I'm wondering whether it is possible to Partition the closed interval $[0,1]$ into closed, countably infinite sets.

The only observations I could make were as follows:

  1. When we remove the endpoints, we effectively end up with $\mathbb R$, and can consider the image of $$\mathscr S :=\{r+\mathbb N\mid r\in [0,1)\}$$

    1. Because every $S\in \mathscr S$ is bounded, it must contain at least one limit point. Because our partition itself must be uncountable, $\bigcup_{S\in \mathscr S} L(S)$ ist uncountable, and by second countability, has uncountably many limit points.

    2. A natural example for a partition into countably infinite sets is $\{r+\mathbb Q\}$, but these elements are not closed – on the contrary, every sets closure is $\mathbb R$ already.

I'm not really sure where to go from there. My intuition says that this won't work, because it would get “too crowded” near the boundary points, but I'm not sure how to make this precise.


Yes. You can build such a partition by transfinite recursion.

Specifically, we construct a sequence of disjoint closed countably infinite subsets $A_\alpha\subset[0,1]$ where $\alpha$ ranges over the ordinals. Having constructed $A_\beta$ for all $\beta<\alpha$, let $S=[0,1]\setminus\bigcup_{\beta<\alpha}A_\beta$. If $S$ is uncountable, then there is some sequence of distinct elements of $S$ which converges to an element of $S$ (if not, every element of $S$ would be isolated in $S$ and $S$ would be discrete, but there is no uncountable discrete subset of $[0,1]$). Pick such a sequence and let $A_\alpha$ consist of the terms of the sequence together with their limit.

If $S$ is countable, we instead halt the recursion, and put each element of $S$ into a different one of the $A_\beta$ for $\beta<\alpha$ (there must have been uncountably many $\beta<\alpha$ so we can do this, and the sets will remain closed after adding a single point).

Since $[0,1]$ is a set, the recursion must halt eventually, and we obtain a partition of $[0,1]$ into countably infinite closed sets.

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  • $\begingroup$ Nice! Gonna verify this when I'm less tired. Slightly OT: how did you come up with this? In particular, in what area do such constructions arise naturally? I honestly didn't expect transfinite recursion to come into play here. $\endgroup$ – Lukas Juhrich Sep 7 '18 at 22:33
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    $\begingroup$ This is just the basic "greedy algorithm": if you want to construct a collection of objects with some properties, try constructing them one by one. If your collection is going to be uncountable, this will involve transfinite recursion. $\endgroup$ – Eric Wofsey Sep 7 '18 at 22:45
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    $\begingroup$ @Luke Transfinite recursion is a very useful hammer - any time you're trying to build a collection of sets with some property, it's a good idea to see if there's a "greedy approach" via transfinite recursion. $\endgroup$ – Noah Schweber Sep 7 '18 at 22:46
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    $\begingroup$ Gosh. I've heard of transfinite induction for years, and always thought, "I should learn what that actually is...", and now with this example, you've given me a quick and practical lesson about (a) what it is, and (b) how it's used. MSE has given me some great opportunities to improve my explanatory skills, but also to learn new mathematics. Thanks! $\endgroup$ – John Hughes Sep 7 '18 at 22:58
  • $\begingroup$ @bof I didn't claim that it is (and that's part of why I called it a hammer in the first place). $\endgroup$ – Noah Schweber Sep 8 '18 at 0:12

It may be worth pointing out that a partition of $[0,1]$ into countably infinite closed sets can be constructed effectively, without using transfinite induction or the axiom of choice.

Let $C$ be the Cantor set. It is well known that we can define a bijection $h:(\frac13,\frac23)\to C.$

If $(a,b)$ is a connected component of $[0,1]\setminus C$ — in other words, $(a,b)$ is one of the "middle thirds" removed from $[0,1]$ in forming the Cantor set — define a bijection $h_{a,b}:(a,b)\to C\cap[2a-b,2b-a].$ No arbitrary choices are needed for this: $[2a-b,2b-a]$ the closed interval of which $(a,b)$ was the deleted middle third, so $C\cap[2a-b,b-2a]$ is similar to the Cantor set, and we can define $h_{a,b}$ as a composition of $h$ with the obvious affine maps: $(a,b)\to(\frac13,\frac23)\to C\to C\cap[2a-b,2b-a].$

Next, I define a map $f:[0,1]\to C.$

If $t\in C,$ let $f(t)=t.$

If $t\in[0,1]\setminus C,$ let $f(t)=h_{a,b}(t),$ where $(a,b)$ is the connected component of $[0,1]\setminus C$ containing $t.$

Finally, for each $x\in C,$ define $A_x=f^{-1}(x).$

Clearly, $\{A_x:x\in C\}$ is a partition of $[0,1],$ and each set $A_x$ is countable, since it contains exactly one point of $C$ and at most one point in each component interval $(a,b)$ of $[0,1]\setminus C.$ A little consideration will show that $A_x$ is also infinite and closed; in fact, it consists of the point $x$ and a sequence of points converging to $x.$

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    $\begingroup$ This is very clever. I was wondering if there was an explicit way to do it. :) $\endgroup$ – Eric Wofsey Sep 8 '18 at 1:26

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