Are all finite strings over some infinite alphabet countable?

Question

Over some infinite alphabet $\Sigma$, can we state that the set of all possible finite strings is countable?

Answer 1

If the alphabet is uncountable then the set of all possible finite strings is uncountable. If you allow infinite length strings then by the diagonalization argument the set is also uncountable.

Assume the alphabet is countable and strings have finite length. Let's assign to each alphabet symbol a natural number, i.e., each symbol corresponds to a natural number and denote a string by a sequence of numbers. For example, if $a$ corresponds to $2$, $b$ to $5$, and $c$ to $11$ then $aabcb$ would be denoted by $2 \ 2 \ 5 \ 11 \ 5$ (I separate them by a single space to avoid confusion). We will prove that the set of all strings is countable.

We group every string of length $n$ whose individual symbols sum to $k$ into the set $C_{n,k}$. For example $000 \in C_{3,0}$, $1192 \in C_{4,13}$, and $1\ 13 \ 3 \ 1 \in C_{4,18}$. For each pair $\langle n,k \rangle$, $C_{n,k}$ is clearly finite and hence is countable. Furthermore, any string $a_1\dots a_n$ does belong to $C_{n,k}$, where $k=a_1+\dots +a_n$ which means every string is included in some set $C_{n,k}$. Since the set of all possible pairs $\langle n,k \rangle$ is countable the union of all possible $C_{n,k}$, i.e, $\bigcup_{\langle n,k \rangle \in N\times N}{C_{n,k}}$ is also countable.

Answer 2

Another way: If $\Sigma=\{\mathbb{a_0, a_1,a_2,...}\}$ is countable, then to show that $\Sigma^*$ is countable we need only produce an injection $f: \Sigma^*\to\mathbb{N}$. But this is easy; e.g., just let $f([\mathbb{a_{i_1},a_{i_2},...,a_{i_n}}])$ be the number whose hexadecimal (or any base $>10$) representation is $\tt{Ai_1Ai_2...Ai_n}$, where $\tt{i_k}$ is the decimal representation of $i_k$. E.g., $f([\mathbb{a_0,a_{12},a_7}]) = \tt{A0A12A7}_{hex}=168432295_{dec}$.