$Y$ 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].
References
 1 B.Ikenaga, Notes on Topology, August 16, 2000, available online http://www.millersv.edu/ bikenaga/topology/topnote.htmlhttp://www.millersv.edu/ bikenaga/topology/topnote.html.
Title  $Y$ is compact if and only if every open cover of $Y$ has a finite subcover 

Canonical name  YIsCompactIfAndOnlyIfEveryOpenCoverOfYHasAFiniteSubcover 
Date of creation  20130322 13:34:07 
Last modified on  20130322 13:34:07 
Owner  mathcam (2727) 
Last modified by  mathcam (2727) 
Numerical id  6 
Author  mathcam (2727) 
Entry type  Theorem 
Classification  msc 54D30 