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

As the title, it is possible to find subsets of $[0,1]$ such that

$\displaystyle [0,1]= \bigcup_{n \in \mathbb N}^{\cdot} C_n $ where $C_n$ are closed and disjoint subset of $[0,1]$ ?

Two observations:

$ \cdot $ if we not consider a countable set of subsets it is sufficient to take every point as a subset.

$ \cdot $ if we study the particular case where the subsets are closed intervals the answer is no, Is not possible; to see that 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 a Sierpiński's Lemma)

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(Presumably you want all the $C_n$ to be nonempty as well.) It's a Baire category argument. A quick google leads back to MO; see this discussion: mathoverflow.net/questions/48970 –  Todd Trimble Nov 14 '12 at 1:00
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marked as duplicate by Andreas Blass, Todd Trimble, David Roberts, Terry Tao, Theo Johnson-Freyd Nov 14 '12 at 2:45

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1 Answer

This is essentially a duplicate of Why are the integers with the cofinite topology not path-connected? for which I gave an answer that included the following:

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.

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Your answer appeared 1 sec. before my comment. –  Todd Trimble Nov 14 '12 at 1:02
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