Talk:Axiom of Choice Implies Zorn's Lemma

A formal interpretation would yield that this result in fact does not require the axiom of choice. And then, as it is included in Zorn's Lemma, shouldn't that be granted the status of the Axiom namespace? For it is in fact an axiom, as much as AC is. Thoughts? Oh, and, nice result. --Lord_Farin 16:47, 6 February 2012 (EST)

Yes it does, you require a choice function on all $A \in \mathbb X$.

The thing is this. There are many results whose truth is equivalent to AoC. No big deal. This just means that if AoC fails to hold, then so do all those results. Zorn's Lemma is one. So yes, we could if we wanted to make Zorn's Lemma the "axiom" and prove AoC from it (in whatever form suits you). But this is generally not done, because ZL is complicated to grasp in one thought. Yes, it can be considered an "axiom", as can all the other results which are equivalent to AoC, but why have lots of axioms saying the same thing? The whole point of axiomatics is that those axioms are minimal in number and size. (Yes I know some of the quoted axioms can be deduced from the others, go and shoot the authors who wrote them up in the first place.)

What I'm trying to say is, because a result is equivalent to an axiom, doesn't make that result an axiom. Yes, the axioms you choose are arbitrary. Pick one of all these equivalent results and make it an axiom and show all the other equivalent results are equivalent to it. Pick one, any one, it doesn't matter, but can we pick a simple one? Like Aoc? --prime mover 17:34, 6 February 2012 (EST)

I hope that the proof of 'AC implies Zorn' does not need AC in the first place. When AC is rejected, this result is still, trivially, true. It's always raining when I play golf.

On second thought, I agree that Zorn needn't be in the axiom namespace. --Lord_Farin 01:01, 7 February 2012 (EST)

"When AC is rejected, this result is still, trivially, true." No, I don't think so. Can you back up your claim? Because I'm about to post up a page stating that Zorn's Lemma implies AoC. --prime mover 01:25, 7 February 2012 (EST)

Come on. We have that $A \implies B \iff \neg A \lor B$. The only way this theorem (Axiom of Choice Implies Zorn's Lemma) could be false would be when AC is true, but Zorn isn't. Negating AC does not eviscerate the theorem that acceptance of AC leads to the truth of Zorn. Which is the only thing this theorem says; hence the name. Hence the very first comment I made. Proving Zorn does require AC, but that is something different entirely. Let me point out that I never play golf... --Lord_Farin 01:33, 7 February 2012 (EST)

Yes I know, I wrote that page myself. If AoC is true, then Zorn's Lemma is true. That is indeed all this theorem is stating. I'm about to post up another theorem that says, "If Zorn's Lemma is true, then AoC is true." I just haven't got to doing it yet. What is your problem again? Because you're not making sense. --prime mover 02:00, 7 February 2012 (EST)

Ah yes, sorry, I understand what you're saying now. Let $A$ be AoC. Let $B$ be "AoC implies Zorn's Lemma". $B$ is true, $A$ may not be. So what do you want this page to read? I'm in over my head. Reword it, restructure, it, delete it completely for all I care because I've come to hate this entire field. --prime mover 02:13, 7 February 2012 (EST)

Well, the page needn't change. It's just that I consider it a tad... careless to say that this result depends on AC. Namely, it doesn't. So, I was bringing this up to discuss what possible alternative phrasings there are. Sorry to hear that you hate logic and axiomatic set theory; these are some of my favourite fields. --Lord_Farin 02:19, 7 February 2012 (EST)

OFFS you know what the F I meant. You meant "This result" to mean "AoC -> ZL" and I fell into your trap by thinking you meant "ZL". It's precisely this sort of misdirection that makes me hate logic. It's not logic I hate, it's Fing logicians. --prime mover 02:25, 7 February 2012 (EST)

If it calms your mind: I wasn't confusing you on purpose. --Lord_Farin 03:01, 7 February 2012 (EST)