Prove that \(\mathbb R\) is not compact.
Solution
1. The family\[ \mathcal U=\{(-n,n):n\in\mathbb N,\ n\ge1\} \]is an open cover of \(\mathbb R\).
Each interval \((-n,n)\) is Euclidean open. Given \(x\in\mathbb R\), choose an integer \(n>|x|\). Then \[ -n<x<n, \] so \(x\in(-n,n)\). Thus the union of the members of \(\mathcal U\) is all of \(\mathbb R\).
2. No finite subfamily of \(\mathcal U\) covers \(\mathbb R\).
Take finitely many members \[ (-n_1,n_1),\ldots,(-n_k,n_k). \] Let \[ N=\max\{n_1,\ldots,n_k\}. \] Since these intervals are nested as their radii increase, \[ (-n_1,n_1)\cup\cdots\cup(-n_k,n_k)=(-N,N). \] The point \(N+1\) does not belong to this union. Hence the chosen finite subfamily does not cover \(\mathbb R\).
3. Therefore \(\mathbb R\) is not compact.
Compactness requires every open cover to have a finite subcover. The open cover in 1 has no finite subcover by 2, so \(\mathbb R\) fails the definition of compactness.