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:
- 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?
- 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.
Best Answer
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.