Proof on showing whether the complement of the set $\mathbb Q\times\mathbb Q$ is open or close set The question is as follows:

Given set A = $\mathbb Q\times\mathbb Q$
Determine whether the complement set $ A^c$ is open or close.

Here are my thoughts :
1/ $\ A^c = \{\mbox{all points }(x,y)\mbox{ where at least one of }x, y \mbox{ is irrational}\}$.
2/ Known that $\mathbb Q$ is neither open or close, its Cartesian product $\mathbb Q\times \mathbb Q$ therefore is neither open or close.
3/ Thus the complement of that Cartesian product is neither open or close
Are my ideas ok ?
Would it be safe to say the following ?
Since $\mathbb Q$ and its complement have density, then for a sequence of points $(x,y)$ in $\ A^c$, we can always find a point $(a,b)$, which is an element of $A$, such that the sequence converges to $(a,b)$.   Thus $\ A^c$ is not closed
On the other hand, $\ A^c$ is not open since we can always find an element in this set, such that either the $1$st or $2$nd coordinate is a rational number.
Please help me on this question
Thanks in advance ^_^
 A: Your first three points are sufficient, and so yes, it's more than "okay": If you know $A = \mathbb Q \times \mathbb Q\;$ is neither open nor closed (I'm assuming that's with respect to $\mathbb R \times \mathbb R),\;$ then its complement $A^C$ (in $\mathbb R \times \mathbb R$) is neither closed nor open.  
There's no reason to say anything more.
A: Your conclusion is correct, and so is most of what you’ve written. This bit, however, is just backwards:

... for a sequence of points $(x,y)$ in  $A^c$, we can always find a point $(a,b)$, which is an element of $A$, such that the sequence converges to $(a,b)$.

What you should say is that if $\langle a,b\rangle$ is any point of $A$, there is a sequence of points of $A^c$ converging to $A$ (and therefore $A^c$ is not closed, and $A$ is not open).
Here’s a way to clean up the exposition without changing any of the ideas:

If $\langle p,q\rangle\in A$, let $\langle x_n:n\in\Bbb N\rangle$ be a sequence of irrational numbers converging to $p$. Then $\big\langle\langle x_n,q\rangle:n\in\Bbb N\big\rangle$ is a sequence in $A^c$ converging to $\langle p,q\rangle$, so $A^c$ is not closed (and $A$ is not open). If $\langle x,y\rangle\in A^c$, let $\langle p_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb Q$ converging to $x$, and let $\langle q_n:n\in\Bbb N\rangle$ be a sequence in $\Bbb Q$ converging to $y$. Then $\big\langle\langle p_n,q_n\rangle:n\in\Bbb N\big\rangle$ is a sequence in $A$ converging to $\langle x,y\rangle$, so $A$ is not closed, and therefore $A^c$ is not open.


If you know about the product topology on $\Bbb R^2$, you can use it instead of sequences. Let $U$ be any non-empty open set in $\Bbb R^2$. Then there are non-empty open sets $V$ and $W$ in $\Bbb R$ such that $V\times W\subseteq U$. $V\cap\Bbb Q\ne\varnothing\ne W\cap\Bbb Q$, so $U\cap A\supseteq(V\times W)\cap A=(V\cap\Bbb Q)\times(W\cap\Bbb Q)\ne\varnothing$, and therefore $U\nsubseteq A^c$. Since $A^c$ does not contain any non-empty open set, $A^c$ is clearly not open. On the other hand, there is an irrational $x\in V\cap\Bbb Q$, and $\{x\}\times W\subseteq A^c$, so $U\nsubseteq A$, $A$ also is not open, and therefore $A^c$ is not closed.
