# Dense and open sets in R^n.

1. Sep 22, 2008

### MathematicalPhysicist

I need to show that a countable intersection of open and dense sets (the sets are open and dense at the same time) in R^n is dense in R^n.

Now I heard someone used in the exam a theorem called berr theorem for which the above statement is an immediate consequence.
We haven't learnt this theorem so I guess there's a simple way to prove this.

Thus far, what I can see is that for n=1, an open subset of R is an open interval (a,b) and a dense set in R is mainly Q or variation of the rationals set, for example the set: Q(sqrt(2)={c| c=a+sqrt(2)*b where a,b in Q} etc, but I don't see how can a set be both dense and open in R, I mean open sets alone aren't dense in R, cause their closure is the closed interval.

Any hints?

2. Sep 22, 2008

### quasar987

Well for instance, The set R\{0} is open and dense in R.

The name is spelled Baire and the theorem is called "Baire category theorem" in case you give up and want to look up the proof.

3. Sep 22, 2008

### MathematicalPhysicist

a funny thing is that this theorem is included in point set topology but we haven't learnt it.
I wonder why on earth I took this course with this lecturer.

4. Sep 22, 2008

### quasar987

Well, it would be silly if you had seen the theorem, since you were asked to prove it. It is a difficult problem, but not impossible. For instance, it is problem 33 of chapter 3 of Elementary Classical by Marsden & Hoffman.

5. Sep 22, 2008

### MathematicalPhysicist

Yes I can see that it's not a tough theorem.

But, from what I understood the only pupil which ansewred this question correctly, only quoted the theorem and didn't prove it.
Usually Iv'e been told that theorems which aren't covered in the course and being used in exam should be proven, I guess as courses get more abstract they don't really care about it.

6. Sep 23, 2008

### mrandersdk

you can proove pretty simple like this:

A set A is dense if and only if $B \cap A$ is nonempty for all non-empty open sets B, (at least that is the definition we used for dense).

Then let U and V be open dense sets, you need to prove that $U\cap V$ is dense, so let let W be any open non-empty set. You need to show that $W \cap (U \cap V)$ is non-empty, but $W \cap (U \cap V) = (W \cap U) \cap V$ and because W and U are open $W \cap U$ is open, and because U is dense it is non-empty, then $(W \cap U) \cap V$ is a intersection of a non-empty open set, with V, and again this is dense, so that intersection must be non-empty. QED.

edit: didn't see you needed countable, this only works for finite, sorry about that.

Last edited: Sep 23, 2008
7. Sep 25, 2008

### tim_lou

Actually, this problem shouldn't be too hard... in fact, if I remember correctly, it was one of the Rudin's problem. One very very important thing you need is completeness. In fact, any complete metric space is Baire.

I guess for a give away hint: you just need to show any open set W contains a point in the countable intersection. By induction, you can always find an open set contained in U1∩U2∩...∩Un and W and , take all those points and you got a bunch of cauchy sequences (easily done by imposing some conditions in the induction, i.e. limit the diameter of the open sets). Take the limit of the cauchy sequence and done.

8. Sep 26, 2008

### MathematicalPhysicist

Yes, Iv'e seen the proof in Munkres.

This is also true for a compact Hausdorff space.