0
$\begingroup$

The problem statement: Calculate the cardinal of the set consisting of all the dense and enumerable subsets of $\mathbb R^2$

The attempt at a solution:

I couldn't go farther than this: If $A$ is the set consisting of all the dense and enumerable subsets of $\mathbb R^2$, then $A\subset B \space$, $\space B=\{ S \subset \mathbb R^2: |S|=\aleph_0\}$. So $|A|\leq |B|={(c^2)}^{\aleph_0}=c^{2\aleph_0}=c^{\aleph_0}$

I have no idea what else to do, could anyone give me suggestions or hints on how could I go on?, for example, I could find a lower bound for the cardinal of $A$. Or maybe there is a bijective function from this set to another set $C$, where the cardinality of $C$ is easy to calculate.

$\endgroup$
5
$\begingroup$

You just need to go one step further:

$$|B|=\mathfrak{c}^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0\cdot\aleph_0}=2^{\aleph_0}=\mathfrak{c}\;.$$

So $|A|\le\mathfrak{c}$, and you’ll be done if you can find $\mathfrak{c}$ different countable dense subsets of $\Bbb R$. What about sets of the form $\Bbb Q^2\cup\text{(something very simple)}\,$?

$\endgroup$
6
  • 1
    $\begingroup$ As an aside, it seems pretty clear from context that the OP meant $(c^2)^{\aleph_0}$ rather than $c^{2^{\aleph_0}}$. $\endgroup$ – Erick Wong Dec 10 '13 at 5:04
  • 1
    $\begingroup$ @Erick: It would have been clear if I’d seen the $^2$. sigh My eyes aren’t what they once were. $\endgroup$ – Brian M. Scott Dec 10 '13 at 5:07
  • $\begingroup$ Yes, sorry for my notation, I meant just what Erick Wong wrote, I've edited the original post. $\endgroup$ – user100106 Dec 10 '13 at 5:08
  • $\begingroup$ @user100106: No problem. I really would have interpreted it correctly if my eyes hadn’t slipped right over the exponent in $\Bbb R^2$, so I’m at least partly responsible for the misunderstanding. $\endgroup$ – Brian M. Scott Dec 10 '13 at 5:09
  • $\begingroup$ @BrianM.Scott Don't worry. Your answer was very useful, I just didn't know that $c^{\aleph_0}=c$, with that information and the other suggestion I can finish it from here. Thanks Brian! $\endgroup$ – user100106 Dec 10 '13 at 5:11

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.