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!
 A: 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 $C_\beta$.
