Timeline for Is a function needed here?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 17, 2021 at 18:18 | history | edited | Ivan Feshchenko | CC BY-SA 4.0 |
added 3 characters in body
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| Apr 14, 2021 at 22:43 | vote | accept | Ivan Feshchenko | ||
| Apr 9, 2021 at 21:45 | history | edited | Ivan Feshchenko | CC BY-SA 4.0 |
deleted 3 characters in body
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| Apr 8, 2021 at 0:36 | answer | added | Dmitri Pavlov | timeline score: 5 | |
| Apr 8, 2021 at 0:25 | comment | added | Alec Rhea | @DmitriPavlov Would you mind posting your comment as an answer to close out the question? | |
| Apr 7, 2021 at 22:57 | comment | added | Dmitri Pavlov | The answers are as follows: these arguments are correct, the indicated function is not needed. Indeed, your argument might as well prove that for all H1,…,Hn, H such that c_F(H1,…,Hn)⩽c, you have ∥Pn⋯P2P1−P0∥⩽g_n(c). Quantifying over a collection of sets is perfectly legitimate even if this collection of sets forms a proper class. There is no need to mention A_n(c) at all, it only creates further confusion. | |
| Apr 7, 2021 at 22:03 | review | Close votes | |||
| Apr 14, 2021 at 4:59 | |||||
| Apr 7, 2021 at 21:53 | comment | added | LSpice |
Also, a TeX note: please do not use repeated periods to simulate dots; that's what the various \dots commands are for. Compare the spacing in $H_1 \cap H_2 \cap \dotsb \cap H_n$ H_1 \cap H_2 \cap \dotsb \cap H_n to that in $H_1 \cap H_2 \cap ... \cap H_n$ H_1 \cap H_2 \cap ... \cap H_n. I have edited accordingly.
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| Apr 7, 2021 at 21:52 | history | edited | LSpice | CC BY-SA 4.0 |
`\dots`
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| Apr 7, 2021 at 21:47 | comment | added | LSpice | As in all your questions, you do not need the axiom of choice to choose one element from a non-empty class; that's what non-empty means. You only need AC to make many choices simultaneously. See, for example, mathoverflow.net/questions/387353/… . | |
| Apr 7, 2021 at 21:41 | history | asked | Ivan Feshchenko | CC BY-SA 4.0 |