Exercise 1.4.2

Question

Let $A \subseteq \mathbf{R}$ be nonempty and bounded above, and let $s \in \mathbf{R}$ have the property that for all $n \in \mathbf{N}, s+\frac{1}{n}$ is an upper bound for $A$ and $s-\frac{1}{n}$ is not an upper bound for $A$. Show $s=\sup A$.

Ulisse Mini · Accepted Answer

This is basically a rephrasing of Lemma 1.3.8 using the archimedean property. The most straightforward approach is to argue by contradiction:

(i): If $s < \sup A$ then there exists an $n$ such that $s + 1 ∕ n < \sup A$ contradicting $\sup A$ being the least upper bound.
(ii): If $s > \sup A$ then there exists an $n$ such that $s - 1 ∕ n > \sup A$ where $s - 1 ∕ n$ is not an upper bound, contradicting $\sup A$ being an upper bound.

Thus $s = \sup A$ is the only remaining possibility.

ulissemini

2022-01-27 00:00

Exercise 1.4.2

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Exercise 1.4.2

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