6
$\begingroup$

From what I understand in ZF we have that (i) implies (ii) implies (iii) implies (iv)

(i) Infinite and well-orderable

(ii) Infinite and $|A\times A|=|A|$

(iii) Dedekind-Infinite (i.e. $|A|+1=|A|$)

(iv) Infinite (i.e. not bijective with any finite ordinal)

I know that there are models where there exist infinite sets that are Dedekind-finite. I also recently learned there are Dedekind-Infinite sets that do not have the property: $|A\times A|=|A|$. Namely

$$X\sqcup \aleph(X)$$

where $X$ is not well-orderable and $\aleph(X)$ is the Hartogs' ordinal number.

TWO QUESTIONS

(1) I am pretty sure that (ii) does not imply (i), but only because of how Tarski's Theorem is proved (requiring all infinite sets to have property ii). Is this an open problem? I also know for example that assuming CH at an infinite cardinal $\mathfrak{p}$ and its power set $2^\mathfrak{p}$ implies $\mathfrak{p}$ can be well-ordered, but it is not known that if CH only at $\mathfrak{p}$ is sufficient. Since CH($\mathfrak{p}$) implies $\mathfrak{p}^2=\mathfrak{p}$, it seems to me that it is an open question.

(2) Between (iii) and (ii) there is a notion of infinity: Infinite and $|A|\times 2=|A|$. Is this notion strictly in between?

If you have a reference that answers or studies these kinds of questions, then please share it with me. Also thanks for your time and patience in writing a response!

$\endgroup$

1 Answer 1

4
$\begingroup$

Your intuition is correct that (ii) does not imply (i). There are models of ZF in which $\mathbb{R}$ is not well-orderable, but $\mathbb{R}$ always satisfies (ii) since $|\mathbb{R}\times\mathbb{R}|=2^{\aleph_0}\cdot 2^{\aleph_0}=2^{\aleph_0+\aleph_0}=2^{\aleph_0}$.

The answer to your second question is also yes. To get a set $A$ which satisfies your condition but not (ii), let $X$ be a set that is not well-orderable, let $Y=X\times\mathbb{N}$, and let $A=Y\sqcup\aleph(Y)$. Then $|A|\times 2=|A|$ since the same is true of $Y$ and $\aleph(Y)$. But if $|A\times A|=|A|$, then $Y$ is well-orderable, which implies $X$ is well-orderable since it injects into $Y$.

To get a set $A$ which satisfies (iii) but not your condition, let $X$ be an infinite Dedekind-finite set and let $A=X\sqcup\mathbb{N}$. Clearly $A$ satisfies (iii). If there were an injection $f:A\times\{0,1\}\to A$, then $f$ could map at most finitely many elements of $X\times \{0,1\}$ to $\mathbb{N}$, since otherwise you could invert $f$ on an infinite subset of $\mathbb{N}$ to get an injection from a countably infinite set to $X$. So there is a finite set $F$ such that $f$ restricts to an injection $g:X\times\{0,1\}\setminus F\to X$. Taking finitely many points of $X\times\{1\}$ to replace all the points of $F\cap X\times\{0\}$, we may assume that $g$ is actually defined on all of $X\times\{0\}$, plus at least one point of $X\times \{1\}$. This would mean $|X|+1=|X|$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .