# Prove or disprove that if $A$ and $B$ are dense, then $A \cdot B$ is dense

There's this test question I'm having difficulty with. Let $A\subseteq \mathbb{R}$ and $B\subseteq \mathbb{R}$ be two sets in $\mathbb{R}$. Prove or disprove that if $A$ and $B$ are dense, then $A \cdot B$ is dense.

Note: If $A$ is dense then $\forall a_1, a_2 \in A,$ such that $a_1 < a_2$ there exists $a_3 \in A$ such that $a_1 < a_3 < a_2$.

Also, we define $A\cdot B$ = $\{a \cdot b \mid a\in A$ and $b \in B\}$.

• What is $A*B$ ? – user171326 Feb 19 '17 at 15:15
• Could you explain what does $A*B$ mean ? – Adren Feb 19 '17 at 15:15
• I've edited to explain it better, thanks for the notice – blz Feb 19 '17 at 15:17
• Hint : if $A$ is dense and $B$ is non-empty then $A*B$ is dense. – user171326 Feb 19 '17 at 15:18
• @N.H.: Let $A = [1, 2]; B = \{0, 2\}$. Where's the point in $A*B$ between $0$ and $2$? – John Hughes Feb 19 '17 at 15:19

Initial hint: Given $a_1 b_1$ and $a_2 b_2$ in $A*B$, you need to find $a_3 b_3$ with $a_1 b_1 < a_3 b_3 < a_2 b_2$. How do you think you should pick $a_3$ and $b_3$?

Hint 2: Once you think this works, there's probably more to do.

OP writes:

Well I've tried to define $a_3=\frac{a_1+a}{2}$ and $b_3= \frac{b_1+b_2}{2}$ so $a_1 b_1<a_3 b_3<a_2b_2$ but this isn't true for $a_1=−1,a_2=0,b_1=1,b_2=5$, for example. How can I find those $a_3$ and $b_3$such that $a_1b_1<a_3b_3<a_2b_2$?

Doing that gives you a number $a_3$ that is between $a_1$ and $a_2$, but how do you know that $a_3 \in A$? (Answer: you don't, for its entirely possible that it's not.) You need some other way to produce $a_3$ and $b_3$.

But as you note, that doesn't quite work -- negative numbers give you a problem. And now you have to think about whether the statement is actually true in cases where $A$ and $B$ have some negative numbers in them, or whether maybe it's false in those cases.

By the way: nothing you described actually used the given fact that $A$ and $B$ are dense, and that's a bad sign, because the claim that $A*B$ is dense is certainly not in general true if $A$ and $B$ are not dense.

Also: You might want to address the question raised in the comments in the original posting --- what definition of "dense" are you using? If it's "Dense in $\Bbb R$," then the claim is much easier to prove.

• Well I've tried to define $a_3 = \frac{a_1 + a_2}{2}$ and $b_3 = \frac{b_1+ b_2}{2}$ so $a_1 b_1 < a_3 b_3 < a_2 b_2$ but this isn't true for $a_1 = -1, a_2 = 0, b_1 = 1, b_2 = 5$, for example. How can I find those $a_3$ and $b_3$ such that $a_1 b_1 < a_3 b_3 < a_2 b_2$? – blz Feb 20 '17 at 8:13
• See post-comment additions. – John Hughes Feb 20 '17 at 12:15

Here is another point of view :

Given $A\subset\mathbb{R}$, we have :

$A$ is dense iff for all $x\in\mathbb{R}$, there exists a sequence of elements of $A$ which converges to $x$.

Now suppose $A,B$ are dense subsets of $\mathbb{R}$ and take $x\in\mathbb{R}$. There exist $(a_n)\in A^\mathbb{N}$ and $(b_n)\in B^\mathbb{N}$ such that $\lim_{n\to\infty}a_n=x$ and $\lim_{n\to\infty}b_n=1$; hence $\lim_{n\to\infty}a_nb_n=x$, which proves that $AB$ is also a dense subset of $\mathbb{R}$.

• How do you know that there exists $(b_n)\in B^\mathbb{N}$ such that $\lim_{n\to\infty}b_n = 1$? What if B = [2,3]? I can't just modify it and choose $(b_n)\in B^\mathbb{N}$ such that $\lim_{n\to\infty}b_n = 2$, (hoping eventually for $\lim_{n\to\infty}a_nb_n = x$) because there won't necessarily be $(a_n)\in a^\mathbb{N}$ such that $\lim_{n\to\infty}a_n = x/2$, for example If I choose A = [1,2]. – blz Feb 20 '17 at 8:46
• @blz: I realize that we don't use same definition of density. What I wrote in my answer concerns subsets of $\mathbb{R}$ which are everywhere dense – Adren Feb 20 '17 at 9:47