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!



Topologically Ordered Families of Disjoint Cantor Sets in $I$?
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 $C\_\beta$.


