# Homework Help: Boundary of the interior of the rationals

1. Sep 28, 2010

### Design

1. The problem statement, all variables and given/known data
S = Set of rational numbers
Boundary(interior(S)) = ?

3. The attempt at a solution
I have no Idea how to do this, I don't know what interior of the rational numbers are. Maybe you guys could give an example of like the interior of the natural numbers or the boundary of the natural numbers etc?

thank you

Last edited: Sep 28, 2010
2. Sep 28, 2010

### Design

Re: Topology

The question is given like this ∂(S^int)
But maybe I wasn't being clear with the notation so I just wrote it in words up there. Please need help on this!

3. Sep 28, 2010

### Office_Shredder

Staff Emeritus
Re: Topology

What's the definition of the interior of a set?

4. Sep 28, 2010

### Design

Re: Topology

X a point in the set is in the interior of the set if there exist radius r such that B(r,x) is a subset of S

r = radius, B(r,x) the open ball?

I don't see how this definition can help me with the set of the rational number tho because this definition is so abstract and it looks like we are talking about R^1

Last edited: Sep 28, 2010
5. Sep 28, 2010

### Dick

Re: Topology

We are talking about R^1. And in R^1 the 'open ball' B(r,x) is the open interval (x-r,x+r). Can that be a subset of the rationals?

6. Sep 28, 2010

### Design

Re: Topology

Umm no that cannot be a subset of the rationals since x-r/x+r can equal a irrational number.
Does this mean that the rational numbers have no interior? So what does that mean?
Also can you give a hint about the boundary part as well :)

7. Sep 28, 2010

### Dick

Re: Topology

What has (x-r)/(x+r) got to do with it? I'm asking you if every number in the interval (x-r,x+r) (that's the set of all numbers s such that x-r<s<x+r) can be rational? If you don't know just say so. Don't make some sort of goofy guess.

8. Sep 28, 2010

### Design

Re: Topology

I didn't mean to mean divide, I meant to say that x-r can equal a irrational number and so can x+r. So no all the numbers in that set can't be rational.

9. Sep 28, 2010

### Dick

Re: Topology

Ok, they can. But that's not the point. If you take B(1/4,1/2) then 1/2+1/4 and 1/2-1/4 are both rational. Does that mean every point in (1/2-1/4,1/2+1/4) is rational?

10. Sep 28, 2010

### Design

Re: Topology

No it doesn't, when you don't have a common denominator for the radius and the center adding each other for ball.

For instance B(1/65, 1/2)?

11. Sep 28, 2010

### Dick

Re: Topology

This is going nowhere fast. Could you look up what 'dense' means in topology and how that applies to the rationals and the irrationals in R^1? That might ring a bell about something you have been told that you need to solve this problem.

12. Sep 28, 2010

### Design

Re: Topology

We actually never covered anything about dense for toplogy. Just one chapter about interior,boundary and closure and an assignment on it. We are moving to limits for Advanced Calculus now (2nd year course).
I'm not really sure how to apply it to the rationals and the irrationals even after looking at a definition. I don't see it how it is related since we haven't established a subset from the rational numbers.

Last edited: Sep 28, 2010
13. Sep 28, 2010

### Inferior89

Re: Topology

Is it possible to chose an r>0 so that every number inside the interval x-r,x+r is rational?

14. Sep 28, 2010

### Design

Re: Topology

Are you asking for all x if so, Its not possible.

15. Sep 28, 2010

### Inferior89

Re: Topology

For any x.

16. Sep 28, 2010

### Design

Re: Topology

Is it not possible to choose an r>0 for any x that will the interval x-r,x+r rational.

Last edited: Sep 28, 2010
17. Sep 28, 2010

### Office_Shredder

Staff Emeritus
Re: Topology

Of course it's possible. But that's not relevant. The question is, does the set of rationals have any interior points? If so, then the whole interval (x-r,x+r) must consist only of rational numbers. Not just the endpoints

18. Sep 28, 2010

### Inferior89

Re: Topology

Read my question again. I didn't ask if x-r and x+r could be made rational but if it is possible to chose a r > 0 so that the interval [x-r,x+r] only contain numbers that is a subset of the rational numbers.

19. Sep 28, 2010

### Design

Re: Topology

The rational numbers do have some interior points. So what your saying is the interior of the rational numbers is the rational numbers where (x-r,x+r) are being satisfied?

I see what your saying now, your taking the ones that can be made rational and making them into a set?

20. Sep 28, 2010

### Office_Shredder

Staff Emeritus
Re: Topology

No, they don't

What do you mean by (x-r,x+r) is satisfied? (x-r,x+r) is an interval in the real numbers