Can an infinite subset of $\omega$ be Dedekind-finite in ZF?

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Can an infinite subset of $\omega$ be Dedekind-finite in ZF?

Say a set $X$ is infinite if for every $y \in \omega$ there is an injection from $y$ to $X$. And say $X$ is Dedekind-finite if there is no injection from $\omega$ to $X$. Equivalently, $X$ is infinite and Dedekind-finite iff its Hartogs number is $\omega$. Is it consistent for a set $X \subset \omega$ to have this property?

I have found references showing that is possible to have $X \subset \mathbb{R}$, but these may be assuming countable choice.

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No: if $X$ is well-orderable and infinite, then $X$ is Dedekind-infinite, and every subset of a well-orderable set is well-orderable.

Specifically:

  • Since $X$ is well-orderable, by transfinite recursion $X$ is in bijection with some ordinal $\alpha$.

  • Since $X$ is infinite, $\alpha\ge\omega$.

  • OK, so look at the preimage of $\omega$ under that bijection ...


A related result in spirit is that no linearly orderable set can be amorphous (= not partitionable into two infinite sets): if $X$ is linearly ordered by $\trianglelefteq$, let $L$ be the set of elements with infinitely many things $\trianglerighteq$ them, and $R$ be the set of elements with infinitely many things $\trianglelefteq$ them.

If $R\cap L\not=\emptyset$, then any element of the intersection induces a partition of $X$ into two infinite pieces; so $R\cap L=\emptyset$. This means that exactly one of $R, L$ is infinite.

Without loss of generality, suppose it's $L$. Since no element of $L$ is in $R$, we know that $L$ - thought of as a linear order in its own right - has the finite predecessor property. But the only infinite linear order with the finite predecessor property is $\omega$ - so $X$ isn't even Dedekind-finite!

Note that this argument doesn't prevent the existence linearly-orderable infinite Dedekind-finite sets; the point is that we used the assumption that $X$ was amorphous in analyzing $R\cap L$.

In a certain sense, this is the main obstacle: while it is not possible to have an amorphous set of reals, it is possible (in ZF) to have an amorphous set of sets of reals.

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No. If $X\subseteq\omega$ then the map $x\mapsto|\{y\in X:y\lt x\}|$ is a bijection from $X$ to some finite ordinal or to $\omega.$