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Dec 1, 2022 at 23:38 history edited verret CC BY-SA 4.0
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Dec 1, 2022 at 0:31 answer added saolof timeline score: 0
Nov 29, 2022 at 17:44 history edited wlad CC BY-SA 4.0
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Nov 29, 2022 at 17:20 history edited wlad CC BY-SA 4.0
Easier to understand than "constancy"
Apr 12, 2020 at 19:05 answer added Andrej Bauer timeline score: 3
Apr 12, 2020 at 7:17 comment added François G. Dorais @AndrejBauer I added an answer (indeed based on a singular cover). I think it might be broken but perhaps there is a way to fix it...
Apr 12, 2020 at 3:54 answer added François G. Dorais timeline score: 0
Apr 11, 2020 at 18:10 comment added Andrej Bauer @FrançoisG.Dorais: I don't have a proof, and in fact now I am worried that some trickery is possible with singular covers.
Apr 11, 2020 at 15:29 comment added Andrej Bauer @FrançoisG.Dorais I would be interested in a reference showing the existence of such a function, too, especially since I think I can prove that it does not exist.
Apr 11, 2020 at 15:25 comment added Andrej Bauer The approximate IVT requires no choice.
Apr 11, 2020 at 9:39 comment added François G. Dorais I should be asleep by now, but unless I am mistaken you can get such a function from the "usual" proof that [0,1] is not compact in Eff. If that's not the case, I'll look back at this later this week.
Apr 11, 2020 at 9:10 comment added wlad @FrançoisG.Dorais Is there a reference for the existence of such functions in the Effective Topos?
Apr 11, 2020 at 9:00 comment added François G. Dorais So what goes wrong with a non-constant yet locally constant function for the definition of pointwize derivative?
Apr 11, 2020 at 8:58 comment added wlad @FrançoisG.Dorais Locally constant but non-constant functions could be useful, but not in the Effective Topos. The constancy principle can be proved from Dependent Choice, which is available in the Effective Topos
Apr 11, 2020 at 8:57 comment added François G. Dorais Non-constant locally constant functions don't exist because $\mathbb R$ is connected, but they do exist in the effective topos, for example, if I remember correctly.
Apr 11, 2020 at 8:55 comment added wlad @FrançoisG.Dorais Classically, if a function has derivative $0$ everywhere then it is constant. Any such counterexample would also be a classical counterexample. So I think you're mistaken
Apr 11, 2020 at 8:53 comment added François G. Dorais The question? Or am I mistaken? (I'm not entirely sure about the definition of derivatives.)
Apr 11, 2020 at 8:52 comment added wlad @FrançoisG.Dorais To?
Apr 11, 2020 at 8:52 comment added François G. Dorais Wouldn't a non-constant yet locally constant function be a counterexample?
Apr 11, 2020 at 8:26 comment added wlad @EmilJeřábek I believe it is constructive, but the proof uses dependent choice
Apr 11, 2020 at 8:23 comment added Emil Jeřábek Is an approximate mean value theorem also nonconstructive?
Apr 10, 2020 at 19:43 comment added Noah Schweber Note for readers more familiar with classical model theory, "open induction" in Andrej Bauer's comment does not mean what it means in the context of weak theories of arithmetic in classical logic (namely, $IOpen$).
Apr 10, 2020 at 19:31 answer added Franka Waaldijk timeline score: 3
Apr 10, 2020 at 18:07 history edited wlad CC BY-SA 4.0
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Apr 10, 2020 at 15:46 history edited wlad CC BY-SA 4.0
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Apr 10, 2020 at 15:44 comment added wlad I see. Thanks hal.inria.fr/hal-01376054/document
Apr 10, 2020 at 15:32 comment added wlad @AndrejBauer looking up "open induction"
Apr 10, 2020 at 15:28 comment added Andrej Bauer Interesting. I think I can prove it using open induction, but not without anything.
Apr 10, 2020 at 14:45 comment added Andrej Bauer I think a better title would be "The constancy principle in choiceless constructive foundations".
Apr 10, 2020 at 12:17 history edited wlad
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Apr 10, 2020 at 11:15 history edited wlad CC BY-SA 4.0
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Apr 10, 2020 at 10:56 history asked wlad CC BY-SA 4.0