I am looking for hint on the question:
If $A, B \subseteq \mathbb{R}$ are bounded intervals with $A \cap B \neq \varnothing$, then $\sup(A \cap B) = \min\{\sup A, \sup B\}$.
So far, I have shown that $\beta = \min\{\sup A, \sup B\}$ is an upper bound for $A \cap B$. Hence, $\sup(A \cap B) \leq \beta$. Where I get stuck is showing this must be the least upper bound. I wish to show that $\sup(A \cap B) \geq \beta$, by supposing $\sup(A \cap B) < \beta$, and finding a contradiction by finding some element $x \in A \cap B$ such that $x > \sup(A \cap B)$. I think I need to use the assumption that $A, B$ (and thus so is $A \cap B$) are intervals, but I am not sure how if this is the case.
This is where I am stuck and would appreciate a hint on how to move forward. Thank you.