TITLE: Dense Open Sets and Cocountability
QUESTION [5 upvotes]: Are all dense open sets in R cocountable?
That is, are all dense open sets in R such that their complements are at most countable?
It would seem like they must be since the closed sets that are uncountable are all intervals, so their complements are not dense.

REPLY [2 votes]: Here is something even more spectacular. Let $\epsilon >0$.  Let $z_n$ be an enumeration of all points in $R^d$ with all rational coordinates. We know this sequence's range is dense in $R^d$.  
About each $z_n$ choose an open ball of small radius $B_n$ whose volume is less than $\epsilon/2^n$.  Now write
$$U_\epsilon = \bigcup_{n=1}^\infty B_n.$$
This is a union of open balls and is therefore open.  Its total Lebesgue measure (volume) is less than $\epsilon$. It is an open dense subset of $R^d$.  An open dense subset of $R^d$, can be very small indeed.