TITLE: Proof that if $a \in A$ and $a$ is upper bound, then $a$ is the supremum of $A$
QUESTION [2 upvotes]: I'm not sure if the following proof is correct because I haven't seen it elsewhere.
We have that $a$ is an upper bound, so by the definition of supremum $sup A \leq a$.
But $a \in A$ so $a \leq sup A$ because $sup A$ is an upper bound of $A$ by definition.
So we conclude that $a = sup A$.

REPLY [1 votes]: Your proof is right!
By definition of supremum $\sup A \leq y  ,\forall y$ upper bound of $A$ . Thus $\sup A \leq a$. Furthermore $\sup A$ is also an upper bound itself which implies that $\sup A \geq a$ since you have for hypotesis that $a \in A$. We have shown that both inequalities hold so it must be $a=\sup A$
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