $Y$. By the definition of the subspace topology, each $U_i$ is of the form $U_i=V_i\cap Y$ for some open set $V_i$ in $X$. Now $U_i\subset V_i$, so ${\{V_i\}}_{i\in I}$ is a cover of $Y$ by open sets in $X$. By assumption, it has a finite subcover ${\{V_i\}}_{i\in J}$. It follows that ${\{U_i\}}_{i\in J}$ covers $Y$, and $Y$ is compact. $\mathrm{\square }$

The above proof follows the proof given in [1].

Title	$Y$ is compact if and only if every open cover of $Y$ has a finite subcover
Canonical name	YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover
Date of creation	2013-03-22 13:34:07
Last modified on	2013-03-22 13:34:07
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	6
Author	mathcam (2727)
Entry type	Theorem
Classification	msc 54D30