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?

  • $\begingroup$ What the negation of $x \ne 2$? You might want to look at your first negation... $\endgroup$ – John Hughes Jul 28 at 4:52
  • $\begingroup$ @John Hughes sir, it will be $x=2$. $\endgroup$ – Shubhadip Ghosh Jul 28 at 4:57
  • $\begingroup$ Negating a clause will not change $\cap$ to $\cup$, nor will it change $\cup$ to $\cap$. E.g. if a set $T$ is defined by $T=A\cap B$ then the negation of the clause $T\ne \phi$ is the clause $T=\phi$, and $T$ $ is $ $ A\cap B$ in both clauses $\endgroup$ – DanielWainfleet Jul 28 at 8:37

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 $$ \exists \epsilon > 0,\ N(a, \epsilon) \color{red}{\cap} S = \emptyset\ \lor\ N(a, \epsilon) \color{red}{\cap} S^\complement = \emptyset $$


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