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