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For Cantor-connectedness I use the following definition:

A $p$-metric space $(X,d)$ is Cantor-connected if for any $\epsilon > 0$, any two points $x, y \in X$ can be connected by an $\epsilon$-chain, i.e., there exist points $x = x_{0}, x_{1}, \cdots, x_{n} = y$ in $X$ such that $d(x_{i-1},x_{i}) \leq \epsilon$ for all $i \leq n$.

I would like to prove the following characterisation of Cantor-connectedness:

A space $(X,d)$ is Cantor-connected if and only if it cannot be partitioned into two sets $A$ and $B$ such that $d(A,B) > 0.$

I already have the following:

We first prove that if $(X,d)$ is Cantor-connected, $X$ cannot be partitioned into two sets $A$ and $B$ such that $d(A,B) > 0.$ Suppose that there exists a partition of $X$ into sets $A$ and $B$ with $d(A,B) > 0.$ Take $a \in A$ and $b \in B$, take $\epsilon > 0$. By the fact that $X$ is Cantor-connected, there exist $a = x_{0}, x_{1}, \cdots, x_{n} = b$ in $X$ such that $d(x_{i-1},x_{i}) \leq \epsilon$ for all $i \in \{1, \cdots,n\}$. But then we get $$d(a,b) \leq \sum_{i=1}^{n}d(x_{i-1},x_{i}) \leq n \epsilon.$$ By arbitrariness of $\epsilon$ we get that $d(a,b) = 0$, hence $d(A,B)=0$. This is a contradiction.

Now we show that if $X$ cannot be partitioned into two sets $A$ and $B$ with $d(A,B) > 0$, then $X$ is Cantor-connected. Suppose that $x, y \in X$ and $\epsilon > 0$ arbitrary. Set $x_{0} = x$. Then there exists $x_{1} \in X$ such that $d(x_{0},x_{1}) \leq \epsilon$. If not, $\{x\}$ and $X \setminus \{x\}$ would form a partition with $d(\{x\}, X \setminus \{x\}) > 0$. Analogously we find $x_{2} \in X$ such that $d(x_{1},x_{2}) \leq \epsilon.$ This will give us $x_{0}, x_{1}, x_{2}, \cdots \in X$ with $d(x_{i-1},x_{i}) \leq \epsilon.$ Now we have to prove that after a finite number of steps, we get $x_{n}= y$ and $d(x_{n-1}, x_{n}) \leq \epsilon.$

Can anyone explain to me why the process in the last step stops after a finite number of steps?

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up vote 3 down vote accepted

I see no reason why your process would stop after finitely many steps. Suppose you are working in $\mathbb{R}^2$, and you are trying to connect $x = (0,0)$ with $y = (1,0)$. It could be that every step just moves along the $y$-axis: something like $x_0 = (0,0)$, $x_1 = (0,\epsilon)$, $x_2 = (0,2 \epsilon )$, etc. (Actually, your process doesn't even ensure that the points you take are distinct!)

To show that the given condition implies Cantor-connectedness, let $x \in X$ and $\varepsilon >0$ be given. Consider the set $$A_{x,\varepsilon} = \{ y \in X : y \text{ can be reached from } x \text{ by finitely many steps each of length } \leq \epsilon \}.$$ Note that if $A_{x,\varepsilon}$ and $X \setminus A_{x,\varepsilon}$ are both nonempty, then $d (A_{x,\varepsilon},X \setminus A_{x,\varepsilon} ) > 0$ (in fact, $\geq \varepsilon$). The condition will then imply that $A_{x,\varepsilon} = X$ for all $x$ and $\varepsilon$, which means that $X$ is Cantor-connected.

I would just like to point out that in the proof of the forward direction you should explicitly state that since $a = x_0$ belongs to $A$ and $b = x_n$ belongs to $B$, then there must be an $i$ such that $x_i \in A$ and $x_{i+1} \in B$, and these are at most $\varepsilon$ apart, meaning that $d(A,B) \leq \varepsilon$.

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Your method of proof is doomed, since it forms a sequence without regard to the point $y$ that you want to get to.

Instead, consider a fixed $x$ and take the set of all points that can be reached from $x$ with an $\epsilon$-chain. If it isn't all of $X$, then what else can you say about this set?

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