Supremum proof real analysis

Question

hope somebody can help me. I have a feeling it is a simple question, but I just cannot figure it out.

Suppose A and B are two non-empty bounded subsets of $\mathbb{R}$ such that $\sup A =\sup B$ and $\sup A$ is NOT in A. Show that $\forall_{a > \in A} \exists_{b \in B} : a<b$

Thanks in advance

Answer 1

Let $K=\sup A$. Let $a$ be some element of $A$ and consider $\epsilon=K-a>0$ (since $K\not\in A$). Assume $\not\exists b\in B$ such that $a<b$. Then $B$ is bounded above by $K-\frac{1}{2}\epsilon$ and so $\sup B\neq K$. Contradiction.

Answer 2

Let $a \in A$ be arbitrary. Show that there is $b \in B$ such that $a < b$.

Proof by contradiction: Assume that $b \leq a$ for all $b \in B$. Then $\sup B \leq a < \sup A$, which contradicts the assumption $\sup A = \sup B$. Note that the last inequality is strict by the assumption $\sup A \not\in A$.

The intuition is that both sets have the same "upper endpoint". If $b \leq a$ would hold for all $b \in B$, then, clearly, the "upper endpoint" of $B$ would be less than that of $A$.