Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The Cantor set is uncountable so I expect countably many of them to be able to cover $\mathbb R$, but the set has measure $0$ so countably many of them also has set of measure $0$ and thus can't cover the real line.

Why/where is my intuition broken?

share|cite|improve this question
Cardinality arguments are great at showing that something is impossible but lousy at showing that something is possible. – Qiaochu Yuan Oct 14 '10 at 21:00
Perhaps it is worth mentioning that ${\mathbb R}$ and ${\mathbb R}\times{\mathbb R}$ have the same size. This means that there are subsets of ${\mathbb R}$ of the same size as ${\mathbb R}$ but such that we need $|{\mathbb R}|$ many of them to cover ${\mathbb R}$. Thus, the dichotomy countable/uncountable tells us little here. – Andrés E. Caicedo Jan 24 '11 at 1:10
up vote 34 down vote accepted

You can't cover a complete metric space with a countable union of closed nowhere dense subsets, by Baire's theorem. Cardinality tells us very little. Even infinite dimensional but separable Banach spaces have the same cardinality as lines, but perhaps here it is more intuitive that you will never cover such a space with countably many lines. Similarly, you can't cover the plane with countably many lines, polygons, conic sections, etc.

To see how Baire's theorem is more fundamental than measure here, note that the same is true for "fat" Cantor sets. That is, you take out smaller intervals to obtain Cantor-like sets with positive measure. These sets are still closed and nowhere dense, so by Baire's theorem the line is not a countable union of such sets, even though comparing measures wouldn't tell you this.

share|cite|improve this answer
Another way to obtain "large" nowhere dense sets is to remove "small" (in measure) dense open sets, which you can do by covering each element of a countable dense subset by smaller and smaller open intervals. – Jonas Meyer Oct 14 '10 at 17:50

It's not just that the Cantor set has measure 0, it's also that the reals are a Baire space and the Cantor set is nowhere dense. E.g. take a fat Cantor set, $C$, and look at $C + \mathbb{Z}$ which has infinite measure but is still be nowhere dense.

UPDATE: There is a related discussion on MO that could be of interest.

share|cite|improve this answer
Obviously our answers were written independently, and it's neat that we both use Baire and fat Cantor sets. – Jonas Meyer Oct 14 '10 at 17:22

Your Answer


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