Prove that the open interval \((0,1)\) considered as a subset of \(\mathbb R\) in the usual topology is not compact.
Solution
For each integer \(n\ge2\), set \[ U_n=\left(\frac1n,1\right)\subseteq(0,1). \]
1. The family\[ \{U_n:n\ge2\} \]is an open cover of \((0,1)\).
Each \(U_n\) is open in the subspace topology on \((0,1)\). Let \(x\in(0,1)\). By the Archimedean property, choose an integer \(n\) with \[ n>\frac1x. \] Then \[ \frac1n<x<1, \] so \(x\in U_n\). Thus the family covers \((0,1)\).
2. No finite subfamily of this cover covers \((0,1)\).
Take finitely many sets \[ U_{n_1},\ldots,U_{n_k} \] and let \[ N=\max\{n_1,\ldots,n_k\}. \] Since \(1/n\) decreases as \(n\) increases, \[ U_{n_i}\subseteq U_N \] for every \(i\). Hence \[ U_{n_1}\cup\cdots\cup U_{n_k}=U_N. \] But \[ \frac1{2N}\in(0,1) \] and \[ \frac1{2N}\notin U_N \] because \(1/(2N)<1/N\). Therefore the finite subfamily does not cover \((0,1)\).
3. Hence \((0,1)\) is not compact.
By 1 there is an open cover of \((0,1)\), and by 2 it has no finite subcover. This is exactly the negation of compactness.