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

Does the unit interval have a finite partition $P$ such that no element of $P$ contains an open interval? I would think that the answer is no, because each element of $P$ would have Lebesgue measure zero, but what if the elements of $P$ are not measurable?

share|cite|improve this question
There are many sets of positive measure, indeed measure $1$, that contain no interval. – André Nicolas Feb 4 '13 at 8:56
That's intriguing, do you have references to a few examples? – Herng Yi Feb 4 '13 at 12:10
Well, there is an example in the answer you accepted. If the complement of a counable dense set is too easy an example, one can produce fancier ones. – André Nicolas Feb 4 '13 at 17:24
up vote 4 down vote accepted

Indeed, you could use the Vitali set for such a partition but that would be overkill. Let $P_1=[0,1]\cap \mathbb Q$ and $P_2=[0,1] \setminus P_1$ then $P_1 \sqcup P_2=[0,1]$ but neither $P_1$ nor $P_2$ contain an open interval because any open interval contains both rational and irrational elements.

share|cite|improve this answer
Thanks for the answer! Haha I feel silly for overlooking this counterexample – Herng Yi Feb 4 '13 at 12:11

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.