Timeline for Locally compact space that is not topologically complete
Current License: CC BY-SA 3.0
10 events
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| Jun 27, 2016 at 10:16 | history | edited | François G. Dorais | CC BY-SA 3.0 |
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| S Jun 27, 2016 at 6:41 | history | suggested | yada | CC BY-SA 3.0 |
A paracompact space $X$ is the disjoint union (or "free union") of $\sigma$-compact spaces. Such a union need not be countable, e.g. an uncountable discrete space is metrizable but not a countable union of $\sigma$-compact subspaces.
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| Jun 27, 2016 at 6:41 | review | Suggested edits | |||
| S Jun 27, 2016 at 6:41 | |||||
| Dec 6, 2013 at 21:23 | comment | added | Hugo Rafael Oliveira Ribeiro | I think that I was a little confused about the assumption that $X$ is $\sigma$-compact. Expressing $X$ as a disjoint union $X = \bigcup_{\alpha \in A} X_{\alpha}$, where $X_{\alpha}$ is clopen and separable, the new metric $d'$ defined by $d'(x,y) = d(x,y)$ if $x,y$ are in the same component and $d'(x,y) = 1$ if $x,y$ are in the differents components is equivalent to $d$ and $(X,d')$ is complete. | |
| Dec 6, 2013 at 19:49 | comment | added | Joseph Van Name | A locally compact metric space is as a topological space a disjoint union of separable spaces, but a locally compact metric space is not necessarily separable since an uncountable discrete space is always locally compact, but not separable. | |
| Dec 6, 2013 at 19:29 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Dec 6, 2013 at 16:58 | vote | accept | Hugo Rafael Oliveira Ribeiro | ||
| Dec 6, 2013 at 16:58 | comment | added | Hugo Rafael Oliveira Ribeiro | So, we can conclude that a locally compact metric space is separable... very cool. Thanks | |
| Dec 6, 2013 at 16:05 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
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| Dec 6, 2013 at 14:44 | history | answered | Joseph Van Name | CC BY-SA 3.0 |