Talk:Tarski's theorem about choice

Knaster-Tarski theorem?
Did the requester really want the Knaster-Tarski theorem?David.Monniaux 20:58, 17 Sep 2003 (UTC)


 * 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)


 * 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)

Question on proof

 * "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)

No set can be greater than every ordinal, that's actually a Theorem of Hartogs-Sierpinski. It constructs for every infinte set B a well-ordered subset C of 2^2^2^B with no possible injection from C into B. Then the proof of Tarski's theorem goes through identically, replacing only β with C and surjective with injective. — Preceding unsigned comment added by 87.171.165.186 (talk) 01:49, 22 June 2016 (UTC)


 * 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)

Simplified proof
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 &omega; and the Hartogs number of S. Then A := S&cup;H will be an infinite set. By hypothesis, there is an injection from A&times;A to A. Thus there is an injection f from S&times;H to S&cup;H.

For any element x&isin;S, there must be at least one element in { f(x,&alpha;) : &alpha;&isin;H } &minus; S &sube; H&minus;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&minus;S. Then we order S according to the ordering pulled back from H&minus;S thru g. This is a well-ordering of S as required.

OK? JRSpriggs (talk) 06:58, 24 August 2019 (UTC)

Compared to the proof in the article, this reverses the direction of the bijection and uses the property of injection rather than surjection. JRSpriggs (talk) 07:13, 24 August 2019 (UTC)

Proof of the converse
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&times;A and A when A is an infinite set.

We will show in ZF (without choice) that for any infinite ordinal &alpha;, &alpha;&times;&alpha; ≈ &alpha;. Since the axiom of choice implies that there is an &alpha; ≈ A, we get A&times;A ≈ &alpha;&times;&alpha; ≈ &alpha; ≈ A.

We proceed by transfinite induction with inductive hypothesis: &omega;≤&alpha; → &alpha;&times;&alpha; ≈ &alpha;. It is well-known that there are many pairing functions which realize &omega;&times;&omega; ≈ &omega;.

If &alpha; is an infinite ordinal which is not an initial ordinal, then there is an ordinal &beta; such that &omega;≤&beta;<&alpha; with &alpha; ≈ &beta;. In this case, &alpha;&times;&alpha; ≈ &beta;&times;&beta; ≈ &beta; ≈ &alpha;.

Otherwise, &alpha;>&omega; is initial.

Now I need to define a specific ordering, <csq, on &alpha;&times;&alpha;. ("csq" stands for "continuous square".) For any &beta;, &gamma;, &delta;, &epsilon; less than &alpha;, let
 * (&beta;,&gamma;) <csq (&delta;,&epsilon;) ←→ max(&beta;,&gamma;) < max(&delta;,&epsilon;) or ( max(&beta;,&gamma;) = max(&delta;,&epsilon;) and ( &gamma;<&epsilon; or ( &gamma;=&epsilon; and &beta;<&delta; ) ) ).

This is a well-ordering on &alpha;&times;&alpha;. So it gives us &alpha;&times;&alpha; ≈csq &zeta; for some ordinal &zeta;. If &zeta;=&alpha;, then we are done.

If &alpha;<&zeta;, then there is some (&beta;,&gamma;) in &alpha;&times;&alpha; at the location corresponding to &alpha; in the <csq ordering. Let &eta; = max(&beta;,&gamma;)+1 which is less than &alpha;. Then &alpha; can be injected into &eta;&times;&eta; ≈ &eta; which can be injected into &alpha;, so the Schröder–Bernstein theorem gives us &alpha; ≈ &eta; contradicting the fact that &alpha; is initial.

If &zeta;<&alpha;, then we can inject &alpha; into &alpha;&times;&alpha; using the map which takes &eta; to (&eta;,0). So &alpha; injects into &alpha;&times;&alpha; ≈ &zeta; which injects into &alpha;. Again using the Schröder–Bernstein theorem gives &alpha; ≈ &zeta; contradicting the fact that &alpha; is initial.

OK? JRSpriggs (talk) 21:42, 26 August 2019 (UTC)

If we extend the analysis of &zeta; to situations where &alpha; is not initial, we find: OK? JRSpriggs (talk) 07:14, 30 August 2019 (UTC)
 * &zeta;<&alpha; never occurs because &zeta; is a strictly increasing and continuous function of &alpha;. See normal function.
 * The class of &alpha; for which &zeta;=&alpha; is closed. (I suspect that it includes all &alpha; of the form &omega;&omega; &nu; .)