# 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?

## 3 Answers

(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

$$h:A\cup B\to\Bbb N:x\mapsto\begin{cases} 2f(x),&\text{if }x\in A\\ 2g(x)+1,&\text{if }x\in B\;. \end{cases}$$

• Thanks! And for B I should have said "If there is a one-one function with domain $B$ and range equal to all of $\mathbb{N}$, then the set $B$ is denumerable " will that be a better definition ? – Q.matin Feb 18 '13 at 8:05
• @Q.matin: It depends on whether you want denumerable to include finite. I prefer the term countable, but I use both to mean finite or countably infinite; that’s compatible with the definition that I used in my answer. If you want it to mean countably infinite, then yes, you want it to be a bijection. (The trick that I suggested will still work, though.) – Brian M. Scott Feb 18 '13 at 8:08
• Thanks again for the help Brian! – Q.matin Feb 18 '13 at 8:10
• @Q.matin: You’re welcome; my pleasure! – Brian M. Scott Feb 18 '13 at 8:13

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 $$A\cup B=\{a_1,b_1,a_2,b_2,\ldots\}.$$

• A lot more clear now, thanks! For part b since A and B are denumerable then there is an injection and from that I have to follow that their union will be a bijection? Also, How hard would you rate these two question in a scale from 1 to 10. 10 being the hardest for a real analysis intro course? – Q.matin Feb 18 '13 at 8:00
• See @Brian M. Scott's answer for an explicit construction of the bijection. He's simply mapping the elements of $A$ into to the even integers, and the elements of $B$ into the odd integers. My answer was merely an attempt to make it intuitively clear that $A\cup B$ is indeed denumerable. I would rate a. as a 1, and b. is a bit harder so maybe a 3. – Stefan Hansen Feb 18 '13 at 8:05
• Thank a lot, even though I was hoping for a higher scale rate which means I have a lot more studying to do! – Q.matin Feb 18 '13 at 8:07
• Again, it's tough for me to rate, because I'm not sure what a real analysis intro course covers. But anyhow, a. is pretty easy and intuitively clear, whereas b. is much more abstract and needs some careful thought. – Stefan Hansen Feb 18 '13 at 8:09

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 lot more clear now, thanks! For part b since A and B are denumerable then there is an injection and from that I have to follow that their union will be a bijection? Also, How hard would you rate these two question in a scale from 1 to 10. 10 being the hardest for a real analysis intro course? – Q.matin Feb 18 '13 at 8:00
• I would rate them as pretty easy on a difficulty scale, but pretty abstract for someone who hasn't seen this stuff (so don't feel bad if it's confusing at first). – Jim Feb 18 '13 at 8:05