[Math] Does the Kleene Closure of an alphabet contain an infinite string

automataformal-languagesregular-language

Suppose we have an alphabet $\Sigma$, does $\Sigma^*$ contain an infinite string? My reasoning is, since $\Sigma^*$ contains an infinite number of strings, one of those strings must have an infinite length, assuming there is no restriction on the length of strings.

My professor says that's wrong, and an automata can only process finite length strings. Is he correct?

As a corollary, suppose $\Sigma$ contains an infinite alphabet. Does $\Sigma^*$ contain an infinite string?

My answer was yes for that too, since a string might contain all elements from the alphabet and since the alphabet is infinite, the string must be too. But my professor says I'm wrong here too.

Can someone explain why he said that and where I'm going wrong here?

Best Answer

Your main mistake is right here:

My reasoning is, since $\Sigma^*$ contains an infinite number of strings, one of those strings must have an infinite length, assuming there is no restriction on the length of strings.

Consider the alphabet $\Sigma=\{a\}$. There is one string of length $0$: the empty string. There is one string of length $1$: $a$. There is one string of length $2$: $aa$. In fact, for each non-negative integer $n$ there is exactly one string of length $n$: $\underbrace{aaa\ldots aaa}_{n\text{ copies}}$.

There are no infinite integers, so each string is finite, but there are infinitely many strings in $\Sigma^*$.

The automata that you’re studying are defined in such a way that they process only finite strings. People have defined and studied automata that process infinite strings, but they are a significantly more advanced topic. In the context of what you’re studying, your professor is correct.

Whether $\Sigma$ is finite or not, the elements of $\Sigma^*$ are by definition the finite strings of members of $\Sigma$, so by definition $\Sigma^*$ does not contain any infinite words.

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