# Proof verification: If $a\in A$ is an upper bound for $A$, then $a=\sup A$

by csch2   Last Updated June 12, 2019 08:20 AM

Prove that if $$a$$ is an upper bound for $$A$$, and if $$a$$ is also an element of $$A$$, then it must be that $$a=\sup A$$.

We are given the following lemma:

Lemma 1.3.8: Assume $$s\in\textbf{R}$$ is an upper bound for a set $$A\subseteq\textbf{R}$$. Then, $$s=\sup A$$ if and only if, for every choice of $$\epsilon>0$$, there exists an element $$a\in A$$ satisfying $$s-a<\epsilon$$.

Proof: It is given that $$a$$ is an upper bound for $$A$$. To verify that $$a=\sup A$$, we use Lemma 1.3.8. We want to show that $$a-\epsilon for some $$a_0\in A,\epsilon>0$$. Since $$a\in A$$, let $$a=a_0$$. Then we get that $$a-\epsilon, which holds for all $$\epsilon>0$$. Therefore, by Lemma 1.3.8, we have that $$a=\sup A$$.

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