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11:07 AM
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Q: Zorn's Lemma And Axiom of Choice

Abcd JHow can I prove Zorn's lemma is equivalent to Axiom of choice?

can you refer me to a proof for ZL $\Longrightarrow$ AC? — Jneven 9 mins ago
Probably if you search a bit online you could find some proofs. Some of them even directly on this site.
I also mentioned a short sketch of the proof in the answer where you commented:
> The proof of ZL $\Rightarrow$ AC is similar to most applications of Zorn's lemma. We want to show that there exists a selector (choice function). This selector is obtained from ZL as maximal partial selector. So the only thing is to show that union of a chain (w.r.t. inclusion) of partial functions, which select one element from each set is again such a function.
But to give at least one specific book it is shown in the solution of Problem 2, Chapter 14 in the book P. Komjáth, V. Totik: Problems and Theorems in Classical Set Theory.
@Jneven There are many sources where this proof can be found. (I'd guess that it should be in most standard textbooks of set theory.) Anyway, we can discuss this in set theory chatroom - I'll post there some suggestions I am able to find. — Martin Sleziak 2 mins ago
There are several books which show proofs of several equivalent forms of AC in the same theorem. So, for example, they might go AC $\Rightarrow$ ZL $\Rightarrow$ TP $\Rightarrow$ AC. (Where TP stands for Teichmüller's Principle.) And other similar variations.
The current version of the article at ProofWiki seems to me more as a sketch of the proof than an actual proof: proofwiki.org/wiki/Zorn%27s_Lemma_Implies_Axiom_of_Choice
The implication e $\Rightarrow$ a of Theorem 9.3 in Andras Hajnal, Peter Hamburger: Set Theory is exactly this implication. This part of the proof of Theorem 9.3 is on page 73.
 
11:54 AM
Maybe you would be also satisfied with the proof ZL $\Rightarrow$ WO, since from the Well-ordering Principle you can get Axiom of Choice immediately.
But if you're not interested in well-orderings, it is more natural to look at ZL $\Rightarrow$ AC. (It's probably easier. Especially for people who did not work much with well-orders.)
 

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