What is the cardinality of set of functions $F:\mathbb{N} \longrightarrow \{0,1,2\}$? Actually I have tried it but I am unable to create a mapping to see the reality as I wants to know that what is the value of 3 to the power aleph null?
 A: Let $F_n$ be set of all functions $\mathbb N \to \{0, 1, \ldots, n\}$.
Cardinality of $F_2$ is continuum (either definition or should be already known).
$|F_3| \geq |F_2|$, as $F_2 \subset F_3$. Analogously, $|F_3| \leq |F_4|$.
To show that $|F_2| = |F_4|$, we can construct bijection between them: for $f \in F_2$ take $g \in F_4$ s.t. $g(k) = 2\cdot f(2k) + f(2k + 1)$ (essentially, $f$ is binary sequence, and if we group bits by $2$, we can think about this as a sequence of quaternary sequence).
So, $|F_2| \leq |F_3| \leq |F_4| = |F_2|$. From Cantor-Bernstein theorem it follows $|F_2| = |F_3|$.
A: This is the cardinality of the set of infinite sequences of characters over the alphabet $\{0,1,2\}$. Each such sequence is the decimal expansion in the ternary number system of a real number in the closed interval $[0,1]$. So the cardinality is $\aleph_1$. A similar argument can be used to show that the cardinality of any function space $\mathbb{N} \rightarrow S$ is $\aleph_1$ if $S$ is a finite, non-empty set.
