Proof
(related to Proposition: Convergent Sequence without Limit Is Not a Compact Subset of Metric Space)
We will show that the following subset is not compact in the metric space of real numbers \(\mathbb R\): \[U:=\left\{\frac 1n:~n\in\mathbb N\setminus \{0\}\right\}\subset\mathbb R.\]
- Set the open real intervals \[U_1:=\left]\frac 12,2\right[\quad\text{ and }\quad U_n:=\left]\frac 1{n+1},\frac1{n-1}\right[\text{ for }n\ge 2.\]
- Thus, \((U_n)_{n\ge 1}\) is an open cover of $U$.
- Moreover, every open set \(U_n\) contains exactly one element of \(U\), i.e. the point \(\frac 1n\).
- Thus, \(U\) cannot be covered by any finite subcover \(U_{n_1},U_{n_2},\ldots,U_{n_k}\).
- Thus, \(U\) is not compact.
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984
