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  • 0 posts edited
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  • 16 votes cast
Aug
19
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 =/
Aug
18
asked Matrix subsets dimension
May
16
awarded  Commentator
May
16
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?
May
16
accepted Basis to a manifold by coordinate balls
May
14
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?
May
14
asked Basis to a manifold by coordinate balls
May
4
awarded  Editor
May
4
comment Equivalent to the Euclidean fifth postulate
Thank you. I missed that.
May
4
revised Equivalent to the Euclidean fifth postulate
added 23 characters in body
May
4
asked Equivalent to the Euclidean fifth postulate
Apr
29
comment Disconnected space - Disjoint Union
Hum... this is an exercise I solved using the other definition. But it's an interesting definition too =]
Apr
28
comment Disconnected space - Disjoint Union
Answering, should V be the union of h[X_i] where i is different from x_0?
Apr
28
accepted Disconnected space - Disjoint Union
Apr
28
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.
Apr
28
comment Disconnected space - Disjoint Union
@Jay X is a general topological space, I can't assume it's even Hausdorff.
Apr
28
asked Disconnected space - Disjoint Union
Apr
26
accepted Box topology on $\prod_{n=1}^\infty\mathbb{R}$
Apr
26
comment Box topology on $\prod_{n=1}^\infty\mathbb{R}$
Hum, understood! Your construction clarified things. Thank you!
Apr
26
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?