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Happy new year, it's time for the Axiom of Choice of Choice Award 2022! Each day you can vote between two equivalent formulations of the Axiom of Choice (see also mastodon.vierkantor.com/@Vierk), until one remains at the end of January.

about ACoCA 2022 

The day numbers refer to the daily AoC thread I made for the previous month, which contain more detailed descriptions: mastodon.vierkantor.com/@Vierk

As standard, "P is equivalent to AC" means "ZF ⊢ P ↔ AC"; feel free to salt & pepper with large cardinal axioms in the case of e.g. large categories.

Axiom of Choice of Choice Award 2022 round 1 match 1: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 2: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 3: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 4: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 5: pick your favourite equivalent of the axiom of choice out of the two options below:

Will Zorn's lemma be eliminated in the first round? Vote now and you may be the one casting the decisive vote!

Axiom of Choice of Choice Award 2022 round 1 match 6: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 7: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 8: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 9: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 10: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 10: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 11: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 12: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 13: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 14: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 1 match 15: pick your favourite equivalent of the axiom of choice out of the two options below:

Will the maximal antichain win, or the union of 2 smaller sets? Cast your vote now to break the tie!

SURPRISE CONTENDER: the final matchup will include an entry from intuitionistic type theory!

Axiom of Choice of Choice Award 2022 round 1 match 16: pick your favourite equivalent of the axiom of choice out of the two options below:

SECOND ROUND: we have seen all the contestants to the ACoCA in action. Congratulations to each entrant who made it to the second round!

Axiom of Choice of Choice Award 2022 round 2 match 1: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 2 match 2: pick your favourite equivalent of the axiom of choice out of the two options below:

Axiom of Choice of Choice Award 2022 round 2 match 3: pick your favourite equivalent of the axiom of choice out of the two options below:

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@Vierkantor 1 is simpler, intuitive, and what most people are used to. But 16 is cooler... :fishthink:

@socks I certainly didn't expect anything like that happening in the first round!

@Vierkantor

So we have to pick one choice from this list?

Is that even always possible? :)

@hhardy01 It's only a finite collection, so yes, totally possible without AC!

@hhardy01 yes, according to me. no, according to some physicists who claim that 0 is not finite, see e.g. en.wikipedia.org/wiki/Finite_n

@hhardy01 sure, it's in bijection with an ordinal ≤ ω (namely itself)

@Vierkantor

What if the number of elements can be computed to be something strange like a surreal number?

@hhardy01 then you're probably using a different definition of "number of elements" than me. namely, to me the number of elements of a set X is the smallest ordinal that X injects into. this exists due to Hartogs' lemma, although you need AC to prove useful properties.

@Vierkantor

Okay we have gone beyond what I even marginally understand so resorting to stackexchange here:

"One can extend the notion of cardinality to include negative and non-integer values by using the Euler characteristic and homotopy cardinality. For example, the space of finite sets has homotopy cardinality $e=\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dotsi$. The idea is to sum over each finite set, inversely weighted by the size of their symmetry group...
-cont-

@Vierkantor

-cardinality-
...John Baez discusses this in detail on his blog. He has plenty of references, as well as lecture notes, course notes, and blog posts about the topic here. The first sentence on the linked page:

"We all know what it means for a set to have 6 elements, but what sort of thing has -1 elements, or 5/2? Believe it or not, these questions have nice answers." -Baez""

math.stackexchange.com/questio

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@Vierkantor glad to see that everyone made the right choice here

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@Vierkantor

You put your [right leg] in,
You put your [right leg] out;
You put your [right leg] in,
And you shake it all about.
You do the hokey pokey,
And you turn yourself around.
That's what it's all about! Yeah!

Babylon 5 Hocky Pocky

youtu.be/vr8tOa_En7A

@Vierkantor

For choice 2:

A wild Law of the Excluded Middle has appeared
But Russell's antinomy
But ZFC

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