Timeline for Uniformly closed ideals of smooth/real analytic functions
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 5 at 0:59 | answer | added | Igor Khavkine | timeline score: 2 | |
| Feb 4 at 14:05 | comment | added | Igor Khavkine | My bad! I got mixed up with what the ambient space was, as you pointed out. | |
| Feb 4 at 13:44 | comment | added | Thomas Kurbach | @IgorKhavkine I think there might be a misunderstanding about the word 'closed' here. A subset $Y$ of a topological space $X$ is (sequentially) closed if for every sequence $x_n\in Y$ convergent in $X$ one has that the limit belongs to $Y$ as well. I am fully aware that there exist sequences of smooth/analytic functions with only continuous limit, however that is not the question here. The question is what $J$ are closed in the subspace topology on $R$ inherited by $C^0(U,\mathbb{R})$. | |
| Feb 4 at 12:03 | comment | added | Igor Khavkine | It seems to me that the example of the ideal $J_X$ proposed in the question is false. Take any nonempty open $V \subset U \setminus X$ with compact closure. Outside of trivial cases, elements of $J_X$ will always converge in the given topology to some continuous non-smooth (non-analytic) functions (which will still vanish on $X$ though). So $J_X$ cannot be closed. By similar reasoning, I think almost no ideal will be closed. | |
| May 31, 2023 at 15:03 | history | edited | Thomas Kurbach | CC BY-SA 4.0 |
added 32 characters in body
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| May 31, 2023 at 14:56 | history | asked | Thomas Kurbach | CC BY-SA 4.0 |