# split perfect set into countable many pairwise disjoint perfect sets

We know that each perfect set can be written as a continuum many pairwise disjoint many perfect set. This will rely on the well know theorem which says:

Let $$X$$ be a nonempty perfect polish space. Then there exists an embedding of $$C$$ (cantor set) into $$X.$$

My question is how can I write a perfect set as a countable many pairwise disjoint perfect sets ? Does anyone have an idea?

• @bof Do you mean $2^{\aleph_0}$ in your first comment? To write $[0,1]$ as a disjoint union of $\mathfrak c$ many perfect sets consider the Peano curve $[0,1]\to [0,1]^2$ and look at the preimages of the vertical segments. Surely they are $\mathfrak c$ many disjoint closed sets. To show that they are perfect note that you can split $[0,1]^2$ into $4^n$ squares each of which is mapped on a segment of length $1/4^n$ in $[0,1]$ by the inverse of the Peano curve, and then as $n\to\infty$ their diameters go to zero – Alessandro Codenotti Jun 30 at 9:10
• @AlessandroCodenotti Thank you. I will have to think about your last sentence. I guess it depends on a specific construction of the curve? – bof Jun 30 at 9:25
• @bof Sorry, I misremembered the fractal I wanted, looking at the images it's actually the Hilbert curve, if you look at the coloured gif on wikipedia you can see the $4^n$ squares I'm talking about after a few iterations. Also my comment about $2^{\aleph_0}$ is wrong, the ZFC result is that $[0,1]$ is not the union of less than $\mathfrak d$ disjoint closed sets. – Alessandro Codenotti Jun 30 at 9:31
• @AlessandroCodenotti I think I see how to partition $[0,1]$ into $\frak c$ many disjoint perfect sets by transfinite induction. If my thinking is right it also shows how to partition $[0,1]\setminus S$ into $\frak c$ disjoint perfect sets whenever $|S|\lt\frak c$. – bof Jun 30 at 17:05
• @AlessandroCodenotti I found another proof of the fact that $[0,1]$ can be partitioned into $\mathfrak c$ copies of the Cantor set $\mathbb C$ in Thearem 1.14 of Bankston & McGovern, Topological partitions, General Topology Appl, 10 (1979),215-229. First they partition $[0,1]$ into countably many copies of $\mathbb C$ and one copy of $\mathbb P$, the space of irrational numbers. Then $\mathbb P$ is easily partitioned into $\mathfrak c$ copies of $\mathbb C$ since $\mathbb P$ is homeomorphic to $\mathbb P\times\mathbb C$. – bof Jul 1 at 9:55

Theorem 1. The perfect set $$[0,1]$$ can not be written as the union of a pairwise disjoint family $$\mathcal F$$ of nonempty closed sets with $$1\lt|\mathcal F|\le\aleph_0$$.

Proof. Since $$[0,1]$$ is connected, we need only consider the case where $$\mathcal F$$ is countably infinite. Let $$\mathcal F=\{F_1,F_2,F_3,\dots\}$$. It is easy to see that there is a closed interval $$I_1\subset[0,1]$$ such that $$I_1\cap F_1=\emptyset$$ while $$I_1\cap F_n\ne\emptyset$$ for infinitely many $$n$$. Similarly there is a closed interval $$I_2\subset I_1$$ which is disjoint from $$F_2$$ but still meets infinitely many $$F_n$$. Continuing in this way we get a nested sequence of closed intervals whose intersection contains a point of $$[0,1]$$ which belongs to no $$F_n$$.

Theorem 2. If $$X$$ is a nonempty zero-dimensional metric space with no isolated points, then $$X$$ can be written as the union of $$\aleph_0$$ pairwise disjoint nonempty closed sets with no isolated points.

Proof. Choose a point $$a\in X$$. Construct a sequence of nonempty clopen sets $$A_1,A_2,A_3,\dots$$ converging to $$a$$; every neighborhood of $$a$$ contains all but finitely many of the $$A_n$$. Partition $$X$$ into the closed sets $$A_2,A_4,A_6,\dots$$ and $$X\setminus(A_2\cup A_4\cup A_6\cup\cdots)$$.

Corollary. If $$X$$ is a nowhere dense perfect subset of $$\mathbb R$$, then $$X$$ can be written as the union of $$\aleph_0$$ pairwise disjoint perfect sets.

• @bf , Thank you. But as you perfect might not be zero dimensional in general. How we can do the countable partition – 00GB Jun 30 at 13:31
• @00GB bof's answer shows that a countable partition is impossible in some cases, such as the perfect set $[0,1]$ – Alessandro Codenotti Jun 30 at 13:37
• @AlessandroCodenotti, you are right. My perfect set is algebraically independent and I know it can be written as countable pairwise disjoint of perfect sets . But how? – 00GB Jun 30 at 14:09
• @00GB It sounds like you asked the wrong question. If you don't know how to write your perfect set as a countable union of disjoint perfect sets, what makes you so sure it can be done? – bof Jun 30 at 17:09
• @00GB By the way, a totally disconnected subset of $]mathbb R$ is zero-dimensional. Is your perfect set zero-dimensional? – bof Jun 30 at 17:11