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 $y=(y_1,y_2,y_3,\cdots)$:

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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.