Find the accumulation point(s) of $A = \{ (x,y) : x \in \mathbb{Q}, y \in \mathbb{N} \}$ 
Find the accumulation point(s) of $A = \{ (x,y) : x \in \mathbb{Q}, y \in \mathbb{N} \} \subset \mathbb{R}^2$ in

Let us denote the accumulation point(s) of $A$ by $A^\prime$.
I am torn between two possible answers, but I think that I am completely overthinking this question.
On the one side, I want to say that $A^\prime = \emptyset$
But on the other side I want to say
$A^\prime = \mathbb{I} \times \mathbb{N}$ where $\mathbb{I}$ denotes the set of irrational numbers.
The reason for this is:
Consider $(a,b)$ where $a \in \mathbb{I}, b \in \mathbb{N}$. We can say, because between every two irrational numbers there is a rational number, that for every $r>0, B^\prime((a,b),r)\cap A \neq \emptyset \implies \mathbb{I}\times \mathbb{N} \subseteq A^\prime$
Can someone PLEASE point me into the right direction and give me the correct reasoning to answer this question?

Your argument is correct, however, it not only holds for irrationals but rationals, hence, $A'=\mathbb{R}\times\mathbb{N}$.
