An exercise from Herrlich

Question

Construct a set $X$ with the following properties:

1. X is well orderable.
2. $|X| \not\lt |\mathbb{R}|.$
3. $|X|\not\gt |\mathbb{R}|.$
4. $|X| = |\mathbb{R}|$ iff $|\mathbb{R}|$ is well orderable.

This is an exercise from Herrlich Axiom of choice. The author gives a hint for this problem suggesting the use of Hartogs function, denote $h(X)$. If $h(X)$ is a successor cardinal, the fourth condition implies putting X as the predecessor cardinal.http://matwbn.icm.edu.pl/ksiazki/fm/fm2/fm2114.pdf”> Thisis the reference to this exercise.

Answer 1

In ZF.

Let $C$ be the class of every ordinal $Y$ such that if there exists an injection $f:Y\to \Bbb R$ then there exists a bijection $g:Y\to \Bbb R.$

$C\ne \emptyset$ because not every ordinal can be injectively embedded into $\Bbb R,$ and any ordinal that can't, is in $C.$

Let $X=\min C.$

$(1).$ If $\Bbb R$ can be well-ordered:

Any well-ordering is order-isomorphic to the $\in$- order of some (unique) ordinal. So let $X'$ be the $\in$-least ordinal that is bijective with $\Bbb R.$ Then $|X'|=|\Bbb R|$ and $X'\in C.$

Let $h:X'\to \Bbb R$ be a bijection. Now if $Y\in X'$ then $h|_Y$ injects $Y$ into $\Bbb R,$ but $Y$ is not bijective with $\Bbb R$ by def'n of $X',$ so $Y\not \in C.$

So $X=\min C\ge X'\in C,$ so $X=X'.$

$(2).$ If $\Bbb R$ cannot be well-ordered:

$(2i).$ Any subset of an ordinal is well-ordered by $\in.$ So there cannot be an injection or bijection $h:\Bbb R \to X,$ else $<^*,$ where $x<^*y\iff h(x)\in h(y),$ would well-order $\Bbb R.$ So $\neg (|\Bbb R|<|X|).$

$(2ii).$ There cannot be an injection $f:X\to \Bbb R,$ else (since $X\in C$) there would be a bijection $g:X\to \Bbb R,$ but then $<',$ where $x<'y\iff g^{-1}(x)\in g^{-1}(y),$ would well-order $\Bbb R$. So $\neg (|\Bbb R|>|X|).$

$(3).$ By $(1)$ and by the "bijection" clause of $(2i)$ we have: $|X|=|\Bbb R|$ iff $\Bbb R$ can be well-ordered.