Topologically Ordered Families of Disjoint Cantor Sets in $I$?

Suppose that we have an uncountable collection $C_\alpha$ of disjoint Cantor Sets contained in the closed unit interval $I$. Suppose we have ordered the indices $\alpha \in [0,1]$ as well. Then is it possible for the collection to satisfy the following property: If $x_n \in C_{\alpha_n}$ and $x_n \rightarrow y$ with $y \in C_\beta$ from the left then every accumulation point of $(\alpha_n)$ is contained in $[0, \beta]$ and similarly if they converge to $y$ from the right then every accumulation point of $(\alpha_n)$ is contained in $[\beta, 1]$.

In other words, if you converge to a point in one of the Cantor Sets $C_\beta$ from the left, then you can only do so via points coming out of Cantor Sets that are less than $\beta$ in the order and vice-versa. Can you order the Cantor Sets in this way that agrees with the index order 'up to the existence of limit points'?

This problem is coming from a general problem concerning admissible ensembles of continua in the plane. But this order theory aspect is very far removed from my expertise, what little I have haha So I need some help! My intuition is that it is impossible.

Thanks!

Such an enumeration does not exist. To derive a contradiction, take any family of Cantor sets $(C_\alpha)_{\alpha\in A}$ in $[0,1]$, indexed by an uncountable subset $A\subset[0,1]$.

Let $\mathcal K[0,1]$ be the space of non-empty compact subsets of $[0,1]$, endowed with the Hausdorff metric $d_H$. It is well-known that the metric space $\mathcal K[0,1]$ is compact and hence second-countable. Since second-countable spaces do not contain uncountable discrete subspaces, the uncountable set $\{(\alpha,C_\alpha):\alpha\in A\}\subset [0,1]\times\mathcal K[0,1]$ contains a non-isolated point $(\beta,C_\beta)$, which allows us to find a sequence $(\alpha_n)_{n\in\omega}$ in $A$ such that $\alpha_n\to\beta$ and $C_{\alpha_n}\to C_\beta$.

Using the Ramsey Theorem, we can find an increasing sequence $(n_k)_{k\in\omega}$ such that the sequence $(\alpha_{n_k})_{k\in\omega}$ is either strictly increasing or strictly decreasing. Replacing $(\alpha_n)$ by this subsequence, we can assume that the sequence $(\alpha_n)_{n\in\omega}$ is either increasing or decreasing.

First assume that the sequence $(\alpha_n)_{n\in\omega}$ is increasing. Let $x_\beta$ be the smallest element of the Cantor set $C_\beta\subset[0,1]$. Since $C_\beta$ has no isolated points, the Cantor set $C_\beta$ contains a strictly decreasing sequence $(y_n)_{n\in\omega}$ of real numbers that converges to the point $x_\beta$. For every $n\in\omega$ select a neighborhood $U_n\subset [0,1]$ of the point $y_n$ such that for any $n<m$ and $u\in U_n$, $v\in U_m$ we have $v<u$ and $|u-y_n|<1/2^n$. Since $C_{\alpha_n}\to C_\beta$, for every $k\in\omega$ there exists a number $n_k$ such that $U_k\cap C_{\alpha_n}\ne\emptyset$ for every $n\ge n_k$. Replacing each $n_k$ by a larger number, we can assume that the seqeunce $(n_k)_{k\in\omega}$ is strictly increasing. Then $U_k\cap C_{\alpha_{n_k}}\ne\emptyset$ for all $k\in\omega$.

Replacing the sequence $(\alpha_n)_{n\in\omega}$ by $(\alpha_{n_k})_{k\in\omega}$, we can assume that $U_k\cap C_{\alpha_k}\ne\emptyset$ for all $k\in\omega$. Then for every $k\in\omega$ we can choose a point $x_{\alpha_n}\in C_{\alpha_n}\cap U_n$. The choice of the sequence $(U_n)_{n\in\omega}$ ensures that the sequence $(x_{\alpha_n})_{n\in\omega}$ is strictly decreasing and converges to $x_\beta$ (whereas the sequence $(\alpha_n)_{n\in\omega}$ is increasing and converges to $\beta$. But this contradicts the choice of the enumeration $(C_\alpha)_{\alpha\in A}$.

By analogy we can consider the case of decreasing sequence $(\alpha_n)_{n\in\omega}$. In this case we should take $x_\beta$ to be the largest element of $C_\beta$.

• Is $(\beta, \omega)$ the same as $(\beta, 1]$? When you say that "the following lemma implies," the implication would definitely prove the theorem. But if you take just one point $x_\alpha$ from each $C_\alpha$, then what if we just let $\alpha = x_\alpha$? Then the lemma will be true, but left/right convergence will agree with the order. Maybe I am just not seeing how to go from the lemma to the result. I am really stumped! – John Samples Aug 7 '18 at 7:57
• @JohnSamples $(\beta,\omega_1)$ is not the same as $(\beta,1]$: $\beta$ is just an index (a countable ordinal) - it is an element of $\omega_1$, which is not order embeddable into $[0,1]$. So, the notation $(\beta,1]$ is incorrect: $\beta$ and $1$ live in different sets. – Taras Banakh Aug 7 '18 at 21:45
• @JohnSamples Aha! I have just realized that the set of your indices $\beta$ is a subset of $[0,1]$! I thought it is $\omega_1$. This changes the situation. Let me think on the problem a bit more. – Taras Banakh Aug 7 '18 at 21:49
• Ok, thanks! I am trying to prove the other direction and see if I can construct such a family. Here are some possibly relevant results that MSE dug up: math.stackexchange.com/questions/2874786/… – John Samples Aug 8 '18 at 3:49
• @JohnSamples I have rewritten my answer to fit the right interpretation of your question. – Taras Banakh Aug 8 '18 at 11:11