I am new to compact sets, and I had some hard time trying to solve this:
Let $X$ be an infinite set; $T$ the discrete topology on $X$ .Prove that $(X,T)$ is not compact.
I know that I may consider the open cover $\{\{x\},x \in X\}$ and prove that it doesn't have a finite subcover,but I don't know how to write it exactly.
Moreover, I seriously can't understant why $X$ itself cannot be considered as a finite subcover?
Any help is appreciated.
Best Answer
Indeed take the cover $\mathcal{U}=\{\{x\}: x \in X\}$. This is clearly a cover of $X$ as every $p \in X$ is in $\{p\} \in \mathcal{U}$ and in fact only in that set.
It's an open cover because the topology is discrete and so all subsets of $X$ are open.
A finite subcover is a finite subset of $\mathcal{U}$ (so these are all singleton sets too) that together cover $X$ too. But this cannot exist: if $\mathcal{F}$ is a finite subset of $\mathcal{U}$ then it consists of finitely many singletons $\{x_1\},\{x_2\},\ldots, \{x_N\}$ for some finite $N$. But $X$ has infinitely many points so there are infinitely many $x \notin \{x_1,\ldots,x_N\}$, say $p$ is one of those. Then $p$ is in none of the sets of $\mathcal{F}$, and so this finite subset of $\mathcal{U}$ is not a cover of $X$.
Hence $\mathcal{U}$ has no finite subcover. In fact, if you think about it, we cannot remove a single element $\{p\}$ from $\mathcal{U}$ or $p$ would no longer be covered at all; every point is only covered once... The set $X$ is not even an element of $\mathcal{U}$ so does not qualify as a possible subcover. A subcover is a selection of open sets from the cover, and $X$ is not one of them, there are just singletons.
To think about: the open cover of all doubletons $\{x,y\}, x \neq y$ would also have worked: a finite subset (say of size $N$) of that cover can only cover $2N$ points at most and so must also miss points of $X$, etc. In this cover every point is covered infinitely many times, and still we cannot reduce it to a finite subcover.
To show non-compactness we only have to find one cover without a finite subcover, because compactness is the strong statement that every open cover of $X$ must have a finite subcover. So one counterexample suffices to kill it.