# Cardinality of the set of either completely multiplicative or multiplicative functions defined on $\mathbb N$ and which take values in $\mathbb N$

Suppose that $S$ is a set of functions such that $f \in S$ if and only if $f$ is either multiplicative or completely multiplicative and every $f \in S$ is defined on the set of natural numbers and takes natural numbers as its values ($f: \mathbb N \to \mathbb N$).

Recall that the function $f$ is completely multiplicative if, for every integers $a$ and $b$ we have $f(ab)=f(a)f(b)$ and $f(1)=1$. Such functions exist, one example is Liouville function.

Recall that the function $f$ is multiplicative if $f(1)=1$ and $f(ab)=f(a)f(b)$ for every coprime integers $a$ and $b$. An example of such a function is Euler`s totient function.

Only curiosity led me to the question of "how big" is the set of either multiplicative or completely multiplicative functions defined on the set $\mathbb N$ and which take values in the set $\mathbb N$.

It is clear that the set $S$ is at least countable because for every $n \in \mathbb N$ the function $f(a)=a^n$ is completely multiplicative.

So the real question is:

Is $S$ uncountable?

To define a totally multiplicative function $f: \mathbb{N}\to \mathbb{N}$, you only need to define its values on the primes numbers, since then $f(n)$ is just given by the prime factors decomposition of $n$.
But conversely, you can see that any value is allowed for the $f(p)$ where $p$ is prime. So $S$ has the cardinality of the set of functions $P\to \mathbb{N}$ where $P$ is the set of prime numbers. And this set is clearly uncountable (it has the same cardinality as $\mathbb{N}^{\mathbb{N}}$).
Another way to say all that is that $\mathbb{N}^*$ is the free abelian monoid on $P$, but you need not know such a notion to conclude.