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.