TITLE: Show that $\sup(A \cup B)=\max\{\sup A,\sup B\}$
QUESTION [7 upvotes]: 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?

REPLY [4 votes]: $s$ 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 \}$.