TITLE: Supremum of an union of bounded sets
QUESTION [9 upvotes]: Given $A$, $B$ are bounded subsets of $\Bbb R$. 
Prove 

$A\cup B$ is bounded. 
$\sup(A \cup B) =\sup\{\sup A, \sup B\}$.

Can anyone help with this proof?

REPLY [15 votes]: Without loss of generality assume that $\sup A\le\sup B$, so that $\sup\{\sup A,\sup B\}=\sup B$, and you simply want to show that $\sup(A\cup B)=\sup B$. Clearly $\sup(A\cup B)\ge\sup B$, so it suffices to show that $\sup(A\cup B)\le\sup B$.
To show that $\sup(A\cup B)\le\sup B$, just prove that $\sup B$ is an upper bound for $A\cup B$, i.e., that $x\le\sup B$ for every $x\in A\cup B$. This isn’t hard if you remember that we assumed at the start that $\sup A\le\sup B$.