Suppose, $A$ is a finite alphabet. $L \subset A^*$ is a language. Let's call $L$ concatenation-free iff $\forall u, v \in L$ we have $uv \notin L$.

Does there exist some function $c: \mathbb{N} \to (0; 1)$, such that for any finite language $L \subset A^*$, there exists a concatenation-free sublanguage $L_0 \subset L$, such that $|L_0| \geq c(|A|)|L|$?

The only thing I currently know about this problem, is that we can take $c(1) = \frac{1}{3}$. That is a direct consequence of Erdos-Sidon theorem, that states:

$\forall A \subset \mathbb{Z}$ $\exists$ a sum-free $A_0 \subset A$, such that $|A_0| \geq \frac{|A|}{3}$

However, I do not know how to deal with $|A| \geq 2$.

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$c(n)=1/3$ works for every $n$. Let $L$ be a finite language, $A$ be the multiset (say, nondecreasing sequence) of lengths of words in $L$. Then there exists a sum-free submultiser (subsequence) $B$ of $A$ of cardinality $\ge |A|/3$. Take $L_0$ to be the set of all words in $L$ whose lengths are in $B$. $L_0$ is concatenation-free and contains $\ge |L|/3$ words.See the paper "Sum-free subsets" by Alon and Kleitman https://documentcloud.adobe.com/link/track?uri=urn%3Aaaid%3Ascds%3AUS%3A038f5fe3-82b8-4e2a-bd71-53e1612d4dc9, Proposition 1.2.

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    $\begingroup$ $A$ should be defined as a multiset (but Erdos–Sidon still works) $\endgroup$ Mar 12 '20 at 15:07
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    $\begingroup$ Replaced "set" by "multiset" twice a @FedorPetrov suggested. $\endgroup$
    – user6976
    Mar 12 '20 at 15:25
  • $\begingroup$ What’s the name of the linked document? I can’t access it. $\endgroup$
    – user76284
    Mar 12 '20 at 16:21
  • $\begingroup$ It is a paper by Alon and Kleitman "Sum-free subsets". You can google it yourself. $\endgroup$
    – user6976
    Mar 12 '20 at 16:30

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