Prove the boundary of rationals is real

1. Aug 13, 2007

kekido

1. The problem statement, all variables and given/known data

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

2. Relevant equations

3. 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?

2. Aug 13, 2007

HallsofIvy

Staff Emeritus
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}.)

3. Aug 13, 2007

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).

4. Aug 13, 2007

kekido

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 $$\partial(Q^c)=R$$