Boundary, closure, and interior of $\{(x,y)\in \mathbb{R}^2 \;|\; x \in \mathbb{Q} \text{ and }y>0\} \subseteq \mathbb{R}^2$ 
Find the closure, boundary, and interior of the following subset of $\mathbb{R}^2$:
  $$
A=\{(x,y)\in \mathbb{R}^2 \;|\; x \in \mathbb{Q} \text{ and }y>0\}
$$

It is clear that the closure is the whole upper place along with the $x$-axis as any point in this set must be a limit point of $A$. What I am puzzling over is the $\text{int } A$ and $\operatorname{Bd}A$.
The way I think of it is that as 
$$
\operatorname{Bd}A=\overline{A} \cap \overline{X-A}
$$
Now $\overline{A}$ is the entire upper plane so $\operatorname{Bd}A$ must be a subset of that. But $X-A$ is just the set of all $(x,y)$ with $x\in \mathbb{Q}^C$ and $y>0$. Then again any point in the upper plane is a limit point of $X-A$ so that $\overline{A}=\overline{X-A}$ then $\operatorname{Bd}A=\overline{A}$. Is this correct? I feel this just a special case of what I assume is the generalization: if $X$ is a topological space with dense subset $A$ then $\operatorname{Bd}A=\overline{A}$.
As for $\operatorname{int}A$, I am a bit more lost. I do not believe that $A$ contains any open sets in $\mathbb{R}^2$ as $A$ should be totally disconnected and singleton sets in $\mathbb{R}^2$ are not open. Then is it the case that $\operatorname{int}A=\emptyset$?
 A: Your intuition that the interior is empty is correct. To show this, you might use that the complement of the interior is the closure of the complement. Based on the work you've already done, I suspect that you can easily determine $\overline{X-A}$. (However, note that $X-A$ includes all points with $y\leq 0$).
It should be noted that $A$ is not totally disconnected, since it contains the "half-lines" $$L_q:=\{(x,y): x = q, y>0\}$$ for each $q \in \mathbb Q$.
A: Note that since $\mathbb{R} \setminus \mathbb Q$ is dense in $\mathbb{R}$, it follows that every open set in $\mathbb R^2$ must include points whose $x$-coordinate is irrational. From this and the fact that no point of $A$ has an irrational $x$-coordinate it is easy to conclude that $\operatorname{int} A = \varnothing$.
As for the boundary, using the "complement of the interior is the closure of the complement" idea from Morgan O's answer, it actually follows that $$\operatorname{Bd} A = \overline{A} \setminus \operatorname{int} A$$
(since $\overline{A} \cap \overline{ \mathbb R^2 \setminus A } = \overline{A} \cap ( \mathbb R^2 \setminus \operatorname{int} A ) = \overline{A} \setminus \operatorname{int}A$). So putting together the ideas from the other parts allows for a solution to this.
