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

Question

An apparently elementary question that bugs me for quite some time:

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

Recall that the open sets in the cofinite topology on a set are the subsets whose complement is finite or the entire space.

Obviously, the integers are connected in the cofinite topology, but to prove that they are not path-connected is much more subtle. I admit that this looks like the next best homework problem (and was dismissed as such in this thread), but if you think about it, it does not seem to be obvious at all.

An equivalent reformulation of (1) is:

(2) The unit interval $[0,1] \subset \mathbb{R}$ cannot be written as a countable union of pairwise disjoint non-empty closed sets.

I can prove this, but I'm not really satisfied with my argument, see below.

My questions are:

1. Does anybody know a reference for a proof of (1), (2) or an equivalent statement, and if so, do you happen to know who has proved this originally?
2. Do you have an easier or slicker proof than mine?

---

Here's an outline of my rather clumsy proof of (2):

Let $[0,1] = \bigcup_{n=1}^{\infty} F_{n}$ with $F_{n}$ closed, non-empty and $F_{i} \cap F_{j} = \emptyset$ for $i \neq j$.

The idea is to construct by induction a decreasing family $I_{1} \supset I_{2} \supset \cdots$ of non-empty closed intervals such that $I_{n} \cap F_{n} = \emptyset$. Then $I = \bigcap_{n=1}^{\infty} I_{n}$ is non-empty. On the other hand, since every $x \in I$ lies in exactly one $F_{n}$, and since $x \in I \subset I_{n}$ and $I_{n} \cap F_{n} = \emptyset$, we see that $I$ must be empty, a contradiction.

In order to construct the decreasing sequence of intervals, we proceed as follows:

Since $F_{1}$ and $F_{2}$ are closed and disjoint, there are open sets $U_{1} \supset F_{1}$ and $U_{2} \supset F_{2}$ such that $U_{1} \cap U_{2} = \emptyset$. Let $I_{1} = [a,b]$ be a connected component of $[0,1] \smallsetminus U_{1}$ such that $I_{1} \cap F_{2} \neq \emptyset$. By construction, $I_{1}$ is not contained in $F_{2}$, so by connectedness of $I_{1}$ there must be infinitely many $F_{n}$'s such that $F_{n} \cap I_{1} \neq \emptyset$.

Replacing $[0,1]$ by $I_{1}$ and the $F_{n}$'s by a (monotone) enumeration of those $F_{n}$ with non-empty intersection with $I_{1}$, we can repeat the argument of the previous paragraph and get $I_{2}$.

[In case we have thrown away $F_{3}, F_{4}, \ldots, F_{m}$ in the induction step (i.e, their intersection with $I_{1}$ is empty but $F_{m+1} \cap I_{1} \neq \emptyset$), we put $I_{3}, \ldots, I_{m}$ to be equal to $I_{2}$ and so on.]

---

Added: Feb 15, 2011

I was informed that a proof of (2) appears in C. Kuratowski, Topologie II, §42, III, 6 on p.113 of the 1950 French edition, with essentially the same argument as I gave above. There it is attributed to W. Sierpiński, Un théorème sur les continus, Tôhoku Mathematical Journal 13 (1918), p. 300-303.

Answer 1

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.

Answer 2

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.

Answer 3

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 $F_1$.

$$*$$

[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.

Answer 4

There is a theorem that in each path-connected top. space any two different points can be joined by an injective continuous curve but I do not remeber if one needs Hausdorff for that theorem. If not, then you could get your statement (1) directly since there cannot be an injective map from [0,1] to Z.