# Negating “$\forall n\in \Bbb N$ and $\forall\varepsilon>0$, $\exists A\in\mathcal A$ such that $n<|a|<n+\varepsilon$ for infinitely many $a\in A$.”

Let $$\mathcal A$$ be a collection of subsets of $$\Bbb R$$.

"$$\forall n\in \Bbb N$$ and $$\forall \varepsilon>0$$, $$\exists A\in\mathcal A$$ such that $$n<|a| for infinitely many $$a\in A$$."

What is the negation of above statement?

Is this the correct negation? "$$\exists n\in \Bbb N$$ and $$\exists\varepsilon>0$$ such that for each $$A\in \mathcal{A}, n<|a| holds for finitely many $$a\in A$$."

Some of my friends say that negation is: "$$\exists n\in \Bbb N$$ and $$\exists\varepsilon>0$$ such that for each $$A\in \mathcal A, n\ge|a|$$ or $$|a|\ge n+\varepsilon$$ holds for infinitely many $$a\in A$$."

Which one of these is correct?

As @Omnomnomnom said in his answer (provided that we intend "finitely many" as "only finitely many"), your statement

$$\exists n\in \Bbb N$$ and $$\exists\varepsilon>0$$ such that for each $$A\in \mathcal A, n<|a| holds for only finitely many $$a\in A$$

is the correct negation. Let us see why (it could be useful for negating similar statements).

The statement you want to negate is

$$\forall n\in \Bbb N$$ and $$\forall \varepsilon>0$$, $$\exists A\in\mathcal A$$ such that $$n<|a| for infinitely many $$a\in A$$

which can equivalently be reformulated as

$$\forall n\in \Bbb N$$ and $$\forall \varepsilon>0$$, $$\exists A\in\mathcal A$$ such that $$\exists \, \alpha \subseteq A$$ infinite such that $$n<|a| for all $$a\in \alpha$$.

It is clear that the negation of the last statement is the following:

$$\exists n\in \Bbb N$$ and $$\exists \varepsilon>0$$ such that $$\forall A\in\mathcal A$$ and $$\forall \alpha \subseteq A$$, if $$\alpha$$ is infinite then "$$n<|a|" does not hold for some $$a\in \alpha$$

which is clearly equivalent to your negation, because it says that it is impossible to find an infinite subset of $$A$$ such that "$$n<|a|" holds for all $$a$$ in this subset.

$$∃𝑛 \in ℕ$$ and $$∃𝜀>0$$ such that for each $$𝐴∈\mathcal{A}$$, $$𝑛≥|𝑎|$$ or $$|𝑎|≥𝑛+𝜀$$ holds for infinitely many $$𝑎∈𝐴$$
is not equivalent to your negation because, according to your friend's negation, it is still possible that "$$n<|a|" does not hold for infinitely many $$a \in A$$, and "$$n<|a|" holds for infinitely many $$a \in A$$ as well.
Your statement "$$\exists n\in \Bbb N$$ and $$\exists\varepsilon>0$$ such that for each $$A\in \mathcal A, n<|a| holds for finitely many $$a\in A$$." is the correct negation.
This is equivalent to the statement "$$\exists n\in \Bbb N$$ and $$\exists\varepsilon>0$$ such that for each $$A\in \mathcal A, n\ge|a|$$ or $$|a|\ge n+\varepsilon$$ holds for all but finitely many many $$a\in A$$." This is close to (but not quite the same as) your friends' statement.