Cardinality of set of well-orderable subsets of a non-well-orderable set

Question

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 $P'(A)$?

Answer 1

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.