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!!

• Not sure if it helps (I would have to think more about infectivity), but it's worth noting that if $f : \Bbb R \to \Bbb R$, then $f(x) + f(-x)$ is even. Nov 3, 2018 at 2:13

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.

• C is short for continuum, denoting the cardinality of $\mathbb{R}$. Thanks a lot for your solution, quite lovely! :D Nov 2, 2018 at 21:37