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In trying to prove that binary space (a homeomorphic space to the better known Cantor Set) is totally disconnected, what traits of the space do I need? Is binary space point-wise open (that would certainly be sufficient)?

As a quick reminder $B$ is the set of infinite sequences of $0$'s and $1$'s where for $x=(x_1,x_2,x_3,\cdots)$ and $y=(y_1,y_2,y_3,\cdots)$:

$$ d(x,y) = \begin{cases} \frac 1 n, & \text{n=min}[i|x_i\neq y_i] \\ 0, & x_i=y_i \end{cases} $$

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I think the most general fact that applies here (and I can think of) is that every ultrametric space is totally disconnected, as balls are closed.

An ultrametric space is a metric space satisfying strong triangle inequality, i.e. $d(x,z)\leq \max(d(x,y),d(y,z))$. It is routine to check that it is satisfied by the metric you cited, and it is not hard to see that it implies that balls are closed.

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  • $\begingroup$ Ok. So open balls are by definition open but also easily shown to be closed. Hence, the space has a basis of clopen sets and is totally disconnected. Do we have to refer to balls for this problem? What about a finite subset of B? Would we then need to use the open balls of the original set and a subspace topology? $\endgroup$ – Musicpulpite Oct 14 '14 at 3:49
  • $\begingroup$ @Musicpulpite: I don't really understand the question. What's wrong with referring to balls? A finite (Hausdorff) space is always discrete (and hence totally disconnected), so the other one seems trivial. A subspace of a Hausdorff space is Hausdorff, and for that matter, a subspace of a totally disconnected space is totally disconnected. Both are mostly straightforward. $\endgroup$ – tomasz Oct 14 '14 at 10:42

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