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Chapter 11: Problem 10

Let \(K \neq \emptyset\) be a compact set in \(\mathbb{R}\). Show that inf \(K\) and sup \(K\) exist and belong to \(K\).

Short Answer

Expert verified
The infimum and supremum of set \(K\) exist and belong to \(K\), due to the property of compactness, which guarantees that \(K\) contains all its limit points, including inf \(K\) and sup \(K\).

Step by step solution

01

Identify compactness

A set \(K\) is compact if every open cover of the set has a finite subcover. This means that if we take any collection of open intervals that collectively cover all of \(K\), then there is a finite subset of these intervals that also cover all of \(K\).
02

Prove that inf \(K\) exists and belongs to \(K\)

By definition, infimum (or inf \(K\)) of a set is the greatest lower bound of that set. Assume that \(\{K_n\}\) is a sequence of real numbers in \(K\) such that \(K_n\) is decreasing and the limit of this sequence is \(L\). By compactness of \(K\), \(K\) contains all its limit points, including \(L\), so \(L\) belongs to \(K\) and \(L = \inf K\), hence inf \(K\) exists and belongs to \(K\).
03

Prove that sup \(K\) exists and belongs to \(K\)

Sup \(K\), the supremum of \(K\), is defined as the least upper bound of the set \(K\). The proof is analogous to the proof for inf \(K\). Suppose \(\{K_n\}\) is a sequence of real numbers in \(K\) such that \(K_n\) is increasing and \(\lim K_n = U\). Due to the compactness of \(K\), \(K\) contains all its limit points; therefore \(U\) belongs to \(K\) and \(U = \sup K\), hence sup \(K\) exists and is an element of \(K\).

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