Talk:Cantor cube

Dimension
The Cantor cube $$2^\omega$$ can be given a topology where its not zero dimensional (the natural topology on the reals, of course, giving you a one-dimensional space). This article, as well as the Cantor space article, fail to discuss the topology imposed on the set: its zero dim only if the discrete topology is used.

Similarly, giving $$2^1=\{0,1\}$$ the topology $$\{\{0\},\{0,1\}\}$$ makes it not Hausdorf (and not discrete). These issues should be clarified as otherwise its misleading.

Similarly, a discussion of topology w.r.t. group operations is required. For example, I have no clue what the group structure is intended for $$2^1=\{0,1\}$$ if the topology $$\{\{0\},\{0,1\}\}$$ is intended, since clearly, this topology is incompatible with the permutation of two objects... linas 16:03, 18 December 2006 (UTC)


 * I'm certain the product topology is meant; I can't honestly recall what the source says, but it makes sense. I'm also pretty sure that I wouldn't have misstated Schepin's theorem by leaving out a restriction on the zero-dimensionality of the spaces. So I'll just slide that in... Melchoir 19:05, 18 December 2006 (UTC)


 * Yeah, OK, it'll do for now; I've been reading literature in ergodic theory that is remarkably sloppy in this way; its irritating, on the one hand, but also good, because it gives me something to work on :-) linas 04:08, 20 December 2006 (UTC)


 * Well, maybe you can add something later on! Especially if you figure out where precisely the Hausdorff condition goes, speaking of sloppiness... Melchoir 04:32, 20 December 2006 (UTC)