# Cardinality of all injective functions from $\mathbb{N}$ to $\mathbb{R}$.

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.

• Zero nothing nadda. – Rene Schipperus Sep 8 '14 at 17:17
• – Martin Sleziak Sep 8 '14 at 18:35

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

• Or just ask it the other way round (injections N -> R). – Martin Brandenburg Sep 8 '14 at 17:19
• Right. I'll claim it was a typo! – Elimination Sep 8 '14 at 17:19
• @Elimination: Alright, a different hint, then. – Asaf Karagila Sep 8 '14 at 17:26
• Can I get an hint for the hint? :) – Elimination Sep 8 '14 at 17:37
• @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. – Asaf Karagila Sep 8 '14 at 17:40

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 $\left|\left(0,1\right)\right| = \aleph$

• 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.) – Asaf Karagila Sep 8 '14 at 17:30