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

I happen to have been thinking about this question recently. The proof I like uses the fact that a nested sequence of open intervals has non-empty intersection provided neither end point is eventually constant. Now one inductively constructs a sequence of such intervals as follows. Each interval is a component of the complement of the union of the first n closed sets, for some n. Then wait till the next closed set intersects that interval. (If it never does, then we're trivially done.) It cannot fill the whole interval, and indeed must miss out an interval at the left and an interval at the right. So pass to one of those subintervals in such a way that your left-right choices alternate. Done.

PS The question (with closed intervals instead of closed sets) was an exercise on the first sheet of Cambridge's Analysis I course last year.

Here's a variant of the argument that gives a slightly stronger result, by making explicit the use of the Baire category theorem that is "hidden" in some of the previous proofs.

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.

Notice that countability was used only in order to apply the Baire category theorem. That theorem might (consistently with ZFC) also hold for some larger cardinals (though of course not for any as large as the cardinal of the continuum). One of the many well-studied cardinal characteristics of the continuum is the minimum number of meager sets needed to cover the real line. I call this characteristic cov(B) (for "covering for Baire"); the "cov" part is standard but other authors replace B by M (for "meager") or K (for "Kategorie"). In any case, the proof above shows (after a little work to make sure the $B$ in the proof is sufficiently similar to the real line for Baire category purposes) that $[0,1]$ cannot be partitioned into fewer than cov(B) pairwise disjoint closed sets. I don't know whether the bound cov(B) can be improved here.

Here is my proof of 2 (although all proofs are similar, I guess).

Assume by contradiction there is such a countable closed partition of $I:=[0,1]$. Let's define inductively a function $f_i$ on each closed set $F_i$ this way:

Let $f_0=0$ and $f_1=1$ be constant. For $i>0$, $f_{i+1}$ is defined to be constant on (the trace on $F_{i+1}$ of) each connected component $J$ of $I\setminus \left( F_0 \cup F_1 \cup\dots\cup F_i\right)$, precisely, on $F_{i+1}\cap J$ it equals exactly the mean of the (already assigned) values taken on the end-point(s) of the interval $J$ . These functions glue together to a continuous, nonconstant $\mathbb{Q}$-valued function $f$, a contradiction.

To show the continuity of $f$, it is useful to observe that it is monotone on each connected component of the co-set of $F_0 \cup F_1$ (actually, increasing on $I$ if we assume, wlog, that $F_0$ is all less than $F_1$.

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[edit] A variation of the above proof would (edit: would not) lead to the following argument: any closed partition of $I$ with more than one class has the cardinality of continuum. Sketch: we may assume that $0$ and $1$ are not in the same class. Define suitably a coarser equivalence relation so that the quotient $I/ \mathcal{R}$ is a totally ordered complete set, so it has continuum many elements (edit: provided it has at least two elements, which in general may not be true) and a fortiori as many are the classes of the initial partition.