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- From: Daniel de Rauglaudre <daniel.de_rauglaudre AT inria.fr>
- To: coq-club AT inria.fr
- Subject: [Coq-Club] well-ordering axiom (aka Zermelo's theorem)
- Date: Mon, 12 Sep 2016 10:58:00 +0200
Hi all,
This time, I have a problem to find a definition of the well-ordering
axiom (for all types, not just nat), aka Zermelo's theorem.
I cannot find a definition with only foralls and exists on the Web.
I made this:
Axiom well_ordering : ∀ A,
∃ (R : A → A → Prop),
∀ P, (∃ x, P x) → ∃ ! y, P y ∧ ∀ z, P z → R y z.
Is it correct?
It seems not to be equivalent to the axiom of choice because the
"∀ ∃!" does not imply the existence of a function, since it is an
axiom (of unique choice), you told me (and I checked it in Coq
library).
All of this is because I am trying to make a formal proof of the
paradox of Banach-Tarski, well known to require the axiom of choice
and I am indeed at the point where I need one of its versions.
I need something to build the set (the type, actually) of all (what
they call) "orbits" on the surface of a sphere. With the above axiom
*and* the axiom of functional choice, I managed to prove that
∃ f : point → point, ∀ x y, same_orbit x y ↔ f x = f y.
I did it but since I use an axiom of my own, I am not sure I am
really allowed to do that.
--
Daniel de Rauglaudre
http://pauillac.inria.fr/~ddr/
- [Coq-Club] well-ordering axiom (aka Zermelo's theorem), Daniel de Rauglaudre, 09/12/2016
- Re: [Coq-Club] well-ordering axiom (aka Zermelo's theorem), Guillaume Melquiond, 09/12/2016
- Re: [Coq-Club] well-ordering axiom (aka Zermelo's theorem), Daniel de Rauglaudre, 09/12/2016
- Re: [Coq-Club] well-ordering axiom (aka Zermelo's theorem), Daniel Schepler, 09/12/2016
- Re: [Coq-Club] well-ordering axiom (aka Zermelo's theorem), Guillaume Melquiond, 09/12/2016
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