# Interior, Derived, and Boundary Sets of A

The following question is from Fred H. Croom's book "Principles of Topology"

In $\mathbb{R}^n$, let $R$ denote the set of points having only rational coordinates and $I$ its complement, the set of points having at least one irrational coordinate. Prove that

1. int$R$ = int$I$ = $\emptyset$.
2. $R^{'}$ = $I^{'}$ = $\mathbb{R}^n$.
3. bdy$R$ = bdy$I$ = $\mathbb{R}^n$.

I have an idea for the first part.

Part1: Since every open ball contains both a rational and irrational, it follows that $R$ as well as $I$ have no interior points. If $x$ were an interior point of $R$, then there would exist $\delta >0$ such that $B(x,\delta) \subset R$. However, we know this cannot happen since there is at least one irrational number in the ball. The same argument works for $I$. Therefore, int$R$ = int$I$ = $\emptyset$.

Does this make sense for the first part? I am rather confused on how to approach the second and third part. Any suggestions?

I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

• $I$ should be having irrational points only. Otherwise it could have non empty interior. – Math Wizard Feb 17 '15 at 7:10

Yes, your part 1 proof is good.

But now you're already 90% of the way to #2. If every ball contains points of both $R$ and $I$, what can you say about the set of limits points of $R$ and $I$, respectively?

And 3 is just another minor variation on this same theme. What's the definition of the boundary of a set? Isn't it exactly what you said about every ball in part 1?

Part 2: Every irrational point is a limit point of some rational points since in any neighborhood of an irrational point, there is at least one rational point. Contrary of it that every rational point is a limit point of some irrational points also holds for the same reason. So

$R'= R^n$ and $I'=R^n$

Part 3: use $∂(A)= \overline{A} - A^o$ and $\overline{A}=A'\cup A$, where

$∂(A)$ is boundary of $A$

$\overline{A}$ is the closure of $A$

$A^o$ is the interior of $A$

$A'$ is the limit point set of $A$.

In this problem:

$∂(R)= \overline{R} - R^o= R \cup R^n-\varnothing=R^n$

$∂(I)= \overline{I} - I^o= I \cup R^n-\varnothing=R^n$