Exercise 5 Prove that the union of 2 compact sets is a compa

Exercise 5. Prove that the union of 2 compact sets is a compact set.

Solution

The union of two compact sets is compact. Let A and B be compact sets, and let C be an open cover of their union. It follows that C is an open cover of A (so it has finite subcover C_1) and C is also an open cover of B (so it has a finite subcover C_2). Now the union of C_1 and C_2 is a finite cover for A union B.

By induction, a finite union of compact sets is compact. However, the same is not true of infinite unions of compact sets: the real line is a countable union of (compact) closed unit intervals.

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) We prove this using the definition of compactness. Let A1, A2, . . . An be compact sets. Consider the union Sn k=1 Ak. We will show that this union is also compact. To this end, assume that F is an open cover for Sn k=1 Ak. Since Ai Sn k=1 Ak, then F is also a cover for Ai . By compactness of Ai , there exists a finite subcover F 0 i for Ai . Consider F 0 = Sn i=1 F 0 i . Since each Fi is finite, and there are only finitely many such i, F 0 is finite as well. Furthermore, since each F 0 i cover Ai , then F 0 covers Sn k=1 Ak. Thus, we have found a finite subcover for the union of our compact sets. Thus, it is compact.