Exercise 1.3.7

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

Lemma 1.3.7 states that if $A \subset \mathbb{R}$ and $s \in \mathbb{R}$ which is also an upper bound of $A$, then $s= \text{sup}A$ if and only if for all $\epsilon > 0$, there exists $a \in A$ such that $s- \epsilon < a$. So it's sufficient to show if $a \in A$ is also an upper bound for $A$ that, $a- \epsilon < a$ for all $\epsilon > 0$. By the definition of equality, we know this condition is met.

So an upper bound of a set which is also an element of the set is the supremum.

I think this is really close. I'm not sure what you mean when you write "By the definition of equality, we know this condition is met". Feels a bit awkward to me.

I guess the key issue is this: Given $\varepsilon>0$, you need an $x\in A$ such that $x>s-\varepsilon$. Since $s \in A$, you can just choose $x=s$!