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.

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    $\begingroup$ For (2), a mathematician would say "Baire Category Theorem" instead of repeating its proof. Let $H_n$ be $F_n$ minus its interior. Then $\bigcup H_n$ is a complete metric space, but it has been written as a countable union of nonempty closed sets with empty interior, impossible. $\endgroup$ Dec 10, 2010 at 22:09
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    $\begingroup$ I'd say "a book of mathematics" rather than "a mathematician" -a methematician is a less deterministic object ;-) $\endgroup$ Dec 10, 2010 at 22:43
  • $\begingroup$ For those interested: the paper by Sierpiński is now available online here: journalarchive.jst.go.jp/english/… $\endgroup$ Dec 12, 2011 at 20:18
  • $\begingroup$ Updated link for the Sierpiński article: jstage.jst.go.jp/article/tmj1911/13/0/13_0_300/_article/-char/… $\endgroup$
    – PatrickR
    Jan 30 at 5:59

4 Answers 4


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.

  • $\begingroup$ I like this argument, this is really a much nicer way to put it, than I did, thanks! $\endgroup$ Dec 11, 2010 at 0:48

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.

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    $\begingroup$ cov(B) can be improved to $\mathfrak d$ (applying basically the same argument). $\endgroup$ Nov 11, 2017 at 18:23
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    $\begingroup$ The idea is to use the characterization of $\omega^\omega$ as a unique Polish zero-dimensional nowhere locally compact space. Given an uncountable partition $\mathcal P$ of $[0,1]$ into closed sets, we simply choose sufficiently large countable subfamily $\mathcal P'$ to make the difference $X=[0,1]\setminus \bigcup\mathcal P'$ nowhere locally compact and zero-dimensional. Then $X$ will be a copy of $\omega^\omega$ and $\mathcal P\setminus\mathcal P'$ will be a cover of $X$ by compact subsets. $\endgroup$ Nov 12, 2017 at 6:04
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    $\begingroup$ @Victor $\mathfrak d$ is the cofinality of the partially (pre)ordered set $\omega^\omega$, see en.wikipedia.org/wiki/Cardinal_characteristic_of_the_continuum $\endgroup$ Oct 22, 2018 at 4:51
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    $\begingroup$ @Victor $\mathfrak d<2^\omega$ is consistent relative to ZFC. It holds, for example, in models obtained by adjoining a lot of random reals to increase the cardinality of the continuum. $\endgroup$ Oct 22, 2018 at 17:43
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    $\begingroup$ @Victor Almost, only the cardinal $\mathfrak d$ should be replaced by the cardinal $\acute{\mathfrak n}$ (see mathoverflow.net/questions/285780). So, a cofinite space $X$ is path-connected if and only if $|X|\ge\acute{\mathfrak n}$. $\endgroup$ Oct 22, 2018 at 23:13

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.

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    $\begingroup$ If I understand things correctly, it seems to me that the claim in your last paragraph seems to contradict this result of A. Miller: math.wisc.edu/~miller/res/cov.pdf, where he claims that the unit interval can be the union of $\omega_1$ many disjoint closed sets even when CH fails. $\endgroup$ Mar 24, 2018 at 18:21
  • $\begingroup$ I think you are perfectly right: it may happen that the only partition in closed intervals which is coarser than $\mathcal{R}$ has only one equivalence class. So the suggested argument could not be applied to conclude $\endgroup$ Mar 25, 2018 at 11:06

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.

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    $\begingroup$ Well, you need something: any function from an interval to an indiscrete space is continuous, so indiscrete spaces are path-connected. They cannot be arc-connected if they are countable. That a path-connected Hausdorff space is arc-connected is a pretty non-trivial theorem. $\endgroup$
    – Todd Trimble
    Apr 17, 2015 at 0:57
  • $\begingroup$ Another nice example: the line with two origins has a path from one origin to the other, but no arc from one to the other. And it seems to be the blatant non-Hausdorffness at those origins that makes it work. $\endgroup$
    – Hew Wolff
    Oct 15, 2023 at 19:36

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