Lets Assume to the contrary that there is a Supremum $S_1 \in A \cap B$ that is larger than the Supremum $S_2 \in A$
If $S_1 \in A \cap B$ then $S_1 \in A$ and since $S_2 \in A$
This is a contradiction because $S_1 \not\gt S_2$
and therefore sup$(A \cap B) \leq$ sup$(A)$
Is this sufficient to disprove this? I could use a counter example but since its contradictory to have a smallest upper bound in a set larger than the smallest upper bound of a set I wanted to go this way.
Best Answer
Don't make things unnecessarily complicated. You don't need a proof by contradiction here.
By definition $S=\sup A$ is an upper bound for all elements of $A$, hence an upper bound for all elements of $A\cap B$. By definition $\sup(A\cap B)$ is the smallest of all upper bounds for the elements of $A\cap B$. Thus $S\ge \sup(A\cap B)$.