$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!!

  • $\begingroup$ 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. $\endgroup$
    – JavaMan
    Nov 3, 2018 at 2:13

1 Answer 1


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.

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

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