Prove that every substring closed language L ⊆ {0, 1}* is regular

Question

For $x,y \in \{0,1\}^*$ a language $L ⊆ \{0, 1\}^*$ is called substring closed, if $y \in L$ and $x \preceq y$ ($x$ substring of $y$) implies $x \in L$.

I want to prove that every substring closed language $L ⊆ \{0, 1\}^*$ is regular.

Is it enough to have a $L$ such that just $x$ in it, then prove that $L$ is regular?

Answer 1

There are some substring-closed languages that are not regular.

Here is an example. Let $C=\{01^n0^n1\mid n\ge1\}$ and $F=\{f\mid \exists c\in C, f\preceq c\}$, i.e., $F$ is the language of all substrings of strings in $C$.

• $F$ is substring-closed, since a substring of a substring of string $c$ is also a substring of $c$.
• The intersection of $F$ and the regular language $\{01w01\mid w\in \{0,1\}^*\}$ is $C$, a non-regular language. So $F$ is not regular.

Answer 2

The result I know holds for SUBSEQUENCE closed languages, and not SUBSTRING closed. The confusion is frequent between english and french, since "subsequence" is translated into "sous-mot" which means litteraly "subword" or "substring"…

To be clear, $u=u_1u_2…u_n$ is a subsequence of $v=v_1v_2…v_m$ if and only if there exists $1\leqslant i_1 < i_2 < … < i_n \leqslant m$ such that $u = v_{i_1}…v_{i_n}$.

The proof I know is a bit long. I will add some details if necessary.

For $L$ any language, denote $\widehat{L}$ the set of words that have a subsequence in $L$: $$\widehat{L} = \{v\in\Sigma^*\mid \exists u\in L, u\preccurlyeq v\}$$

The proof goes as such:

• prove that for any sequence $(u_n)_{n\in\mathbb{N}} \in \left(\Sigma^*\right)^{\mathbb{N}}$, there exists two indexes $i<j$ such that $u_i\preccurlyeq u_j$;
• show that for any language $L$, there exists a finite language $F$ such that $\widehat{F} = \widehat{L}$;
• prove that if $L = \widehat{L}$, then $L$ is regular;
• conclude that if $L$ is a subsequence closed language, then it is regular.

Each step uses the result of the previous one.