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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 $a-\epsilon<a_0$ for some $a_0\in A,\epsilon>0$. Since $a\in A$, let $a=a_0$. Then we get that $a-\epsilon<a$, which holds for all $\epsilon>0$. Therefore, by Lemma 1.3.8, we have that $a=\sup A$.

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

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