Let $\Sigma$ be a finite nonempty alphabet. Is the set of all DFAs over $\Sigma$ countable?
I know the set of all regular languages is countable, however, it is impossible to build an injection from the DFAs to the regular languages since a regular language isn't necessarily accepted by only a single DFA.
There are several ways to prove this. Here is a formal one:
You can encode the set of DFAs over $\Sigma$ as words over the constant alphabet $\Sigma' = \{0, 1, \#\}$. This can be done as follows. Let $A = \langle \Sigma, Q, q_0, \delta, F\rangle$ be a DFA over $\Sigma$. We define the encoding of $A$, denoted $\langle A\rangle$, as follows: $$ \langle A\rangle = \langle \Sigma\rangle \#\# \langle Q\rangle \#\# \langle q_0\rangle \#\# \langle \delta\rangle\#\# \langle F\rangle,$$ where $\langle C\rangle$ is an encoding of $C$ over $\Sigma'$, for all $C\in \{ \Sigma, Q, q_0, \delta, F \}$, to be defined below.
We distinguish between several cases:
$C = \Sigma$: we can encode the letters in $\Sigma$ as increasing binary numbers over $\{ 0, 1\}$. For example, if $\Sigma = \{ a_0, a_1, \ldots, a_7\}$, then we encode $a_i$ as the binary representation of the number $i$, and so $\langle a_i \rangle = \text{the binary representation of the number $i$}$. Then, we take $$\langle \Sigma \rangle = \langle a_0\rangle\# \langle a_1\rangle \# \cdots \# \langle a_7\rangle = 000\# 001 \# \cdots \#111$$ so the $\#$'s seprate between different letters.
$C = Q$: also here, we can encode the states in $Q$ as increasing binary numbers.
$C = q_0$: already encoded in the previous item. Let's say we encoded $q_0$ as the number 0 in binary.
$C = \delta$: the encoding of $\delta$ can be simply a list of the transitions in $\delta$ separated by $\#\#\#$: $$ \langle \delta\rangle = \langle t_1\rangle \#\#\#\langle t_2\rangle\#\#\# \cdots \#\#\# \langle t_k \rangle,$$ where $t_i$ is the $i$'th transition of $A$. So it remains to show how we encode a general transition. A transition $t = q \xrightarrow{a} s$ can be encoded as $\langle t\rangle=\langle q \rangle \# \langle a\rangle\# \langle s\rangle$. Note that we already encoded states and letters, so the latter is well-defined.
$C = F$: since we already encoded the set of states $Q$, $\langle F\rangle$ can be simply encoded as the list of numbers describing the states in $F$ separated by $\#$'s.
What we've in total is a description of the DFA $A$ as a finite word over the constant alphabet $\Sigma'$. As the set $\Sigma'$ is countable, then there are at most $\aleph_0$ DFAs over $\Sigma$.
Note: the format of the encoding can be different. Its just one naive way to do it. The idea is to encode a DFA by an object that you know has countably many instances.