Let $A,B$ not empty,bounded subsets of $\mathbb{R}$.Show that $$\sup(A \cup B)= \max \{\sup A, \sup B \}.$$
That's what I have done so far: Let $x\in A \cup B \Rightarrow x \in A \text{ or } x\in B \Rightarrow x\leq \sup(A) \text{ or } x\leq \sup(B) \Rightarrow x\leq \max \{\sup(A),\sup(B)\}$. But,how can I continue?
You're nearly there. You've shown that every $x \in A \cup B$ satisfies $x \leq \max\{\sup(A),\sup(B)\}$. Therefore, by definition (or an elementary property, depending on your definition) of $\sup$, $\sup(A \cup B) \leq \max \{ \sup(A), \sup(B) \}$.
Conversely, every element of $A$ is also in $A \cup B$, so for exactly the same reason, $\sup A \leq \sup(A \cup B)$. Symmetrically, $\sup B \leq \sup(A \cup B)$. By definition of $\max$, $\max \{ \sup(A), \sup(B)) \leq \sup(A \cup B)$.
$s$ is an upper bound for $A \cup B$ iff it is an upper bound for $A$ and for $B$, so $s \geq \sup A$ and $s \geq \sup B$.
The least upper bound, $\sup A \cup B$ is then the least such $s$ - which is clearly $ \max \{ \sup A, \sup B \}$.
