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Chapter 11: Problem 7

Find an infinite collection \(\left\\{K_{n}: n \in \mathbb{N}\right\\}\) of compact sets in \(\mathbb{R}\) such that the union \(\bigcup_{n=1}^{\infty} K_{n}\) is not compact.

Short Answer

Expert verified
An infinite collection of compact sets in real numbers, where each set \(K_n\) is defined as [n, n+1], is such that the union of all these sets is not compact.

Step by step solution

01

Define the compact sets

Let's define the sets \(K_n\) as \([n, n+1]\) for each \(n\) in natural numbers. Each of these sets is compact because it is a closed and bounded subset of real numbers.
02

Define the union of the sets

We define \(\bigcup_{n=1}^{\infty} K_{n}\) as the union of all such sets. This, obviously, is the set of all real numbers greater than or equal to 1.
03

Show that the union is not compact

To show that the union of all \(K_n\) sets is not compact, we will prove that it cannot be covered by a finite number of open sets. Assume \(U = \{ U_i : i \in I \}\) is an open cover for the \(\bigcup_{n=1}^{\infty} K_{n}\). Since each \(K_n = [n, n+1]\) is compact, there exists a finite subcover \(U_{n} = \{ U_{i} : i \in I_{n} \}\), such that \(K_n \subseteq \bigcup_{i \in I_{n}} U_{i}\). However, this cover is not finite as it depends on \(n\), thereby making the union of all \(K_n\) sets not compact.

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Most popular questions from this chapter

Show that if \(f: \mathbb{R} \rightarrow \mathbb{R}\) is continuous, then the set \(\\{x \in \mathbb{R}: f(x)=k\\}\) is closed in \(\mathbb{R}\) for each \(k \in \mathbb{R}\).

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