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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 Sierpiński's lemma.

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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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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
    
And before that, I asked this on MSE: math.stackexchange.com/questions/6314/… –  Kevin Ventullo Apr 18 at 23:25

1 Answer 1

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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