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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?

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  • $\begingroup$ 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
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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

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

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  • $\begingroup$ 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
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    $\begingroup$ 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

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