Prove the boundary of rationals is real

[kekido]

Homework Statement

Let Q be the set of all rational numbers
Prove bd(Q)=R

Homework Equations

The Attempt at a Solution

Let x be a real number, then since the interval |x-r| contains both rationals and irrationals for arbitrary small r, so R is the boundary of Q. Is that right?

 

[HallsofIvy]

Slow down a bit. What, exactly, is the definition of "boundary" and "boundary point". If you can answer that you should be able to answer your own question. (I might point out that, strictly speaking, |x-r| is not an interval, it is a number! You mean either the interval (x-r, x+r) or the set {y| |x-y|< r}.)

 

[CompuChip]

I would assume that

• The boundary is the closure minus the interior
• A point is in the closure if every epsilon-ball around it has non-empty intersection with the set
• A point is in the interior if there is an epsilon-ball which fits entirely in the set

From these definitions it is quite straightforward to prove what the boundary of |R is, given (or proven) that the rationals are dense.

So kekido, your idea is right, but if you want to do it rigorously you should put a little more work into it (show that it satisfies the definitions yadayada andsoforth).

 

[kekido]

Slow down a bit. What, exactly, is the definition of "boundary" and "boundary point". If you can answer that you should be able to answer your own question. (I might point out that, strictly speaking, |x-r| is not an interval, it is a number! You mean either the interval (x-r, x+r) or the set {y| |x-y|< r}.)

Ok, here "boundary" is the set of all boundary points, i.e., \partial(Q)=R

You're right, the interval should be like what you said. I was being sloppy here.

Since the boundary point is defined as for every neighbourhood of the point, it contains both points in S and S^c, so here every small interval of an arbitrary real number contains both rationals and irrationals, so \partial(Q)=R and also \partial(Q^c)=R