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Suppose, with the negation of Axiom of Choice, we have a non-well-orderable set $A$, and its power set $P(A)$, let $P'(A)$ be $\{x \in P(A): x \text{ is well orderable}\}$

Is there an injection from $P'(A)$ into $A$?

What can we say about the cardinality of $P'(A)$?

Metta World Peace
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1 Answers1

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Let me denote this set as $\mathcal W(A)$, which is a notation I recall seeing before (although not where, at the moment).

Clearly if $A$ can be well-ordered then $\mathcal W(A)=\mathcal P(A)$ and therefore we cannot prove that $\mathcal W(A)<\mathcal P(A)$, or that $|A|=|\mathcal W(A)|$. But it is also clear that if $A$ cannot be well-ordered then $\mathcal W(A)\neq\mathcal P(A)$.

  • It is consistent that $|\mathcal W(A)|<|\mathcal P(A)|$:

    Suppose that $A$ is [infinite and] amorphous (cannot be written as a disjoint union of two infinite sets) then every well-orderable subset is finite. However $\mathcal P(A)$ is Dedekind-finite, so the cardinality of all finite subsets of $A$ is strictly smaller than that of $\mathcal P(A)$ (in fact it is exactly half!), and so it is consistent that the inequality is sharp.

  • It is provable that $|A|<|\mathcal W(A)|$:

    1. Tarski, A. On well-ordered subsets of any set. Fundamenta Mathematicae 32, pp. 176–183 (1939);
    2. Truss, J. K. The well-ordered and well-orderable subsets of a set. Mathematical Logic Quarterly 19, pp. 211–214 (1973).

    You could also find a paragraph in L. Halbeisen's Combinatorial Set Theory, Chapter 5, Related Results no. 33.

Asaf Karagila
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    Zermelo gave a proof that $|A|<|\mathcal P(A)|$ that is essentially different from Cantor in that it avoids diagonalization. You can see it here: http://math.stackexchange.com/questions/284812/a-reverse-diagonal-argument/ That proof only uses $\mathcal W(A)$, so in fact Zermelo's argument gives the inequality $|A|<|\mathcal W(A)|$. – Andrés E. Caicedo Jan 27 '13 at 16:13
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    Stevo Todorcevic developed a nice theory of partition calculus of partially ordered sets in "Partition relations for partially ordered sets", Acta Math. 155 (1-2), (1985), 1–25. At the beginning of the paper, Stevo reproves $|A|<|\mathcal W(A)|$ as a consequence of very general results in his partition calculus setting. – Andrés E. Caicedo Jan 27 '13 at 16:16
  • Andres, thank you for the addition! – Asaf Karagila Jan 27 '13 at 16:23