# Negation of Statement (Definition of Boundary Point in Real Analysis) using First Order Logic

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?

• What the negation of $x \ne 2$? You might want to look at your first negation... Commented Jul 28, 2019 at 4:52
• @John Hughes sir, it will be $x=2$. Commented Jul 28, 2019 at 4:57
• 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 Commented Jul 28, 2019 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 $$\exists \epsilon > 0,\ N(a, \epsilon) \color{red}{\cap} S = \emptyset\ \lor\ N(a, \epsilon) \color{red}{\cap} S^\complement = \emptyset$$