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Chapter 2: Problem 9

Show that if \(A\) and \(B\) are bounded subsets of \(\mathbb{R}\), then \(A \cup B\) is a bounded set. Show that \(\sup (A \cup B)=\sup \\{\sup A, \sup B\\}\)

Short Answer

Expert verified
The union of two bounded sets, A and B, is also a bounded set with supremum equal to the maximum of the suprema of A and B.

Step by step solution

01

Establish Boundedness

A set is bounded if it has both a least upper bound (supremum) and greatest lower bound (infimum). Thus, if A and B are both bounded subsets of \(\mathbb{R}\), that means they both have an upper and lower bound. Since the union of the sets, \(A \cup B\), consists of all the points in either A or B, it must also have a least upper bound and greatest lower bound, thus it must also be bounded.
02

Infer About Supremum

If \(C = A \cup B\) then any element \(c\) in \(C\) is in either \(A\) or \(B\). Hence \(c \leq \sup A\) or \(c \leq \sup B\). Therefore, in either case, \(c\) is less than or equal to the larger of \(\sup A\) and \(\sup B\), so we conclude that \(c \leq \max\{ \sup A, \sup B \}\). So, \(\max\{ \sup A, \sup B \}\) is an upper bound for \(C\).
03

Conclude Supremum Equality

However, note that \(\max\{ \sup A, \sup B \}\) is either \(\sup A\) or \(\sup B\), and since \(\sup A\) is an upper bound for \(A\) and \(\sup B\) is an upper bound for \(B\), it follows that \(\max\{ \sup A, \sup B \}\) is always in \(C = A \cup B\). Therefore, as it is an element of \(C\) and an upper bound for \(C\), it is the least upper bound, or supremum of \(C\). Therefore \(\sup (A \cup B)=\max \{\sup A, \sup B\}\).

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