Bounded above and denumerable sets 
a.Let $A$ and $B$ be subsets of $[0,\infty)$ that are bounded above.
  Define the set $AB$ to be $AB = \{ab : a \in A; b \in B\}$. Prove that
  $AB$ is bounded above, and that $\sup(AB) \leq \sup(A) \sup(B)$.
b. Let $A$ and $B$ be two denumberable sets, such that $A\cap B =
\emptyset$. Show that $A\cup B$ is denumerable by exhibiting bijection
  between $A\cup B$ and $\mathbb{N}$.

For a. to be a subset of $[0,\infty)$ then it must be contained in $[0,\infty)$, but how will $[0,\infty)$ be bounded above? I know it is bounded below since it contains $0$ but I do not know why it ois bounded above since $\infty$ is not bounded. 
For b. to be dumerable then there is an injection of domain $B$ in range of $\mathbb{N}$. Thus if the intersection of $A$ and $B$ is nonempty then will that be a cut? 
 A: (a) The set $[0,\infty)$ isn’t bounded above; it’s the sets $A$ and $B$, which are subsets of $[0,\infty)$, that are assumed to be bounded above. In other words, you’re given that there are real numbers $x_A$ and $x_B$ such that $a\le x_A$ for each $a\in A$, and $b\le x_B$ for each $b\in B$. Since we’re dealing only with non-negative numbers here, we can multiply these inequalities to show that $ab\le x_Ax_B$ for each $a\in A$ and $b\in B$; this shows that $x_Ax_B$ is an upper bound for $AB$, proving that $AB$ is bounded. You still have to prove that $\sup AB\le\sup A\sup B$, however. For that it’s useful to realize that if $S$ is a set bounded above by some number $x$, and $s\le x$ for all $s\in S$, then by definition $\sup S\le x$. Apply this with $S=AB$.
(b) If you’ve defined denumerable set to be one that admits an injection into $\Bbb N$, start with injections $f:A\to\Bbb N$ and $g:B\to\Bbb N$, and consider the function
$$h:A\cup B\to\Bbb N:x\mapsto\begin{cases}
2f(x),&\text{if }x\in A\\
2g(x)+1,&\text{if }x\in B\;.
\end{cases}$$
A: The set $[0, \infty)$ itself is not bounded above, the sets $A$ and $B$ which are contained in $[0, \infty)$ but different from $[0, \infty)$ are the ones that are bounded above.
By denumerable I think the question means countably infinite, so for $A$ and $B$ there are bijections $f\colon A \to \mathbb N$ and $g\colon B \to \mathbb N$.  To answer this question you should define a map $h\colon A \cup B \to \mathbb N$ which is also a bijection.
Edit: If your professor defined denumerable as having an injection $A \hookrightarrow \mathbb N$ then you will not in general be able to exhibit a bijection $A \cup B \hookrightarrow \mathbb N$.  Because the question asks for a bijection $A \cup B \rightarrow \mathbb N$ then it must be the case that "denumerable" means there is a bijection $A \to \mathbb N$.  Otherwise the question is incorrect.
A: It should be read that $A$ and $B$ are subsets of $[0,\infty)$ and $A$ and $B$ are bounded from above (not that $[0,\infty)$ is bounded from above). This just means that there exists a $K_A>0$ and $K_B>0$ such that
$$
a\leq K_A\text{ for all } a\in A,\quad\text{and}\quad b\leq K_A\text{ for all } b\in B.
$$
In b. let $A=\{a_1,a_2,\ldots\}$ and $B=\{b_1,b_2,\ldots\}$, then
$$
A\cup B=\{a_1,b_1,a_2,b_2,\ldots\}.
$$
