Is there a language like L in which $\overline {L^*} = \overline L^*$?

Question

Assume that for every language L over the alphabet $\Sigma$, we define $L^*$ , $\overline L$ , $\Sigma^*$ & $L^n$ like this :

$L^n$ means joining L to itself n times.

For an alphabet like $\Sigma$, $\Sigma^*$ is the set of all strings made with 0 or more letters of $\Sigma$.

$L^*=L^0 \cup L^1 \cup L^2 \cup \cdots$

$\overline L = \Sigma^*-L$

Is there any language that holds this ?
$\overline {L^*} = \overline L^*$

Thanks in advance.

Answer 1

So you have:

$\overline {L^*}$ = all the gibberish strings that are not created by concatenating only legal words

${\overline L}^*$ = all the concatenations of gibberish strings

Since the latter may include legal words (think of a language where "a" is not a language string, but "aa" is), while the former must not contain any legal word (as a legal word is a concatenation with arity 0), a required condition is that your language must not have legal words that are composed by concatenating non-legal words.

A trivial way to achieve it is to have the language being defined over a strict subset of the alphabet:

If say alphabet = {0, 1}, and your L is "strings made only of 0", then your get:

$\overline {L^*}$ = all the strings with at least a 1

${\overline L}^*$ = all the strings with at least a 1

Note that this analysis breaks the moment you consider empty strings, as per the question you reported in the comment. This is because the language star always include the empty, and its complement doesn't. So, the formal answer is no, it's not possible, but if you ignore the empty string issue, it becomes achievable.

Answer 2

No, there can't be such a language.

By definition ${\overline L}^*$ always contains the empty string.

But $\overline{L^*}$ is the complement of something that contains the empty string, and so it doesn't itself contain the empty string.