Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

Can someone please help clear up my confusion? Thanks!

share|improve this question
1  
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
1  
You would indeed. –  Brian M. Scott Jan 31 '13 at 3:25
add comment

1 Answer 1

up vote 2 down vote accepted

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 $A$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.