Sets for which only some open coverings have finite subcoverings

Question

I am about halfway through my first course in analysis and have a question about compact sets. Recall that set $S$ in a metric space is said to compact if and only if every open covering of $S$ contains a finite subcovering of $S$.

That every is necessary is strange to me. My question: what sets $Q$ both have open coverings with finite subcoverings and have open coverings without finite subcoverings?

Answer 1

Really, there's a lot of them. For example, every open set $U \subseteq M$ has that property since $U$ is a cover itself and has finite subcover (itself). However, open sets are usually not compact.

As a matter of fact, any subset of $M$ which is not compact has that property, since $M$ itself is an open cover for any subset $U \subseteq M$; and $M$ is the corresponding finite open subcover. However, if $U$ is not compact, then there necessarily exists some open cover which does not have a finite subcover.

Answer 2

Think about examples as $$\mathbb R=\bigcup\limits_{n\in\mathbb{N}}\;(-n,n)$$