Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
add comment

3 Answers

up vote 5 down vote accepted

No, the cardinality of their complements can be as big as that of all of the real numbers. For an explicit example, the complement of the Cantor set is dense and open. It consists of all of the "middle third" open intervals removed from $[0,1]$ together with the unbounded intervals $(-\infty,0)$ and $(1,\infty)$. The Cantor set is a closed uncountable set whose connected components are points.

The complement of a dense open set can be "big"in respects other than cardinality. Let the rational numbers be listed as $r_1,r_2,r_3,\ldots$, and for each $k$ let $I_k$ be an open interval of length $1/2^k$ containing $r_k$. Then $U=I_1\cup I_2\cup I_3\cup\cdots$ is a dense open subset of $\mathbb{R}$, but $U$ is a union of intervals the sum of whose lengths is $1/2 + 1/2^2 + 1/2^3+\cdots=1$. So in a sense the complement of this dense open set must take up "most of the space" in the real numbers.

share|improve this answer
add comment

No; the complement of the Cantor set is a dense open set in $\mathbb{R}$ that is not cocountable.

The Cantor set is closed and uncountable, so its complement is open. By the by, the Cantor set is a closed set that is uncountable but not an interval; in fact, the Cantor set has empty interior, so it cannot contain any intervals. Every open neighborhood of every point in the Cantor set contains both points of the Cantor set and points not in the Cantor set, so the complement is dense.

share|improve this answer
add comment

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.