Timeline for Can you write $\mathbb R^2$ as a disjoint union of two totally disconnected sets?
Current License: CC BY-SA 3.0
15 events
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
Commonmark migration
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| Feb 27, 2015 at 5:19 | comment | added | Todd Trimble | "I could justify the mentioned nature of a closed subset of $\mathbb{R}^2$, which separates the $\mathbb{R}^2$ --it's not trivial but the great Alexander duality is an overkill." Since you say it's not trivial, would you add that detail so that people like me can understand your answer? Thanks. | |
| Feb 27, 2015 at 4:06 | comment | added | Włodzimierz Holsztyński | @ToddTrimble -- it seems that you got enlightened by Alexander duality, period. Making a connection between the two answers is false (and even unfair, but never mind). And yes, I could justify the mentioned nature of a closed subset of $\ \mathbb R^2$, which separates the $\ \mathbb R^2$--it's not trivial but the great Alexander duality is an overkill. | |
| Feb 27, 2015 at 1:35 | comment | added | Kristal Cantwell | It is undeleted now. | |
| Feb 27, 2015 at 0:34 | comment | added | Todd Trimble | @WłodzimierzHolsztyński Well, before Emil enlightened me by referring to Kristal's answer (now unfortunately deleted), the line "which disconnects sets $E_0, F_0$, thus $X$ is not totally disconnected" was a mystery to me. Do you have a way of justifying that inference that is different from Kristal's (which invokes Alexander duality)? It would be a good idea to explain in your answer just what special feature of $\mathbb{R}^2$ you are using. | |
| Feb 26, 2015 at 19:08 | comment | added | Włodzimierz Holsztyński | I don't see here anything referring to any anything in the context of the two answers. | |
| Feb 26, 2015 at 16:20 | comment | added | Włodzimierz Holsztyński | Thank you @ToddTrimble for pointing out to my math-typos. I already fixed the 3 mentioned. I hope that's it. I am very prone to typos (and worse :-) | |
| Feb 26, 2015 at 16:16 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
math typos (harmless but nasty)
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| Feb 26, 2015 at 12:26 | comment | added | Todd Trimble | Ahhh... thank you @EmilJeřábek. But that should probably be edited in for clarity. | |
| Feb 26, 2015 at 12:24 | comment | added | Emil Jeřábek | @ToddTrimble: I believe the place is “... which disconnects sets $E_0\:F_0$, thus $X$ is not totally disconnected”, which summons the property of $\mathbb R^2$ referred to in Kristal Cantwell’s answer. | |
| Feb 26, 2015 at 12:01 | comment | added | Todd Trimble | I think you want $X_0$, not $X$ in the second and third bullet points. Do you want the codomain of $\phi$ to be $\mathbb{R}^2$, or just $\mathbb{R}$? I'm probably missing something obvious, but where are you actually using the fact that we're working with $\mathbb{R}^2$, since the proof must break down somewhere if we replace $\mathbb{R}^2$ with $\mathbb{R}$? | |
| Feb 26, 2015 at 11:02 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typos
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| Feb 26, 2015 at 4:52 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
math typo (missing X:=).
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| Feb 26, 2015 at 3:58 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
typo
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| Feb 26, 2015 at 3:24 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |