Is $\mathbb{N}\times\mathbb{Q}$ neither an open set nor a closed set? The subset of the Euclidean space $R^2$ that I'm having trouble with determining whether it's closed or not is $A:=\mathbb{N}\times\mathbb{Q}$.
I'm trying to use the proposition that $\mathbb{Q}$ is dense in $\mathbb{R}$. I currently proved that $A$ is not open, but when you think about $A^c$, the problematic parts are (n,m) ($n\in \mathbb{N}$, $m\notin Q$). 
Whenever you take a open ball from this point, you can always find (n,p) ($n\in \mathbb{N}$, $p\in\mathbb{Q}$) within the ball since $\mathbb{Q}$ is dense in $\mathbb{R}$. 
So I thought that any open ball based on this point can't be in $A^c$. From here I concluded that A is not closed either. Is my proof correct?
 A: Yup, that's correct!
Another way to show that it's not closed, which might be easier, is the following: can you think of a sequence of points in $\mathbb{N}\times\mathbb{Q}$ which converges to some point in $\mathbb{R}^2$ which isn't in $\mathbb{N}\times\mathbb{Q}$? This boils down to finding a sequence of rationals which approach an irrational, and at the end of the day is basically identical to your proof, but might be intuitively easier.

As a side note: you might find it interesting to know that there are classifications of sets in topological spaces which go beyond just open and closed; the most common one is the Borel hierarchy. We start with open and closed sets at the bottom layer of this hierarchy; the next layer consists of countable unions of closed sets and countable intersections of open sets (called "$F_\sigma$" and "$G_\delta$", or "$\Sigma^0_2$" and "$\Pi^0_2$," respectively). It turns out that $\mathbb{N}\times\mathbb{Q}$ is $F_\sigma$ but not $G_\delta$. And the hierarchy goes beyond this, and in fact classifies all sets you can build from open sets and closed sets via countable unions and intersections (it turns out there are a lot of these).
The general study of classifying subsets of topological spaces beyond just open and closed is generally viewed as part of descriptive set theory.
A: Since $A$ is closed if and only if $\overline{A}=A$ everything boils down to finding the closure of $\mathbb N\times \mathbb Q$. But $$\overline{\mathbb N\times \mathbb Q}=\overline{\mathbb N}\times \overline{\mathbb{Q}}=\mathbb N\times \mathbb R\neq \mathbb N\times \mathbb Q$$ and therefore $\mathbb N\times \mathbb Q$ is not closed. 
To check $\overline{\mathbb N}=\mathbb N$ notice $$\overline{\mathbb N}=\mathbb N\cup \mathbb N^\prime$$ where $\mathbb N^\prime$ is the set of limit points of $\mathbb N$ whichy is empy.
