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

Question

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 $n<|a|<n+\varepsilon$ 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 $A\in \mathcal{A},
n<|a|<n+\varepsilon$ 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?

Answer 1

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 $A\in \mathcal A, n<|a|<n+\varepsilon$ 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 $n<|a|<n+\varepsilon$ 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 $n<|a|<n+\varepsilon$ 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 "$n<|a|<n+\varepsilon$" 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 "$n<|a|<n+\varepsilon$" holds for all $a$ in this subset.

Now, your friend's negation

$∃𝑛 \in ℕ$ 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 "$n<|a|<n+\varepsilon$" does not hold for infinitely many $a \in A$, and "$n<|a|<n+\varepsilon$" holds for infinitely many $a \in A$ as well.