Let $A,B\subseteq \mathbb{R}$ be a non-empty intervals and bounded from above.
If $A\cap B\neq \emptyset $ prove that it is bounded from above and that $Sup(A\cap B)=min\{sup(A),Sup(B)\}$
$A,B$ are bounded from above therefore there are $M_1,M_2$ such that $a\leq M_1$ and $b\leq M_2$.
for $x \in A\cap B$ $(x\in A \wedge x\in B)$, so $x\leq M_1 \wedge x\leq M_2$.
W.L.O.G let say that $A\leq B$ if so than $x\leq M_1\leq M_2$, but there is only one Supremum and by definition it is the least upper bound therefore $Sup( A\cap B)= M_1$ and if $B\leq A$ the proof is the same.
Is it a valid proof?
Best Answer
Since $A=(a_0, a_1)$ and $B=(b_0, b_1)$ with $a_1, b_1 < \infty$ and $a_0, b_0 \ge -\infty$, write out the intersection explicitly:
$$A\cap B = (\max(a_0, b_0), \min(a_1, b_1))$$ The supremum of an interval is its right endpoint so $$\sup A\cap B = \min(a_1, b_1) = \min(\sup A, \sup B)$$
Note that it's irrelevant for this proof if the intervals have open or closed ends at any side.