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 $b \in B\}$.

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

Post-comment addition:

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.

  • $\begingroup$ 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 $a_1 b_1 < a_3 b_3 < a_2 b_2$? $\endgroup$ – blz Feb 20 '17 at 8:13
  • $\begingroup$ See post-comment additions. $\endgroup$ – 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 $\mathbb{R}$.

  • $\begingroup$ 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]. $\endgroup$ – blz Feb 20 '17 at 8:46
  • $\begingroup$ @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 $\endgroup$ – Adren Feb 20 '17 at 9:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.