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Chapter 11: Problem 5

Prove, using Definition \(11.2 .2\), that if \(K_{1}\) and \(K_{2}\) are compact sets in \(\mathbb{R}\), then their union \(K_{1} \cup K_{2}\) is compact.

Short Answer

Expert verified
We have proved that \(K_{1} \cup K_{2}\) is compact as we showed it is a bounded and closed set in \(\mathbb{R}\) and thus satisfies the criteria for the Heine-Borel Theorem (Definition \(11.2.2\)).

Step by step solution

01

Prove Union is Bounded

Firstly, the definition of a compact set stipulates that it must be bounded. In this case, because both sets \(K_{1}\) and \(K_{2}\) are compact, they are both bounded. The union of two bounded sets remains bounded. This is because the maximum and minimum values of the union are simply the maximum of the each of the sets' maximums, and similarly for the minimum values. Thus, \(K_{1} \cup K_{2}\) is bounded.
02

Prove Union is Closed

Secondly, to prove the union of two compact sets is compact, it needs to be demonstrated that \(K_{1} \cup K_{2}\) is closed. The definition of a compact set is that it must be closed. Both \(K_{1}\) and \(K_{2}\) are compact, and therefore each one of them is closed. The union of two closed sets in \(\mathbb{R}\) is also closed, as per the properties of closed sets in real analysis.
03

Appy Heine-Borel Theorem

Now that we've confirmed \(K_{1} \cup K_{2}\) is both bounded and closed, we can apply the Heine-Borel Theorem (referred to as Definition \(11.2.2\)). According to this theorem, if a set is both bounded and closed in \(\mathbb{R}\) then it must be compact. Consequently, the union \(K_{1} \cup K_{2}\) is indeed compact.

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