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

Question

Assume that $b$ is supremum of set $A$, where $b > a$. Let $x \in A$. So we have $x \leq a < b$. Since b is supremum for set A. So let us choose $\epsilon = b-a $ and so we must have some $x$ such that $x > b - \epsilon$, so we have $x > b - (b - a)$, which results in $x > b$ which is a contradiction as b is suprermum for set A

Also b cannot be less than a.

Is my proof correct ?

Thanks

Answer 1

Note you state "$x > b - (b - a)$, which results in $x > b$ which is a contradiction as b is suprermum for set A". However, $x > b - (b - a)$ gives $x \gt a$, but that is still OK since $a$ is supposed to be an upper bound and, thus, you still have a contradiction. Nonetheless, otherwise, your proof looks fine to me, except you may wish to explain that

Also b cannot be less than a.

is due to the supremum $b$, by definition, being greater than or equal to all elements of $A$, including $a$.