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

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 $$a=\sup A$$.

• This seems correct to me. You can also prove this by using the definition that sup$A$ is the least upper bound for $A$. – Rick Jun 12 at 8:18
• @Rick Thank you. I will accept that if you post it as an answer so this question may be closed – csch2 Jun 12 at 8:29

Ok, here we go. Suppose $$b$$ is the supremum of $$A, b \ne a$$. Then, since $$a$$ is an upper bound, $$b < a$$. But we also have that $$a \in A$$. So, $$a \le b$$. This gives us a contradiction because of $$b . Hope this helps.