Find the closure of $A =\{ (x,y): x,y \in \mathbb{Q} \}$ 
Question:
Find the closure of $A =\{ (x,y): x,y \in \mathbb{Q} \}$ in $\mathbb{R}^2$

My attempts:
Let $\bar{A}$ denote the closure of $A$ and $A^\prime$ denote the accumulation point(s) of $A$.
We know that
$$\bar{A} = A \cup A^\prime$$
Now we can find $$A^\prime = \mathbb{R}^2$$
And hence
$$\bar{A} = A \cup \mathbb{R}^2 = \mathbb{R}^2$$
Is this correct?

Yes I believe you are correct. You can intuitively think of the closure of A as the set of all points for which you can construct an open ball, in which the open ball contains points in $\mathbb{Q}$, this is all of $\mathbb{R}$.
