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I want to prove (or disprove) the following conjecture:

Let $\Sigma$ be a finite alphabet, and let $L \subseteq \Sigma^*$ satisfy that there are no $v, w \in L$ such that $v$ is a proper substring of $w$. Then $L$ is finite.

I have the intuition that this is true, haven't been able to find a counter example, but am unable to prove it. If this is true, how can we prove it? If this is false, how can we construct such infinite $L$?

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  • $\begingroup$ Note that you wrote $L \in \Sigma ^*$ in the question, but it should be $L \subseteq \Sigma ^*$. Set $\Sigma^*$ is a set of words, not a set of languages. $\endgroup$
    – Stef
    Dec 21, 2022 at 9:19

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Assuming that by substring you mean consecutive substring, take $$ L= \{10^n1 : n \ge 0\}. $$

On the other hand, if you refer to nonconsecutive substrings, or subsequence ordering, then Higman's Lemma states that any set of incomparable strings is indeed finite.

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