# Compact sets are closed?

I feel really ignorant in asking this question but I am really just don’t understand how a compact set can be considered closed.

By definition of a compact set it means that given an open cover we can find a finite subcover the covers the topological space.

I think the word “open cover” is bothering me because if it is an open cover doesn’t that mean it consists of open sets in the topology? If that is the case how can we have a “closed compact set”?

I know a topology can be defined with the notion of closed sets rather than open sets but I guess I am just really confused by this terminology. Please any explanation would be helpful to help clear up this confusion. Thank you!

I think that what you’re missing is that an open cover of a compact set can cover more than just that set. Let $X$ be a topological space, and let $K$ be a compact subset of $X$. A family $\mathscr{U}$ of open subsets of $X$ is an open cover of $K$ if $K\subseteq\bigcup\mathscr{U}$; it’s not required that $K=\bigcup\mathscr{U}$. You’re right that $\bigcup\mathscr{U}$, being a union of open sets, must be open in $X$, but it needn’t be equal to $K$.
For example, suppose that $X=\Bbb R$ and $K=[0,3]$; the family $\{(-1,2),(1,4)\}$ is an open cover of $[0,3]$: it’s a family of open sets, and $[0,3]\subseteq(-1,2)\cup(1,4)=(-1,4)$. And yes, $(-1,4)$ is certainly open in $\Bbb R$, but $[0,3]$ is not.
Note, by the way, that it’s not actually true that a compact subset of an arbitrary topological space is closed. For example, let $\tau$ be the cofinite topology on $\Bbb Z$: the open sets are $\varnothing$ and the sets whose complements in $\Bbb Z$ are finite. It’s a straightforward exercise to show that every subset of $\Bbb Z$ is compact in this topology, but the only closed sets are the finite ones and $\Bbb Z$ itself. Thus, for example, $\Bbb Z^+$ is a compact subset that isn’t closed.