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Chapter 2: Problem 7

Let \(S \subseteq \mathbb{R}\) be nonempty. Show that \(u \in \mathbb{R}\) is an upper bound of \(S\) if and only if the conditions \(t \in \mathbb{R}\) and \(t>u\) imply that \(t \notin S\).

Short Answer

Expert verified
The fact that \(u\) is an upper bound of \(S\), and the fact that for all \(t \in \mathbb{R}\) such that \(t > u\), \(t \notin S\), are shown to be equivalent to each other, which means that any real number \(t\) greater than the upper bound \(u\) of \(S\) cannot belong to \(S\).

Step by step solution

01

Proof of First Condition

An upper bound of a set \(S\) is a number \(u\) in which every element \(x \in S\) satisfies the inequality \(x \leq u\). This is the definition of an upper bound.Start by assuming that \(u\) is an upper bound of \(S\). So, by definition, every element in \(S\) is less than or equal to \(u\). This means that if you have a number \(t\) that is strictly greater than \(u\) (\(t>u\)), then it cannot belong to the set \(S\). This is because all elements of \(S\) must be less than or equal to \(u\), and \(t\) exceeds this limit.
02

Proof of Second Condition

Now consider the second condition, where for \(t \in \mathbb{R}\) such that \(t>u\), \(t \notin S\). Assume this condition is true. So, for any \(t\) that is greater than \(u\), \(t\) is not in \(S\). As \(t\) can be any number, you can say that all numbers greater than \(u\) do not belong to \(S\). When you put it into other words, all numbers that do belong to \(S\) must be less than or equal to \(u\). Therefore, this condition by its nature defines \(u\) as an upper bound of \(S\), thus proving the equivalence of the two conditions.

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