I'm having trouble intuitively understanding what it means for a set to be compact. I know that by definition a set is compact if for every open cover of the set there exists a finite subcover.
But I don't understand how an infinite set can be compact,because how can an infinite set be covered by a finite number of things? And when is an infinite set not compact?
Thanks in advance
Best Answer
I'll take almost your example : $$U = \bigcup_{n\in\mathbb N^*} (\frac1n, 1)$$
cover (0,1) and is an "open cover".
Let's take only a finite set of $\{ (\frac1n, 1) | n\in\mathbb N^*\}$, say $\{(\frac 1n,1)| 0< n <100\}$. (100 is purely arbitrary, in order to make things the more concrete that I can).
If you consider: $$U_{100} = \bigcup_{n=1}^{100} (\frac1n, 1),$$ the number $\frac1{101}$ is not in $U_{100}$.
And whatever the number of subsets you're taking, you'll be able to find a number which is not in you subcover (do you want the formal proof of that ?).
EDIT (proof of that):
Let $C$ be a finite subset of $E=\{ (\frac1n, 1) | n\in\mathbb N^*\}$ (note that it's possible to have "holes", for example $\{(\frac1{100},1), (\frac12,1)\}$ is a finite subset of $E$).
Let $N$ the greater integer such that $(\frac1{N},1)\in C$ (there is effectively a "greater integer such that" beacause $C$ is finite). The number $\frac1{N+1}$ does not belong to the union of the elements of $C$.
Thus we've proved that whatever the finite subset of E we choose, this subset doesn't cover $(0,1)$. END OF EDIT
For your other question, @user2345215 pointed out that a "thing" can be itself infinite, so there'is no contradiction to be infinite and to be cover by a finite number of things (each thing can be infinite).