What is the negation of the statement
$$\forall ~~\epsilon > 0,~~~~ N(a,\epsilon)\cap S \neq \emptyset \land N(a,\epsilon)\cap S^{\complement} \neq \emptyset~.$$
At first I changed $~\forall~$ to $~\exists~$, then I negate $~~N(a,\epsilon)\cap S \neq \emptyset~~$, it will be $~~N(a,\epsilon)\cup S =\emptyset~~$.
After that $~\land~$ will be $~\lor~$, in the last step $~~N(a,\epsilon)\cap S^{\complement} \neq \emptyset~~$ will be $~~N(a,\epsilon)\cup S^{\complement} =\emptyset~~$.
So negation of the whole statement will be $~~\exists ~~\epsilon>0,~~~~ N(a,\epsilon)\cup S =\emptyset\lor N(a,\epsilon)\cup S^{\complement} =\emptyset~~$.
Is this correct negation of the particular statement?
The negation of a statement like $\mathscr{A} \neq \mathscr{B}$, where $\mathscr{A}$ and $\mathscr{B}$ are any permissible formulas, is just $\mathscr{A} = \mathscr{B}$. $\underline{\textbf{Note that neither $\mathscr{A}$ nor $\mathscr{B}$ changes; only the $\ \neq\ $ changes to a $\ =\ $}}$.
So the negation of $N(a, \epsilon) \cap S \neq \emptyset$ is just $N(a, \epsilon) \cap S = \emptyset$. The $\ \cap\ $ does not change to $\ \cup\ $.
Similarly the negation of $N(a, \epsilon) \cap S^\complement \neq \emptyset$ is just $N(a, \epsilon) \cap S^\complement = \emptyset$. The $\ \cap\ $ does not change to $\ \cup\ $.
Hence your final answer should be $$ \exists \epsilon > 0,\ N(a, \epsilon) \color{red}{\cap} S = \emptyset\ \lor\ N(a, \epsilon) \color{red}{\cap} S^\complement = \emptyset $$