The classical result of Sierpinski says that a compact connected Hausdorff space cannot be written as a countable union of nontrivial pairwise disjoint closed sets.

In $\mathbb{R}^n$, this implies that closed, bounded and connected subset cannot be decomposed in a countable union of nontrivial pairwise disjoint closed sets. Can we construct a connected subset of Euclidean space with countable decomposition of countably many pairwise disjoint closed sets? This is a question that I stumbled a few days ago.

Give this problem a try, can you construct such a set? It’s a lot of fun :). Post a comment if you come up with sth nice!

We begin with my approach, struggle, and a historical note. In the last section, “Sun set” I describe my construction.

Approach and historical note

First, I’ve found an example in $\mathbb{R}^3$. It’s quite easy but wordy, so here are pictures of it (third dimension, here depth, is only to make bridges to avoid intersections we see on 2D projection of this set).

Here are the details of this construction, if you are interested in it.

I didn’t manage to simplify this example to $\mathbb{R}^2$ since I needed an extra dimension to create “bridges” to avoid crossings that destroy pairwise disjointness. Some clever trick is needed to make a similar example on the plane.

I was dissatisfied with my 3D example. I didn’t see any reason why such an example needed to be in three dimensions. I wanted something simpler, something on a plane. I tried out some ideas and found an example in the plane. This has a different feel (more chaotic) than the first example. But whatever works!

I call it sun set and describe it in the coming section. Basically, it’s just a union of line segments with one endpoint on the unit sphere, and the other getting closer to $0$ in each segment.

A few days later I found that this question existed in the early days of topology (I was surprised to see it so late!). Anna Mullikin, in the 1922 paper “Certain Theorems Relating to Plane Connected Point Sets”, constructs a Mullikin nautilus. It is formed by taking a union of infinitely many polygonal chains. The first three are pictured below (screenshot from the original paper).

I was delighted to see this neat example. She has done something I couldn’t — her example has a similar feel to my first example in $\mathbb{R}^3$. But, with a clever shape, she obtained the needed limit points to “connect” parts without the necessity of my overkill “3d bridges”.

Sun set

Let $D = \{d_n : n=0,1,\dots\}$ be a dense subset of unit sphere with $d_n \neq d_m$. For each $n=1,2, \dots$ define the line segment $$ A_n = \{ \lambda d_n : 2^{-n} \leq \lambda \leq 1 \}. $$ And let $A_0 = \{ \lambda d_0 : 0 \leq \lambda \leq 1 \}$. Finally, let $\text{Sun} = \bigcup_n A_n$. It clearly is a union of closed, pairwise disjoint, connected sets.

Theorem. $\text{Sun}$ is connected.

Proof. Let $U,V$ be open disjoint sets with $\text{Sun} = U \cup V$. Without loss of generality assume that $U$ contains $0\in A_0$. Since $A_0$ is connected, it must be contained in $U$. It suffices to show that $A_1, A_2, \dots$ live in $U$ as well.

First, note that there is $N$ for which ball $B = B(0, 2^{-N})\cap A$ is contained in $U$. Consequently, for each $k>N$ an endpoint of $A_k$, namely $2^{-k}d_k$, is close enough to the origin to be contained in $B \subset U$. Hence, all $A_k$ with $k>N$ are in $U$ ( $A_k$’s are connected).

It remains to show that $A_1, A_2, \dots, A_N$ are also in $U$. Pick any $1\leq t \leq N$. By construction $d_t \in A_t$. Aiming at contradiction assume that $d_t$ sits in $V$. Then there is a neighborhood of $d_t$ with no points in common with $U$. This is impossible because $U$ contains $\{d_k : k>N \} \subset D$ which is dense in the unit sphere since it’s just $D$ with finitely many points thrown out. Hence $d_t$ belongs to $U$, and so $A_t$ is in $U$ ( $A_n$’s are connected).

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