# Does there exist a countable partition of [0,1] by disjoint closed subsets? [duplicate]

Possible Duplicate:
Why are the integers with the cofinite topology not path-connected?

As in the title, is it possible to find closed, disjoint subsets $C_n$ of $[0,1]$ such that $[0,1] = \bigcup_{n \in \mathbb N} C_n$?

Two observations:

1. If we do not demand a countable set of subsets, it is sufficient to take every point as a subset.

2. If we consider the particular case where the subsets are closed intervals, then it is not possible. To see that, it is sufficient and not difficult to show that the partition considered is a perfect subset of $\mathbb R$ and so it is not countable. This particular case is even known as SierpiĆski's lemma.

Suppose, toward a contradiction, that we had a partition of $[0,1]$ into countably many closed sets $C_n$; I'll write $B_n$ for the boundary of $C_n$ and $B$ for the union of the $B_n$'s. Observe that, if $p\in B_n$ then each open interval around $p$ meets $B_m$ for some $m\neq n$. (Proof: As $p$ is in the boundary of $C_n$, the interval contains a point $q$ that is not in $C_n$ and hence is in some other $C_m$. If $q\in B_m$ we're done, and otherwise we find a point in $B_m$ between $q$ and $p$.) This observation means that each $B_n$, considered as a subset of $B$, has empty interior. But $B$ is a closed subset of $[0,1]$ (because its complement is the union of the interiors of the $C_n$'s) and therefore a complete metric space. By the Baire category theorem, it cannot be covered by countably many closed sets $B_n$ with empty interiors, so we have the desired contradiction.