Timeline for Is every connected subgroup of a Euclidean space closed?
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 5 at 0:08 | comment | added | Moishe Kohan | @TimothyChow: I am not the right person for such questions, I am a topologist, not a logician. All I can say is the the proofs of existence of connected subgriups use AC. | |
| Oct 5 at 0:02 | comment | added | Timothy Chow | So is it consistent with ZF that there are no dense connected proper subgroups of $\mathbb{R}^2$? | |
| Oct 4 at 2:35 | history | edited | Moishe Kohan | CC BY-SA 4.0 |
added 128 characters in body
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| Oct 3 at 11:39 | comment | added | KP Hart | Jones' paper is quite informative too on what is possible with discontinuous additive functions and their graphs. | |
| Oct 3 at 2:36 | history | edited | YCor | CC BY-SA 4.0 |
added AMS link since DOI points to JSTOR which wants money from you
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| Oct 3 at 1:29 | comment | added | Terry Tao | Thanks for this nice answer! At least I don't feel too embarrassed that I could not see a simple way to answer my student's question either way, it appears to be a somewhat subtle question (in particular, dependent on choice, and on the precise type of connectedness assumed). | |
| Oct 3 at 1:28 | vote | accept | Terry Tao | ||
| Oct 3 at 1:19 | comment | added | Moishe Kohan | @LSpice: it is a nice exercise to check that every Lie subgroup of $\mathbb R^n$ is closed (just look at the exponential map). | |
| Oct 3 at 0:53 | comment | added | Will Sawin | @LSpice In a general group, there can be non-closed path-connected subgroups, as Yemon Choi points out. mathoverflow.net/questions/479975/… In $\mathbb R^n$, tehre are no non-closed path-connected subgroups. The definition used in the paper is "a continuous isomorphic image of a connected Lie group". In $\mathbb R^n$, such things must be closed subgroups by a simple argument. | |
| Oct 2 at 23:56 | comment | added | LSpice | Ah, thanks. What does it mean in this context? And does the cited result imply, directly or indirectly, that there are no non-closed, path-connected subgroups, or only that there are no dense, proper such? | |
| Oct 2 at 23:54 | comment | added | Moishe Kohan | @LSpice: Normally, it does not imply closed, but opinions on this differ. | |
| Oct 2 at 23:53 | comment | added | LSpice | I can never remember the terminology; does "Lie subgroup" imply closed? If so, then the conclusion can be upgraded to the non-existence of non-closed, path-connected subgroups of $\mathbb R^n$ (the natural generalisation of the question asked), not just the non-existence of dense such subgroups, right? | |
| Oct 2 at 23:30 | history | answered | Moishe Kohan | CC BY-SA 4.0 |