Timeline for Is there a normal space that is not uniformly normal
Current License: CC BY-SA 3.0
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| S Sep 25, 2013 at 19:34 | history | suggested | CommunityBot | CC BY-SA 3.0 |
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| Sep 25, 2013 at 19:29 | review | Suggested edits | |||
| S Sep 25, 2013 at 19:34 | |||||
| Sep 25, 2013 at 16:57 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Sep 25, 2013 at 14:01 | comment | added | user31967 | yes, but you wrote conversely. "every uniformly continuous map f:X→[0,1] is continuous" | |
| Sep 25, 2013 at 13:33 | comment | added | Joseph Van Name | $sin(x^{2})$ is continuous but not uniformly continuous. | |
| Sep 25, 2013 at 1:29 | comment | added | user31967 | what do you mean by "X is normal and every uniformly continuous map f:X→[0,1] is continuous."? I think every uniformly continuous is continuous. | |
| Sep 24, 2013 at 22:30 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Sep 24, 2013 at 21:50 | comment | added | user31967 | "unique compatible uniformity" has a lot to say. for example the space is totally bounded. | |
| Sep 24, 2013 at 21:42 | comment | added | Joseph Van Name | Paracompact uniform spaces are generally not uniformly normal. | |
| Sep 24, 2013 at 21:34 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Sep 24, 2013 at 21:25 | comment | added | user31967 | And I think intuitively that every paracompact Hausdorff uniform space is uniformly normal | |
| Sep 24, 2013 at 21:23 | comment | added | user31967 | And because you used proximities it may be true that a space is uniformly normal iff its precompact reflection is. | |
| Sep 24, 2013 at 21:22 | comment | added | user31967 | I think intuitively uniform normality has close connection to divisibility | |
| Sep 24, 2013 at 21:20 | comment | added | user31967 | You answered what I could ask (but I did not because uniform normality was just a raw idea to reach an example that could help construct a counterexample for something that hardly is related to normality) . The last proposition is very good for characterizing uniform normality (although for proving it I'll never resort to proximities, because they proximity spaces have less generality). btw, I have some ideas that may complete this characterzation: | |
| Sep 24, 2013 at 21:15 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Sep 24, 2013 at 20:56 | history | answered | Joseph Van Name | CC BY-SA 3.0 |