# Dense and open sets in R^n.

by MathematicalPhysicist
Tags: dense, sets
 P: 3,128 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?
 PF Patron HW Helper Sci Advisor P: 4,755 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.
 P: 3,128 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.
PF Patron
HW Helper
 P: 230 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.