Let M be an $\epsilon$-NFA and let $S\subseteq Q$. Prove $\epsilon (S)= \epsilon (\epsilon (S))$.

I would like to prove this by contradiction but I don't know if my idea is correct.

Definition of $\epsilon -closure$: $\epsilon : 2^Q \rightarrow 2^Q $

a) $S \subseteq \epsilon (S)$ Base case

b) If $q \in \epsilon (S)$ then $\delta(q,\epsilon )\subseteq \epsilon (S)$ Recursive case

c) and nothing else is in $\epsilon (S)$

Prove i) That $\epsilon (S) \subseteq \epsilon (\epsilon (S))$.

By contradiction, suppose $\exists x \in \epsilon (S)$ such that $x \notin \epsilon (\epsilon (S))$.

Using definitions of $\epsilon (S)$ , we know $S \in \epsilon (S)$. So $x \in S \in \epsilon (S)$, and $\delta(x,\epsilon ) \subseteq \epsilon (S)$.

Using definitions again, we know $S \in \epsilon (S)$, so $x \notin \epsilon (\epsilon (S))\notin \epsilon (S)$. Also $\delta (x, \epsilon) \nsubseteq \epsilon (\epsilon (S))$

We have a contradiction because $x \notin \epsilon (S)$ and we said $x \in \epsilon (S)$.

Therefore i) is true.


Your definition of $\epsilon$-closure is quite problematic. Here is a better formulation:

$\epsilon(S)$ is the intersection of all sets $T \subseteq Q$ such that (i) $T \supseteq S$ and (ii) if $q \in T$ then $\delta(q,\epsilon) \subseteq T$.

Here is a series of claims which imply $\epsilon(S) = \epsilon(\epsilon(S))$.

Claim 1. $\epsilon(S) \supseteq S$.

Proof. $\epsilon(S)$ is the intersection of sets containing $S$, and so contains $S$.

Claim 2. If $q \in \epsilon(S)$ then $\delta(q,\epsilon) \subseteq \epsilon(S)$.

Proof. If $q \in \epsilon(S)$ then $q$ belongs to all sets $T$ in the definition. Property (ii) implies that all of these sets contain $\delta(q,\epsilon)$, and so $\epsilon(S)$ contains $\delta(q,\epsilon)$.

Claim 3. $\epsilon(\epsilon(S)) \supseteq \epsilon(S)$.

Proof. Follows directly from Claim 1.

Claim 4. $\epsilon(\epsilon(S)) \subseteq \epsilon(S)$.

Proof. $\epsilon(S)$ satisfies the $T$-conditions for $\epsilon(S)$: (i) is trivial, and (ii) follows from Claim 2.

Claim 5. $\epsilon(\epsilon(S)) = \epsilon(S)$.

Proof. Follows from Claim 3 and Claim 4.

  • $\begingroup$ I don't really know how to put into words that $\epsilon (\epsilon (S)) \subseteq \epsilon (S)$. Is saying "Since $S \subseteq \epsilon (S)$ then $\epsilon (\epsilon (S)) \subseteq \epsilon (S)$ enough? $\endgroup$
    – Mandy
    Oct 14 '19 at 21:49
  • $\begingroup$ It's not convincing. What I wrote is just a sketch, which needs expansion. $\endgroup$ Oct 14 '19 at 21:51

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