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Next: Problem 4 Up: 18.100B, Fall 2002 In-class Previous: Problem 2

Problem 3

Prove that in any metric space, a finite union of compact subsets is compact.

Proof. [Solution] Let $ K_i,$ $ i=1,\dots,N$ be a finite collection of compact sets in a metric space $ X$ and let $ K=\bigcup_i K_i$ be their union. An open cover $ U_a,$ $ a\in A$ of $ K$ covers each of the $ K_i.$ By the assumed compactness there is a finite subcover $ U_{a_{i,j}}$ such that $ K_i\subset \bigcup_{j=1}^{N(i)}U_{a_{i,j}}$ for each $ i.$ All these sets taken together give a finite subcover of $ K$ which is therefore compact. $ \qedsymbol$

Richard B. Melrose 2002-10-19