# Is an alphabet necessarily a regular language over it?

The definition of regular expressions says they are closed under union, concatenation, and Kleene star. Single symbols also form regular expressions/languages.

Since union is between two languages, does that mean that when the alphabet is infinitely countable, the language which is exactly the alphabet is not a regular language?

Thanks.

• Most definitions of regular languages are made for finite alphabets, and the definitions break down in a number of ways when you try to make the alphabet infinite. (+1) in the hopes people more knowledgeable than me find this, but I don't know of any canonical way to solve this problem. My answer would have to be "regular languages over countable alphabets aren't well defined". Jun 21, 2020 at 23:42

If $$A$$ is an infinite alphabet, then the set $$A$$ is not a rational subset of $$A^*$$. However, it is a recognizable subset of $$A^*$$. Indeed, let $$F = \{1, a, 0\}$$ be the finite monoid defined by $$aa = a0 = 0a = 0$$ and let $$f: A^* \to F$$ be the monoid morphism defined by $$f(c) = a$$ for each $$c \in A$$. Then $$f(A) = \{a\}$$ and $$f^{-1}(a) = A$$, since$$f(1) = 1$$ and $$f(u) = 0$$ if the length of $$u$$ is $$\geqslant 2$$.