Cardinal of the set of real functions

Question

We know that the cardinal of natural numbers is $\aleph_0$, and the cardinal of real numbers is $\mathfrak c$. Is it correct that the cardinal of real functions is $2^{\mathfrak c}$?

Answer 1

You know that it has at least that cardinality. Consider the map:

$$ \Psi : 2^{\mathbb{R}} \rightarrow \{f : \mathbb{R} \rightarrow \mathbb{R} \} $$

given by $\Psi(S) = \chi_S(x)$ where $\chi_S(x) = 1$ if $x \in S$ and $\chi_S(x) = 0$ if $x \not\in S$. Note that $\Psi$ is injective, so you know that $2^{ |\mathbb{R}|} \leq |\{f : \mathbb{R} \rightarrow \mathbb{R} \}|$. Now you just need to get an upper bound on $ |\{f : \mathbb{R} \rightarrow \mathbb{R} \}|$.