Cantor space has a binary tree structure?

Question

I know that every point in the Cantor set has a binary tree representation. i.e., there exist closed intervals, $\{ I_{n}^j \}_{n\in \mathbb{N}, 1\leq j\leq 2^n}$, such that

$$ \mathcal{C}= \bigcap_{n=1}^\infty \Big( \cup_{j=1}^{2^n} I_{n}^j \Big), $$

and $\overline{a}\in \mathcal{C}$ corresponds to an appropriate intersection of intervals. Can we do the same for any subspace $K\subset\mathbb{R}$ which is a Cantor space? A Cantor space is a space homeomorphic to any non-empty, compact, perfect, totally disconnected and metrizable space. I saw that one can associate a Cantor scheme to any Cantor space. However, I was wondering whether this means or implies that such a set $K$ can be written as a similar intersection of union of intervals corresponding to a tree?

I'm afraid that I'm not that familiar with descripitive set theory, so I'm not completely certain of what I've assumed as true. I would appreciate any corrections or insights.

Answer 1

Since there has been no answer so far, I will give a partial answer that I have found. That is that every Cantor set has a tree structure with local degree conditions. i.e.:

Given a cantor set $K\subseteq R$, there exists a countable collection of closed intervals with a partial order of inclusion, $\{ I_\alpha \}_{\alpha}$, such that:

• $\cap_{n\in \mathbb{N}_0} \cup_{\vert \alpha\vert=n} I_\alpha=K $, where $\vert \alpha\vert$ is the depth from the root at each step.

• If $\alpha\preceq \beta$, then $I_\alpha\supset I_\beta$.

Given $\delta>0$, denote $K^{(\delta)}=\overline{ \cup_{x\in K} B_{\delta}(x) }$. Then, I have the following useful facts:

1. For $\delta_2> \delta_1$, $K^{(\delta_1)}\subseteq K^{(\delta_2)}$.
2. $K=\cap_{\delta>0} K^{(\delta)}$.
3. $K=[\inf K,\sup K]\setminus \Big( \cup_{j=1}^\infty G_j \Big)$, where $G_j$ are open ontervals and $\lambda(G_j)\to 0$ and $\lambda$ is the Lebesgue measure.
4. For every $\delta>0$, $K^{(\delta)}$ is a finite union of intervals, and the number of intervals is decrasing with respect to $\delta$.

If we denote $\eta(K;\delta)$ as the number of connected components of $K^{(\delta)}$, then $\eta(K;\delta) \overset{\delta \to 0^+}{\to} \infty$. If we assume that $G_j$ are enumareted in such a way that $\lambda(G_j)$ is decreasing, then $G_j\subseteq K^{(\delta)}$ if and only if $\lambda(G_j)\leq \frac{1}{2}\delta$. Denoting by decreasing order the points of discontinuity for $\eta(K;\delta)$ as $\{\nu_m\}_{m=1}^\infty$, we get that

• $K^{(\nu_m)}\supseteq K^{(\nu_{m+1})}$ for each $m$.
• $K^{(\nu_m)}=\sqcup_{\ell=1}^{\eta(K;\nu_m)} I_{\ell,m}$.
• $K^{(\nu_0)}=[\inf K-\nu_0, \sup K + \nu_0]$ where $\nu_0 > \nu_1$.
• $\cap_{m=0}^\infty K^{(\eta_m)}=K$.

We can then construct a graph $G=(V,E)$, with the vertices $\cup_{m=0}^\infty \big\{ I_{\ell,m}\times\{m\} \big\}_{\ell=1}^{\eta(K;\eta_m)}$, with edges between $I_{\ell,m}\times \{m\}$ to $I_{\ell',m+1}\times \{ m+1 \}$ if and only if $I_{\ell',m+1}\subseteq I_{\ell,m}$. For $\alpha= \vert I_{\ell,m}\times \{m\}$, we have $\vert \alpha \vert=m$.

I would appreciate if someone finds an error and points it out. Also, I have a feeling that one can revise this construction slightly, to get a binary tree structure?