Let $A$ and $B$ be two non-empty bounded above subsets of $\mathbb{R}$ such that $\text{sup}B < \text{sup}A$. (1)
(a) Prove that there exists a real number $a$ such that $a \in A$ and $a$ is an upper bound of $B$.
(b) Is the result still true if we replace hypothesis (1) with the weaker hypothesis $\text{sup}B \leq \text{sup}A$?
(a) So I know that, you need the approximation property for suprema to show that there is an element in $A$ which is greater than $\text{sup}B$ but less than $\text{sup}A$. But I am unsure how to apply this and therefore hence prove the first statement.
Just follow the definitions.
$\sup A$ is the least upper bound of $A$. What does that mean? What is the definition.
Def: 1) $\sup A$ is an upper bound of $A$. And if $y < \sup A$ then $y$ is not an upper bound of $A$.
Well, $\sup B < \sup A$ so $\sup B$ is not an upper bound of $A$.
Well what does that mean?
If $y$ is an upper bound of $A$ means $y \ge a$ for all $a \in A$. SO if $y$ is not an upper bound of $A$ that means that there is an $a \in A$ so that $y < a$.
So $\sup B$ is not an upper bound of $A$ so there is an $a \in A$ so that $\sup B < a$.
Now what can we say about $a$ in terms of the elements of $B$?
....
As for b)
If $\sup B = \sup A$ and $a\in A$ is an upper bound of $B$ that would mean $a \ge \sup B$ because $B$ is the least upper bound.
But $a \in A$ so $a \le \sup A$ because $\sup A$ is an upper bound of $a$.
So we have $\sup B \le a \le \sup A = \sup B$ so $a =\sup B = \sup A$.
Does that have to happen? Is $\sup A \in A$ always?