# Interior, Exterior Boundary of a subset with irrational constraints

Im quite new to topology and recently came across a problem which asked me to state the interior, exterior and boundary of a subset $A$ of a metric space $X$ where $A = \{x \text{ irrational}, y>0\}$ for some $(x,y) \in \mathbb{R^2}$.

The boundary at $y=0$ seems pretty obvious, but the $x$ part is quite bothersome for me.

I want to say that the interior($A$) = $\emptyset$ since you cant have any open balls along each vertical line because we can find a rational number arbitrarily close to a given rational (im not 100% on this)

Then similarly, the exterior would also be empty by just switching the words rational and irrational. If this is the case, all of A must be a boundary [since $A$ is the disjoint union of the int, ext, bd], but this seems silly to me.

Consider in $\Bbb R$ the set $\Bbb Q$ of rationals. Both $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$ are dense in $\Bbb R$, so every nbhd of every point of $\Bbb R$ intersects both $\Bbb Q$ and $\Bbb R\setminus\Bbb Q$. Thus, the boundary of $\Bbb Q$ is all of $\Bbb R$ (and so is the boundary of $\Bbb R\setminus\Bbb Q$). Your set, being $\Bbb R^2$, is a little more complicated, but the same principles apply. In particular, while all of $A$ is part of its boundary, the boundary of $A$ also includes a whole lot of points not in $A$. – Brian M. Scott Jan 31 '13 at 3:03
@BrianM.Scott After some more thought, I want to say that the boundary is $y\ge 0$ while the exterior is $y<0$. Would I be correct in thinking this? – MSEoris Jan 31 '13 at 3:23
Hint: A point is in the boundary of $A$ if every open ball around that point contains members both of $A$ and of its complement. Every real number $x$ (whether rational or irrational) has both rationals and irrationals arbitrarily close to it.
You are correct that the interior is empty. But the exterior is not empty. For example, $(0,-1)$ is in the exterior, because any sufficiently small ball around it contains no points of $A$.