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.
Is this a correct reasoning?