# Cardinality of real bijective functions/injective functions from $\mathbb{R}$ to $\mathbb{R}$

As specified here, the cardinality of the set of bijective functions on $$\mathbb{N}$$ is precisely $$2^{\aleph_0}$$. I was wondering if one can prove a parallel result regarding the bijective function on $$\mathbb{R}$$ (or maybe even the injective functions from $$\mathbb{R}\mapsto \mathbb{R}$$).

I tried to think about an injective function from the set of all $$\mathbb{R}\mapsto \mathbb{R}$$ functions to the set of injective or bijective functions, but I quickly ran out of ideas. Are the two sets even from the same cardinality?

Note: I included the real-analysis tag because I was thinking that maybe for this specific question a relatively complex construction using real functions might be necessary.

For each $$x>0$$ pick $$\epsilon_x\in\{-1,1\}$$ and define $$f(x)=\epsilon_{|x|}x$$ for $$x\ne0$$ and $$f(0)=0$$. Then $$f\circ f$$ is the identity, so $$f$$ is a bijection. This way we get $$|\mathscr{P}(\Bbb R)|$$ bijections from $$\Bbb R$$ to $$\Bbb R$$. There are only $$|\mathscr{P}(\Bbb R)|$$ functions from $$\Bbb R$$ to $$\Bbb R$$, so that's the cardinality of the set of bijections.
• Why there are $|\mathscr{P}(\Bbb R)|$ bijections $\Bbb R$ to $\Bbb R$ ?
Upper bound is cardinality of general function $$R \to R$$: $$R^R=2^R$$; lower bound is $$2^R=R$$ as well (by consider each slot $$x>0$$ as having $$x, -x$$ two choices and that $$f(-x)=-f(x)$$; easy to verify each combination of such choices is bijection).