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 Aug19 comment Matrix subsets dimension Sorry, I've put three "n²-1". Actually, there are just two. I didn't understand it too, but this is how it is written =/ Aug18 asked Matrix subsets dimension May16 awarded Commentator May16 comment Basis to a manifold by coordinate balls Is S really contained in B (line 6)? You meant S contained in the image of B via (phi), no? May16 accepted Basis to a manifold by coordinate balls May14 comment Basis to a manifold by coordinate balls I forgot to ask something else... I have to show that there is a coordinate ball in S^n whose closure is equal to all S^n. The open ball of radius 1 is what we're looking for, isn't? May14 asked Basis to a manifold by coordinate balls May4 awarded Editor May4 comment Equivalent to the Euclidean fifth postulate Thank you. I missed that. May4 revised Equivalent to the Euclidean fifth postulate added 23 characters in body May4 asked Equivalent to the Euclidean fifth postulate Apr29 comment Disconnected space - Disjoint Union Hum... this is an exercise I solved using the other definition. But it's an interesting definition too =] Apr28 comment Disconnected space - Disjoint Union Answering, should V be the union of h[X_i] where i is different from x_0? Apr28 accepted Disconnected space - Disjoint Union Apr28 comment Disconnected space - Disjoint Union Oh, how could I let this pass? I just have to expose an homeomorphis for the first implication. I didn't realize that the two (or more) spaces I would need are just those I know from the definition of conectedness of X. Thank you for opening my eyes. Apr28 comment Disconnected space - Disjoint Union @Jay X is a general topological space, I can't assume it's even Hausdorff. Apr28 asked Disconnected space - Disjoint Union Apr26 accepted Box topology on $\prod_{n=1}^\infty\mathbb{R}$ Apr26 comment Box topology on $\prod_{n=1}^\infty\mathbb{R}$ Hum, understood! Your construction clarified things. Thank you! Apr26 comment Box topology on $\prod_{n=1}^\infty\mathbb{R}$ The set O doesn't contain any term of the sequence, right? Because every x_i doesn't belong to O_i. This way, by the definition, we can't have a sequence converging to zero, is it? I think your argument for the second part still works if we have a finite product, doesn't it?