Question
How many "disconnections" are there in the Cantor set? Countably many, or uncountably?
What do I mean?
I've deduced that an uncountable, totally disconnected set can be made from a connected uncountable set, by only introducing countably-many "disconnections". This seems at best to be counterintuitive, or at worst (possibly more likely) I've made a mistake or misconception.
Can you resolve my cognitive dissonance? Does the Cantor set have countably many or uncountably many "disconnections"?
Let the count of the disconnections be the smallest cardinality of a set of pairs $\{(x_n,y_n):x,y\in\mathcal C\}$ such that $\forall n: x\sim y$ is sufficient to connect together the totally disconnected set.
My example
This is the example I'm getting my intuit from:
Let the Cantor set $\mathcal C$ be the standard totally disconnected and totally-ordered subset of $[0,1]$
Let $x\sim y$ if $x<y$ and there is no $z\in\mathcal C$ such that $x<z<y$ (or if $y<x$ and there is no $z\in\mathcal C$ such that $y<z<x$)
Then there are countably many pairs $x\sim y$ which are sufficient to totally reconnect (or reconnect) an uncountable set.
Just think about that for a minute - a totally ordered, totally disconnected, uncountable space, in which we can reconnect countably many disconnections and arrive at an uncountable, totally-connected space. What gives?
(previous version of this answer was incorrect)
Yes, it is. If there are two non-empty non-intersecting open in quotient topology sets, then there preimages wrt $\sim$ are open sets such that any removed interval has both ends in the same set. Let this preimages be $U$ and $V$. Let $u \in U$, $v \in V$, $u < v$. Let $c = \sup \{U \cap [0, v]\}$. $c \in \mathcal C$ (because $\mathcal C$ is closed) and has elements from $U$ in any left neighbourhood of it, so it doesn't have neighbourhood disjoint from $U$, so $c \in U$ itself.
If $c$ is left end of some removed interval, then right end of the same interval should be also in $U$, contradicting definition of $c$.
If $c$ is not left end of any removed interval, then it has points from $\mathcal C$ in any right neighbourhood, but such points are from $V$ by definition of $c$ - so $c$ has no neighbourhood disjoint from $V$, and thus $U$ isn't open.
I guess part of why it can seem strange is because of how connected component behave when we have infinite number of them. When there are finitely many of them, each connected component is an open set, so we can essentially work with each of them separately. When there are infinitely many connected components - for example, already in $\mathbb Q$ - this is no longer the case, and we can no longer split the space in disjoint union of open sets where each open set is connected.