Cardinality of sets made by functions

Question

What is the cardinality of the set of all functions $h : \mathbb N\rightarrow \{0,1,\ldots,9\}$ so that there are exactly $3$ elements $a \in \{0,1,\ldots,9\}$ for which $h^{-1}[ \{ a \}]$ is infinity? And why would that be so? I tried thinking as in the cartesian plane all kinds of functions that could increase and decrease along the values $\{0,1,\ldots,9\}$ in the axis of ordinates so there could be another kind of function that could show the paths possibles along de abscissa axis. I also think that because of the above it is true to say $\{0,1,\ldots,9\}^{\mathbb N}\subsetneq\mathcal P(\mathbb N \times \mathbb N$).

Answer 1

If $\kappa$ is the cardinality of this set, then:

• $\kappa\le 10^{\aleph_0}=2^{\aleph_0}$ because the set of all functions $h:\mathbb N\to\{0,1,\ldots,9\}$ has cardinality $10^{\aleph_0}=2^{\aleph_0}$.
• $\kappa\ge 3^{\aleph_0}=2^{\aleph_0}$ because we can look just into the subset of functions $h$ such that:

$$h(4k)\in\{1,2,3\}\text{ (any of them)}$$ $$h(4k+1)=1$$ $$h(4k+2)=2$$ $$h(4k+3)=3$$

for all $k\in\mathbb N$.

So, $\kappa\le 2^{\aleph_0}$ and $\kappa\ge 2^{\aleph_0}$, so, by the Schröder–Bernstein theorem, we have $\kappa=2^{\aleph_0}$.