Timeline for Embedding into $C\times [0,1]$
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2019 at 6:41 | comment | added | YCor | @D.S.Lipham I can believe that a few authors used this... but "most authors"..., I'll check when I'm in a library...! | |
| Mar 9, 2019 at 6:20 | comment | added | D.S. Lipham | @YCor I have seen that terminology, but only rarely. Most authors (including Engelking!) use "hereditarily disconnected" for singleton components, and "totally disconnected" for singleton quasi-components. | |
| Mar 9, 2019 at 5:48 | history | edited | YCor |
edited tags
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| Mar 9, 2019 at 5:48 | comment | added | YCor | Your definition of "totally disconnected" is usually called 'totally separated". Using another terminology might be confusing, since "totally disconnected" (= connected components are singletons) is usually weaker. | |
| Mar 9, 2019 at 5:00 | history | edited | D.S. Lipham | CC BY-SA 4.0 |
deleted 728 characters in body
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| Mar 2, 2019 at 22:42 | vote | accept | D.S. Lipham | ||
| Sep 17, 2018 at 16:31 | answer | added | Taras Banakh | timeline score: 4 | |
| Apr 20, 2018 at 11:40 | comment | added | Henno Brandsma | So @AmirSagiv he seems to be using small inductive dimension $\operatorname{ind}(X)$. | |
| Apr 16, 2018 at 5:57 | comment | added | Wlod AA | "One-to-one"? What if |X|=2? | |
| Apr 15, 2018 at 21:53 | comment | added | Henno Brandsma | @AmirSagiv All the usual dimension functions coincide in separable metric spaces. | |
| Apr 15, 2018 at 21:52 | comment | added | Henno Brandsma | Your notion of totally disconnected is non-standard. Normally this means that the components of $X$ are singletons. For you it's the quasicomponents that are singletons. | |
| S Apr 15, 2018 at 20:29 | history | suggested | Amir Sagiv |
relevant subject tag
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| Apr 15, 2018 at 20:11 | comment | added | Amir Sagiv | Thanks! So everything here is inside $\mathbb{R}^n$? | |
| Apr 15, 2018 at 18:17 | comment | added | D.S. Lipham | @AmirSagiv for example, dimension $1$ means there is a basis of open sets with zero-dimensional boundaries. | |
| Apr 15, 2018 at 18:03 | review | Suggested edits | |||
| S Apr 15, 2018 at 20:29 | |||||
| Apr 15, 2018 at 18:02 | comment | added | Amir Sagiv | What definition of dimension are you using? | |
| Apr 15, 2018 at 17:54 | history | asked | D.S. Lipham | CC BY-SA 3.0 |