I need to decide which of these three sets are equipotent:

$M_1=\{(n_1,n_2,n_3)\in\mathbb{N}\times\mathbb{N}\times\mathbb{N}\ |\ n_1+n_2=n_3\}$

$M_2 = \{M\in P(\mathbb{Z})\ |\ 0\in M\}$

$M_3 = \cup _{a\in\mathbb{Z}}\{x\in\mathbb{R}\ |\ a\leq x < \frac{2a+1}{2}\}$

I want to prove (or disprove) the equipotency by finding injections to and from $\mathbb{N}$, $P(\mathbb{N})$ and $\mathbb{R}$ (Cantor-Schroeder-Bernstein).

I've already proven that $M_1$ is equipotent to $\mathbb{N}$:

1) $M_1\rightarrow\mathbb{N}$, $(n_1,n_2,n_3)\mapsto 2^{n_1}\cdot 3^{n_2}\cdot 5^{n_3}$

2) $\mathbb{N}\rightarrow M_2, n\mapsto (n,n,2n)$

I'm stuck finding injections like this for $M_2$ and $M_3$.

It already seems that $M_2$ is equipotent to $P(\mathbb{N})$ and $M_3$ is equipotent to $\mathbb{R}$, but what are the corresponding injections?

  • $\begingroup$ You don't need actual functions to use CSB theorem, so : did you really mean you can use it? $\endgroup$ – DonAntonio Nov 30 '16 at 13:50
  • $\begingroup$ @DonAntonio: Yes, I can use it. But how do I use CSB without giving two injections? $\endgroup$ – de_dust Nov 30 '16 at 13:53
  • $\begingroup$ Just using inequalities between cardinalities of well known sets and usinmg arithmetic of cardinals, of course. This is, I believe, the greatest thing about this theorem $\endgroup$ – DonAntonio Nov 30 '16 at 13:54
  • $\begingroup$ @DonAntonio: Can you give an example please? $\endgroup$ – de_dust Nov 30 '16 at 13:55
  • $\begingroup$ @de For example: $$\aleph_0=|\Bbb N|\le |\Bbb N\times\Bbb N|\le\aleph_0\cdot\aleph_0=\aleph_0\implies|\Bbb N\times\Bbb N|=\aleph_0$$ $\endgroup$ – DonAntonio Nov 30 '16 at 13:58

Since for any set $\;X\in P(\Bbb N)\;$ (for me the naturals do not contain zero) , we have that $\;X\cup\{0\}\in M_2\;$ , so we have that

$$\mathfrak c=|P(\Bbb N)|\le|M_2|\le|P(\Bbb Z)|=\mathfrak c\implies |M_2|=\mathfrak c$$

  • $\begingroup$ Unfortunately, we defined the naturals including zero. I guess it's a little more tricky this way, since the first inequality doesn't hold. $\endgroup$ – de_dust Nov 30 '16 at 20:12
  • 1
    $\begingroup$ Well, then just define $\;\Bbb T:=\Bbb N\setminus\{0\}\;$ and you get exactly the same as, of course, $\;|\Bbb T|=|\Bbb N|\;$ $\endgroup$ – DonAntonio Nov 30 '16 at 20:15

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.