Talk:Tarski's theorem about choice

message: Did the requester really want the Knaster-Tarski theorem?David.Monniaux 20:58, 17 Sep 2003 (UTC)
replies: text: I don't think so.  I think they're two completely different things.  The Mathworld page on the Tarski theorem shows it as having to do with arithmetic operations on real numbers while the Knaster-Tarski theorem has to do with sets of points in a lattice. -- Ortonmc 21:23, 17 Sep 2003 (UTC)

text: I'm no longer sure about this.  After a bit more poking around, it appears there may be more than one theorem called "Tarski's Theorem", the Knaster-Tarski theorem being one of them. -- Ortonmc 21:52, 17 Sep 2003 (UTC)
thread_title:  Knaster-Tarski theorem? 

message: :"Since the collection of all ordinals such that there exist a surjective function from B to the ordinal is a set..."
Why is that?  It smells like it should be the replacement axiom, but I don't see it.--80.109.80.78 (talk) 17:33, 23 January 2014 (UTC)
replies: text: You could take the Hartogs number of the powerset of B. Since any surjection from B to an ordinal determines an injection from the ordinal to the non-empty elements of the powerset of B, this should be an upper bound on the desired ordinal. JRSpriggs (talk) 06:23, 24 August 2019 (UTC)
thread_title:  Question on proof 

message: The proof in the article seems unnecessarily complicated to me. Here is a simpler proof.
We show the well-ordering theorem and thus the axiom of choice. Let S be any set. We wish to show that it can be well-ordered.
Let H be the maximum of ω and the Hartogs number of S. Then A := S∪H will be an infinite set. By hypothesis, there is an injection from A×A to A. Thus there is an injection f from S×H to S∪H.
For any element x∈S, there must be at least one element in { f(x,α) : α∈H } − S ⊆ H−S, otherwise we would have an injection from H to S which is contrary to the choice of H. Let g(x) be the least such element (in the ordering of H). g is an injection from S to H−S.  Then we order S according to the ordering pulled back from H−S thru g. This is a well-ordering of S as required.
OK? JRSpriggs (talk) 06:58, 24 August 2019 (UTC)
thread_title:  Simplified proof 

message: As the article says, "the opposite direction was already known". However, since it was not entirely obvious to me, here is a proof that the axiom of choice implies that there is a bijection between A×A and A when A is an infinite set.
We will show in ZF (without choice) that for any infinite ordinal α, α×α ≈ α. Since the axiom of choice implies that there is an α ≈ A, we get A×A ≈ α×α ≈ α ≈ A.
We proceed by transfinite induction with inductive hypothesis: ω≤α → α×α ≈ α. It is well-known that there are many pairing functions which realize ω×ω ≈ ω.
If α is an infinite ordinal which is not an initial ordinal, then there is an ordinal β such that ω≤β<α with α ≈ β. In this case, α×α ≈ β×β ≈ β ≈ α.
Otherwise, α>ω is initial.
Now I need to define a specific ordering, <csq, on α×α. ("csq" stands for "continuous square".) For any β, γ, δ, ε less than α, let 
:(β,γ) <csq (δ,ε) ←→ max(β,γ) < max(δ,ε) or ( max(β,γ) = max(δ,ε) and ( γ<ε or ( γ=ε and β<δ ) ) ).
This is a well-ordering on α×α. So it gives us α×α ≈csq ζ for some ordinal ζ. If ζ=α, then we are done.
If α<ζ, then there is some (β,γ) in α×α at the location corresponding to α in the <csq ordering. Let η = max(β,γ)+1 which is less than α. Then α can be injected into η×η ≈ η which can be injected into α, so the Schröder–Bernstein theorem gives us α ≈ η contradicting the fact that α is initial.
If ζ<α, then we can inject α into α×α using the map which takes η to (η,0). So α injects into α×α ≈ ζ which injects into α. Again using the Schröder–Bernstein theorem gives α ≈ ζ contradicting the fact that α is initial.
OK? JRSpriggs (talk) 21:42, 26 August 2019 (UTC)
thread_title:  Proof of the converse