I'm thinking since $|\mathbb{R}|>|\mathbb{N}| \times| \mathbb{ Z} |\times |\mathbb{ Q}|$ then there can't be an injection, but is that true?
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4Yes, there cannot be such an injection. – Aphelli Dec 11 '19 at 22:34
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2Yes, of course uncountable sets can't inject into countable sets. – Dzoooks Dec 11 '19 at 22:34
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1Look for Bernstein-Schroder theorem! – HelloDarkness Dec 11 '19 at 22:43
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1@HelloDarkness Although learning the Schröder-Bernstein theorem is a good idea in general, I don't see that it has much to do with this question. – Andreas Blass Dec 12 '19 at 01:31
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@AndreasBlass I guess, just thought of it because it lets me lsee injections as a "$\leq$" in cadnilaty, which, in this case, we know it be false. – HelloDarkness Dec 12 '19 at 02:19
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https://math.stackexchange.com/q/2357899/631742 – Maximilian Janisch Dec 12 '19 at 09:58
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Define $N:=\mathbb{N}\times \mathbb{Z}\times \mathbb{Q}$. Assume there exists such an injection $f: \mathbb{R} \rightarrow N$. Then the function $g: \mathbb{R} \rightarrow f(\mathbb{R}), g(x)=f(x)$ is a bijection (check this). As $N$ is countable (as it is the cartesian product of finitely many countable sets) and $g$ is a bijection, we would get that $\mathbb{R}$ is countable as well, which gives the desired contradiction.
Severin Schraven
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