How does one compute the cardinality of the set of functions f:R→R (not necessarily continuous)?

**Answer**

All you need is a few basics of cardinal arithmetic: if κ and λ are cardinals, none of them zero, and at least one of them is infinite, then κ+λ=κλ=max. And cardinal exponentiation satisfies some of the same laws as regular exponentiation; in particular, (\kappa^{\lambda})^{\nu} = \kappa^{\lambda\nu}.

The cardinality of the set of all real functions is then

|\mathbb{R}|^{|\mathbb{R}|} =\mathfrak{c}^{\mathfrak{c}} = (2^{\aleph_0})^{2^{\aleph_0}} = 2^{\aleph_02^{\aleph_0}} = 2^{2^{\aleph_0}} = 2^{\mathfrak{c}}.

In other words, it is equal to the cardinality of the power set of \mathbb{R}.

With a few extra facts, you can get more. In general, if \kappa is an infinite cardinal, and 2\leq\lambda\leq\kappa, then \lambda^{\kappa}=2^{\kappa}. This follows because:

2^{\kappa} \leq \lambda^{\kappa} \leq (2^{\lambda})^{\kappa} = 2^{\lambda\kappa} = 2^{\kappa},

so you get equality throughout. The extra information you need for this is to know that if \kappa, \lambda, and \nu are nonzero cardinals, \kappa\leq\lambda, then \kappa^{\nu}\leq \lambda^{\nu}.

In particular, for any infinite cardinal \kappa you have \kappa^{\kappa} = 2^{\kappa}.

**Attribution***Source : Link , Question Author : Benji , Answer Author : Arturo Magidin*