Negating "$\forall n\in \Bbb N$ and $\forall\varepsilon>0$, $\exists A\in\mathcal A$ such that $n<|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 $n<|a|<n+\varepsilon$ 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?

 A: 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

$∃ \in ℕ$ and $∃>0$ such that for each $∈\mathcal{A}$, $≥||$ or $||≥+$ holds for infinitely many $∈$

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. 
A: 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 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.
