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Talk:Cantor cube

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Dimension

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The Cantor cube 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 the topology 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 if the topology is intended, since clearly, this topology is incompatible with the permutation of two objects... linas 16:03, 18 December 2006 (UTC)[reply]

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)[reply]
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)[reply]
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)[reply]