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We know $L^+=L^\star\backslash\{\epsilon\}$. How do we interpret that $\{\epsilon\}^+=\{\epsilon\}$?

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    $\begingroup$ "We know that $L^+ = L^*\setminus\{\varepsilon\}$." No we don't. $\endgroup$ – David Richerby Apr 8 '17 at 23:14
  • $\begingroup$ @DavidRicherby I meant when $\epsilon\notin L$ and $L$ is an alphabet set $\endgroup$ – T.... Apr 9 '17 at 20:28
  • $\begingroup$ Then I suggest you edit your question to make it clearer. If you write $L$, everybody is going to assume you're talking about a language, not an alphabet. And when you write $\{\epsilon\}^+$, you are talking about applying $(\cdot)^+$ to a language, rather than the alphabet consisting of just the symbol $\epsilon", right? $\endgroup$ – David Richerby Apr 9 '17 at 20:51
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Your first definition only holds if $\epsilon \notin L$.

The Kleene star and plus are defined using recursive sets $V_i$. We have:

$$V_0 = \{\epsilon\}$$ $$V_1 = L$$ $$V_{i+1} = \{ vw : v \in V_i \wedge w \in L\}$$

Then:

$$L^* = \bigcup\limits_{i=0}^{\infty} V_i$$ $$L^+ = \bigcup\limits_{i=1}^{\infty} V_i$$

Note that $L^+$ omits $V_0 = \{\epsilon\}$, but does not subtract $\epsilon$ from $L$.

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