Htykuut

How does one compute the cardinality of the set of functions $f:mathbb{R} to mathbb{R}$ (not necessarily continuous)?

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

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 $2leqlambdaleqkappa$, then $lambda^{kappa}=2^{kappa}$. This follows because:
$$2^{kappa} leq lambda^{kappa} leq (2^{lambda})^{kappa} = 2^{lambdakappa} = 2^{kappa},$$
so you get equality throughout. The extra information you need for this is to know that if $kappa$, $lambda$, and $nu$ are nonzero cardinals, $kappaleqlambda$, then $kappa^{nu}leq lambda^{nu}$.

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

I guess that you know that $|mathbb{N}| = |mathbb{N}timesmathbb{N}|$ and thus $|mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$

This means that $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})|$. Since $fcolonmathbb{R}tomathbb{R}$ is an element of $P(mathbb{R}timesmathbb{R})$ you have that $mathbb{R}^mathbb{R}$ (all the functions from $mathbb{R}$ to itself) is of cardinality less or equal to the one of $P(mathbb{R}timesmathbb{R})$ which in turn means that $|mathbb{R}^mathbb{R}|le |P(mathbb{R})|$.

Now, since $|P(mathbb{R})| = |2^mathbb{R}|$ which is the set of all functions from $mathbb{R}$ to ${0,1}$, and clearly every function from $mathbb{R}$ into ${0,1}$ is in particular a function from $mathbb{R}$ into itself, we have:
$$|P(mathbb{R})| = |2^mathbb{R}| le |mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})| = |P(mathbb{R})|$$

So all in all we have that $|mathbb{R}^mathbb{R}| = |P(mathbb{R})| = |2^mathbb{R}|$.

This answer is based on, but differs slightly from, user Asaf Karaglia's above.

First, observe that by definition, ${text{all real functions of real variable}}:= {f: ; f: mathbb{R}tomathbb{R}} := mathbb{R}^mathbb{R}$.

The question is about $|{text{all real functions of real variable}}|$, so examine an arbitrary real function of real variable: $f,colon,mathbb{R}tomathbb{R}.$

By inspection, $f,colon,mathbb{R}tomathbb{R} := {(r, f(r)) : r in mathbb{R}} quad subseteq quad P(mathbb{R} times mathbb{R})$.

Thus, $color{green}{|mathbb{R}^{mathbb{R}}| le |P(mathbb{R}timesmathbb{R})|}$.

Before continuing, let's try to simplify $|P(mathbb{R}timesmathbb{R})|$. Observe that $|mathbb{R}| = |mathbb{R}^k| , forall , k in mathbb{N}$. Its proof by mathematical induction requires the induction hypothesis of $|mathbb{R}| = |mathbb{R}^2|$, one proof of which is : $|mathbb{N}| = |mathbb{N}timesmathbb{N}| implies |mathbb{R}| = |2^{mathbb{N}}| = |2^{mathbb{N}timesmathbb{N}}| = |2^mathbb{N}times 2^mathbb{N}| = |mathbb{R}timesmathbb{R}|$.

Verily, $mathbb{R} neq mathbb{R}^2$. Howbeit, for infinite sets $A,B$: $|A| = |B| Longrightarrow require{cancel} cancel{Longleftarrow} |P(A)| = |P(B)|$.

(The converse is discussed here.)

Thus, $|P(mathbb{R})| = |P(mathbb{R}timesmathbb{R})| implies color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$. Now scrutinise $|P(mathbb{R})|$:

● $color{#A9057D}{|P(mathbb{R})| = |2^{mathbb{R}}|}$, where $2^{mathbb{R}} := {f : ; f: mathbb{R} to {0,1}}$,

● Every $f: mathbb{R} to {0,1}$ is a particular case of a function from $mathbb{R}$ to $mathbb{R}$, thus $color{#EC5021}{2^{mathbb{R}} subsetneq mathbb{R}^mathbb{R}}$.

Altogether, $color{#A9057D}{|P(mathbb{R})| =} color{#EC5021}{|2^mathbb{R}| le} color{green}{|mathbb{R}^mathbb{R}| le |P(mathbb{R}timesmathbb{R})|} = |P(mathbb{R})|$

$implies |P(mathbb{R})| qquad qquad quad leq |mathbb{R}^mathbb{R}| leq |P(mathbb{R})| implies color{#A9057D}{underbrace{|P(mathbb{R})|}_{= |2^mathbb{R}|}} = |mathbb{R}^mathbb{R}| $.

This is irrelevent here, still it is 'relevent'. The cardinality of set of all continuous function from $mathbb{R}$ to $mathbb{R}$ $(C(mathbb{R},mathbb{R}))$ is $2 ^ mathbb{N_0} = mathfrak{c}$ because any such function is determined by its value on rationals. hence #$(C(mathbb{R},mathbb{R}))$ = # $mathbb{R}^mathbb{Q}$ which has cardinality $2^mathbb{N_0}$.