Dropping epsilon closure changes the accepted language

Let $Q=\{p,q\}$, $\mathrm{\Sigma }=\{a\}$, start $p$, and $F=\{q\}$, with $p\mathrm{\ne }q$. Put $\delta (p,\epsilon )=\{q\}$, $\delta (q,a)=\{p\}$, $\delta (p,a)=\mathrm{\varnothing }$, and $\delta (q,\epsilon )=\mathrm{\varnothing }$. All four domain entries are specified subsets of the finite state set, so this is an epsilon-NFA. Saturation from $\{p\}$ gives $\{p,q\}$ and then stays there; thus $E(\{p\})=\{p,q\}$.

In the correct construction the empty word reaches $\{p,q\}$, whose intersection with $F$ is $\{q\}$, so it is accepted. In the first defective construction the empty word leaves the start at $\{p\}$, whose intersection with $F$ is empty, so it is rejected. This mismatch occurs before any letter transition, even though all post-letter closures in that variant are retained.

In the second defective construction the start is correctly $\{p,q\}$, but on the single letter $a$ its successor union is $\delta (p,a)\cup \delta (q,a)=\mathrm{\varnothing }\cup \{p\}=\{p\}$. Omitting closure leaves that set, disjoint from $F$, so the variant rejects $a$. The correct construction instead closes it to $E(\{p\})=\{p,q\}$, which meets $F$; equivalently the original NFA has the accepting path $p\stackrel{\epsilon }{\to }q\stackrel{a}{\to }p\stackrel{\epsilon }{\to }q$.

Each defective rule still gives a total DFA on the finite power set: its letter outputs are subsets and each of its starts is a subset. The failures are therefore language failures, not missing transitions. The empty word distinguishes the first variant and $a$ distinguishes the second, refuting both universal preservation claims.