It's not like we can say that, because $\mathbb{N}$ is a subset of $\mathbb{R}$, the restriction of a bijection between $\mathbb{R}$ and $\mathbb{R}^2$ is going also to be a bijection between $\mathbb{N}$ and $\mathbb{N}^2$ (although such a bijection may exist). There are other ways to prove that $\mathbb{N} \cong \mathbb{N}^2$, but is it possible to show this using the fact that $\mathbb{R} \cong \mathbb{R}^2$?
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1Can you conclude ${a,b}$ is in bijection with ${a,b} \times {a,b } = {(a, a), (a,b), (b,a), (b,b)}$? – Nov 09 '19 at 13:05
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No, you cannot show $\mathbb{N}^2 \simeq \mathbb{N}$ from $\mathbb{R}^2 \simeq \mathbb{R}$, as there is no reason to think the witnessing bijection for the former equivalence sends integer pairs to integers.
You can deduce $\mathbb{R}^2 \simeq \mathbb{R}$ from $\mathbb{N}^2 \simeq \mathbb{N}$ if you know that $\mathbb{R} \simeq 2^{\mathbb{N}}$:
$$\mathbb{R} \simeq 2^{\mathbb{N}} \simeq 2^{\mathbb{N}^2} \simeq 2^{\mathbb{N}} \times 2^{\mathbb{N}} \simeq \mathbb{R}^2$$
using the obvious bijections. But not the other way around. But there are even explicit formulas for $\mathbb{N}^2 \simeq \mathbb{N}$, see here, e.g., so there is no need for arguments from a harder to show fact.
Henno Brandsma
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But is there no way to show that the existence of infinitely many bijections between $\mathbb{R}$ and $\mathbb{R}^2$ implies that there exists one that also happens to be a bijection between $\mathbb{N}$ and $\mathbb{N}^2$ – 11122345791216212837 Nov 09 '19 at 13:42
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1It might be worth adding that in general $\mathcal P(A) \simeq \mathcal P(B) \implies A \simeq B$ is independent of ZFC. – Mees de Vries Nov 10 '19 at 06:49
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@MeesdeVries That's interesting. Can you give a source for that? – 11122345791216212837 Nov 11 '19 at 20:17
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1@11122345791216212837 I only know of these slides: http://homepages.math.uic.edu/~jbaldwin/asl07.pdf But I'm sure Google could help you out further. – Mees de Vries Nov 11 '19 at 21:27
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@11122345791216212837 It is well-known (though rather advanced set theory) that you can have (in ZFC) that $(|\Bbb R|=)2^{\aleph_0} = 2^{\aleph_1}$ while of course $\aleph_0 < \aleph_1$, e.g. in a model of Martin's Axiom for $\aleph_2$. – Henno Brandsma Nov 11 '19 at 21:55
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@HennoBrandsma Isn't $|\mathcal{P}(\mathbb{R})| = 2^{\aleph_1}$? – 11122345791216212837 Nov 11 '19 at 22:34
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@11122345791216212837 No, not necessarily. Not in that model of ZFC. – Henno Brandsma Nov 11 '19 at 22:41