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Let $L\subseteq \Sigma^*$ such that $\{\epsilon\}\not\in L$. Then for any $S\subseteq \Sigma^*, S=SL\implies S=\emptyset$.

So we suppose $S=SL$ and $S\ne\emptyset$. Then $\exists w\in S$ such that $0\le|w|\le |v|$ for some $v\in S$. Now $|w|\ne 0$, since $|w|=0\implies\{\epsilon\}=w\in S=SL\implies \{\epsilon\}=xy$ for some $x\in S$ and $y\in L\implies x=y=\{\epsilon\}\in L$ which contradicts that $\{\epsilon\}\not\in L$.
Thus $w\ne\{\epsilon\}$ and $\{\epsilon\}\not\in S$. So we have
\begin{align} w \in S &\implies w\in SL\\ &\implies w=sl\;\text{for some}\; s\in S\; \text{and}\; l\in L\\ &\implies |w|=|s|+|l| \end{align}
Here I'm stuck now. I think if I can show that $|l|=0$, then we can arrive at a contradiction.

Your help is highly appreciated.

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You should consider (one of) the word $w$ of minimal length in $S$, and show that for this word in particular, you can reach the conclusion that $|l| = 0$.

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  • $\begingroup$ So taking $|w|=1$, I can say that either $|s|=0$ or $|l|=0$. But $|s|=0$ is impossible, since $\{\epsilon\}\not\in S$ and hence $|l|=0$, a contradiction. Am I going right? But there is no guarantee that $|w|=1$. $\endgroup$ – Manjoy Das May 26 at 20:49
  • $\begingroup$ No need to do so much cases. Take $|w| = \min\{|u|, u\in S\}$. Then $w = sl$, but $s\in S$, so $|s| \geq |w|$, but we also have $|s| \leq |w|$ (since $w = sl$). $\endgroup$ – Nathaniel May 26 at 21:10
  • $\begingroup$ This leads to $|l|=1$. I am not getting $|l|=0$. $\endgroup$ – Manjoy Das May 26 at 21:13
  • $\begingroup$ No no, this leads to $|l| = 0$, since $|w| = |s|$. $\endgroup$ – Nathaniel May 26 at 21:15
  • $\begingroup$ ahh, my bad... I'm thinking of concatenations. Thanks!! $\endgroup$ – Manjoy Das May 26 at 21:17

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