# How is $L^* - \{\epsilon \} \neq L^+$?

I was asked which among the following is true:

1. $$\Sigma^*-\{\epsilon\} = \Sigma^+$$
2. $$L^* - \{\epsilon \} = L^+$$

As I can see, both $$\Sigma^*$$ & $$L^*$$ are sets. I thought both were true because of set difference, but the answer lists only the first option as correct and the second as false. How so?

If $$\varepsilon \in L$$, then necessarily $$\varepsilon \in L^+$$ (and the converse as well). This is because $$L$$ itself is contained in $$L^+$$ and $$L^+$$ is defined as the union over the powers $$L^i$$ of $$L$$ for $$i \in \mathbb{N}_+$$. Note $$L^+$$ is not defined as $$L^\ast \setminus \{ \varepsilon \}$$; this is a common mistake.
Hence, $$\varepsilon \not\in L^+$$ holds if and only if $$\varepsilon \not\in L$$. This is the case for the alphabet $$\Sigma$$, hence why it is correct in the first case.
• Can't $\sum = \{\epsilon\}$ or $\epsilon \in \sum$? – Mr. Sigma. Dec 20 '18 at 13:17
• @Mr.Sigma. Well, it depends on what $\Sigma$ is. Usually it is meant as an alphabet, so defining it like that would mean $\varepsilon$ is a symbol; I have no clue how you would separate that from the empty word $\varepsilon$ (though I could imagine Machiavellian examiners abusing such things). If $\Sigma$ is meant as a language, then, sure, $\Sigma = \{ \varepsilon \}$ is possible (and not ambiguous). Note $\Sigma$ cannot be just a "set" since it must contain words; otherwise, the definition for $\Sigma^+$ simply does not work because $\Sigma^i$ is undefined (regarding concatenation). – dkaeae Dec 20 '18 at 13:20
• @Mr.Sigma. With "null" you mean the empty word $\varepsilon$? As I have said, I cannot imagine using $\varepsilon$ being used as a symbol except in the (very) restricted case of formulating trick questions. At any rate, $\Sigma$ should be given explicitly; if this is not the case, then $\varepsilon$ can safely be assumed not to be an element of it (since words should not be elements of alphabets, which are sets of symbols). – dkaeae Dec 20 '18 at 13:44
• 2. really doesn't have a yes/no answer because $L$ isn't specified, but what probably is meant there is whether $\forall L: L^* - \{\epsilon \} = L^+$, in which case the answer is no, as dkaeae explains. – reinierpost Dec 20 '18 at 14:40