Suppose there are sets $S\subseteq Q, T\subseteq Q$ such that $T=\epsilon (T),S=\epsilon (S)$.
Prove $\epsilon(S\cap T)\subseteq S \cap T$
Definition of $\epsilon$- closure for epsilon NFA is:
$\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 $\epsilon (S)$
And also, S is a set of all states in epsilon-NFA.
My proof:
Is this a correct reasoning?
Since your definition of $\epsilon$-closure isn't really a definition, it is impossible to prove anything using it. Instead, let me use the following definition: the $\epsilon$-closure of a set $S \subseteq Q$ consists of all states $x \in Q$ which are reachable from a state in $S$ by a (possibly empty) $\epsilon$-path (which is a path consisting of $\epsilon$-transitions).
Suppose that $\epsilon(S) = S$ and $\epsilon(T) = T$, and let $q \in \epsilon(S \cap T)$. Thus $q$ is reachable by a state $r \in S \cap T$ via an $\epsilon$-path. Since $r \in S \cap T$, in particular $r \in S$, and so, by definition, $q \in \epsilon(S) = S$. Similarly, $q \in \epsilon(T) = T$. Therefore $q \in S \cap T$.
