Are only the closed subspaces of a compact space compact?

Question

I have read a Wikipedia article about Stone-Čech compactification and have a question about one particular sentence:

"... $[0, 1]^C$ is compact since $[0, 1]$ is. Consequently, the closure of $X$ in $[0, 1]^C$ is a compactification of $X$."

Does that imply that generally, if a space A is compact, then any closure of a topological space in the A is also compact? So the closure is what is important and open subspace of compact space wouldn´t have to be compact?

Do yu have any examples of open subspaces of compact spaces, s.t. the subspaces are not compact? Thank you.

Answer 1

Yes, if $X$ is compact and $A \subseteq X$ then $\overline{A}=\operatorname{Cl}(A)$ is a closed subset of a compact space and these are always compact in the subspace topology, as is classic.

If $X$ is Hausdorff and $A \subseteq X$ is compact in the subspace topology, then $A$ is closed in $X$; this is also classical. So in $[0,1]^C$ (which is compact and Hausdorff) closed and compact subsets are the same.

In general, an arbitrary (non-closed) subspace of a compact space need not be compact, e.g. the open subspace $(0,1)$ of the compact $[0,1]$ is not compact (but homeomorphic to $\Bbb R$).

Answer 2

Answer to the first question is yes. You can find that here: closed subset is compact

Second has a counter example: Take $A=[0,1]$ and $S=(0,1)$. $ S\subseteq A$ is open. Show that it is not compact.