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What is the cardianlity of: $$ A = \left\{ f:\mathbb{N}\to\mathbb{R} : \text{f is injective} \right\} $$

Trying to prove it using Cantor–Bernstein–Schroeder theorem, I have the obvious side: $$A \subseteq f:\mathbb{N}\to\mathbb{R}$$

Hence, $$\left|A\right| \le \aleph$$

I need to find an injection from a set with cardinality of $\aleph$ to $A$, but couldn't think of a proper one. It's tricky.

Any idea?

Thanks.

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HINT: Prove that $\{f\in A\mid\operatorname{range}(f)\subseteq\Bbb N\}$ has size $\aleph$.

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    $\begingroup$ Or just ask it the other way round (injections N -> R). $\endgroup$ – Martin Brandenburg Sep 8 '14 at 17:19
  • $\begingroup$ Right. I'll claim it was a typo! $\endgroup$ – Elimination Sep 8 '14 at 17:19
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    $\begingroup$ @Elimination: Alright, a different hint, then. $\endgroup$ – Asaf Karagila Sep 8 '14 at 17:26
  • $\begingroup$ Can I get an hint for the hint? :) $\endgroup$ – Elimination Sep 8 '14 at 17:37
  • $\begingroup$ @Elimination: Certainly. For every infinite set $D\subseteq\Bbb N$ consider the function which fixes pointwise $\Bbb N\setminus D$, partitions $D$ into pairs and switches all those pairs. $\endgroup$ – Asaf Karagila Sep 8 '14 at 17:40
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For each $a\in(0,1)$ we'll define a function $f(n) = a+n$. $f$ is clearly in $\mathbb{N}\to\mathbb{R}$ and since it's monotone it is also injective. Moreover, if $a_1\ne a_2$ then $f_1(n) \ne f_2(n)$.

Hence, the set of all functions described is subset of $A$ and it's cardinality is $\left|\left(0,1\right)\right| = \aleph$

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  • $\begingroup$ Yeah, that's also works. (Simpler than my hint; although proving that the set of injections from $\Bbb N$ to itself has size $\aleph$ gives you more information.) $\endgroup$ – Asaf Karagila Sep 8 '14 at 17:30

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