Cardinality of the set of all even functions $f : \mathbb{R} \rightarrow \mathbb{R} $ $f : \mathbb{R} \rightarrow \mathbb{R} $ even, let $S$ be $\{f : \mathbb{R} \rightarrow \mathbb{R}: f(-x)=f(x), \forall x\in\mathbb{R}\}$
Now, since $S$ is a subset of the set of all functions from $\mathbb{R}$ to $ \mathbb{R}$ we know that $k(S) \leq 2^C$.
I'm really not sure how to go about this. I suspect I should find some injection from either $P(\mathbb{R})$ or the set of all functions, to S.
I figured I could work with some mapping like $f \to f\circ g $ where $g(x) = x^2$ because that way I could render any function even. The problem with that is that it's not an injection so I can't apply the Cantor-Schröder–Bernstein theorem.
I would appreciate any hint whatsoever!!
 A: I don't know what $C$ means in your question. Also I don't think composing with $x^2$ will work because this is not injective.
Instead you can argue as follows:
First, prove (unless you already know) that the there is a bijection $b:\mathbb{R}\rightarrow[0,\infty]$.
Then let $f\in S$ and consider $f|_{[0,\infty]}$ this is a map from $[0,\infty]\rightarrow\mathbb{R}$. Now compose this function with $b$ you get a map 
$$\mathbb{R} \overset{b}{\longrightarrow}[0,\infty]\overset{f|_{[0,\infty]}}{{\longrightarrow}\mathbb{R}}$$
Call this map $\varphi:S\rightarrow \mathbb{R}^\mathbb{R}$ where $\mathbb{R}^\mathbb{R}$ denotes the set of functions from $\mathbb{R}\rightarrow\mathbb{R}$.
It is left to show that $\varphi$ is a bijection.
It is 1:1, for if $f,g\in S$ such that $\varphi(f)=\varphi(g)$ then $f,g$ agree on $[0,\infty]$ but since $f(-x)=f(x)$ and $g(-x)=g(x)$ you have that $f=g$.
It is onto, for if $f:\mathbb{R}\rightarrow\mathbb{R}$ we can compose $f$ with $b^{-1}$ to get a map from $[0,\infty]\rightarrow \mathbb{R}$ (first apply $b^{-1}$ and then $f$) applying $\varphi$ on this map we will get $f$ again. In other words $f$ is in the range of $\varphi$ and so it is also onto.
