Sufficient condition for a subset to be compact in a Hausdorff topological space I know that if $X$ is a Hausdorff topological space and $A$ is compact in $X$, then $A$ is closed in $X$. My question is that if $A$ is a closed set in $X$ (where $X$ is Hausdorff), what extra condition is needed to ensure that $A$ is compact in $X$? 
 A: There’s really not much that can be said in general: it depends very much on the Hausdorff space $X$. Some examples:


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*If $X$ is compact, then every closed subset of $X$ is compact. 

*If $X$ is $\mathbb{R}^n$ for some $n$, a closed subset of $X$ is compact if and only if it’s bounded in the usual metric.

*If $X$ is an infinite space with the discrete topology, a closed subset of $X$ is compact if and only if it’s finite.  

*If $X$ is the space $\omega_1$ of countable ordinals with the order topology, a closed subset of $X$ is compact if and only if it’s countable.

*If $X$ is the Sorgenfrey line, every compact subset of $X$ is countable, but $X$ has closed subsets, even bounded ones, that are not compact, e.g., $A=\{1\}\cup\{1-2^{-n}:n\in\omega\}$: $\{[1-2^{-n},1-2^{1-n}):n\in\omega\}\cup\{[1,2)\}$ is an open cover of $A$ with no finite subcover.
A: Brian has already given you many examples explaining that compactness of $A$ strongly depends on properties of the Hausdorff space $X$. It is worth to add to his list the following one:

Let $(X,\rho)$ be a metric space (thus Hausdorff) and $A\subseteq X$. Then $A$ is compact if and only if $A$ is complete and totally bounded.

This is one of the most important characterisations of compact subsets of metric spaces.
Many interesting properties and other characterisations you can find in the Wikipedia article: http://en.wikipedia.org/wiki/Compact_space
