If $A$ and $B$ are bounded subsets of $\mathbb{R}$, show that $\sup{A\cup B}=\sup\{\sup A, \sup B\}$. If $A$ and $B$ are bounded subsets of $\mathbb{R}$, show that $\sup{A\cup B}=\sup\{\sup A, \sup B\}$.
I've already shown that $A\cup B$ is bounded since both $A$ and $B$ are bounded. And I know by the Completeness Theorem that exists $\sup A$ and $\sup B$.
Any help would be greatly appreciated.
 A: Often, to do this sort of problem, you will find it convenient to prove equality by proving that the LHS is both $\leq$ and $\geq$ the right.
For instance: for any $x\in A$, we know that $\sup A\geq x$; similarly, for any $x\in B$, we know that $\sup B\geq x$. So, for every $x\in A\cup B$, we have that 
$$
x\leq\max\{\sup A,\ \sup B\}=\sup\{\sup A,\ \sup B\}.
$$
So, $\sup\{\sup A,\ \sup B\}$ is an upper bound on $A\cup B$; hence it must be true that it is at least the least upper bound:
$$
\sup A\cup B\leq\sup\{\sup A,\ \sup B\}.
$$
Can you see how to prove the other side?
A: On the one hand, you should be able to see/show that if $$x\ge\sup\bigl\{\sup A,\sup B\bigr\},$$ then $x$ is an upper bound for $A$ and an upper bound for $B,$ from which it follows (why?) that $x$ is an upper bound for $A\cup B,$ so that $$x\ge\sup(A\cup B).$$ It follows (why?) that $$\sup\bigl\{\sup A,\sup B\bigr\}\ge\sup(A\cup B).$$
On the other hand, you should be able to see/show that if $$x\ge\sup(A\cup B),$$ then $x$ is an upper bound for $A\cup B,$ from which it follows (why?) that $x$ is an upper bound for $A$ and an upper bound for $B,$ so that $x\ge\sup A$ and $x\ge\sup B,$ and so $$x\ge\sup\bigl\{\sup A,\sup B\bigr\}.$$ It follows (why?) that $$\sup(A\cup B)\ge\sup\bigl\{\sup A,\sup B\bigr\}.$$
