Proof
(related to Proposition: Closed Subsets of Compact Sets are Compact)
- Let $(X,d)$ be a metric space and let \(C\subset X\) be a compact subset of \(X\).
- It has to be shown that any closed subset \(A\subset C\) is also compact.
- Let $U_i)_{i\in I}$ be an open cover of $A$.
- Since \(A\) is closed, the set \(X\setminus A\) is open.
- Thus, $(X\setminus A)\cup\bigcup_{i\in I}U_i=X$ is an open cover of $C$.
- Since \(C\) is compact, it has by definition of compact sets a finite subcover $$C\subset (X\setminus A)\cup U_{i_1}\cup U_{i_2}\cup\ldots\cup U_{i_k}$$ for some \(i_1,i_2,\ldots,i_k\in I\).
- Because \(A\subset C\), also \(A\) has a finite subcover.
- Thus, \(A\) is compact.
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References
Bibliography
- Forster Otto: "Analysis 2, Differentialrechnung im \(\mathbb R^n\), Gewöhnliche Differentialgleichungen", Vieweg Studium, 1984
