# Finite Partitions of the Unit Interval

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?

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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

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.