1
$\begingroup$

Is there a clear inconsistency with the following?

There exists a countable transitive model of Zermelo set theory, such that for all external bijections between sets the images and preimages of sets under them are sets.

Formally:

$\exists M: (M \models \mathsf Z) \land |M|=\omega \land \forall x \in M \, (x \subset M) \land \\ \forall f: \exists x,y \in M \, (f:x \to y \land \operatorname {bijection}(f)) \to \forall a \in M \, ( f[a]\in M \land f^{-1}[a] \in M)$

Where: $f[a]=\{f(x) \mid x \in a\} \\ f^{-1}[a]= \{x \mid f(x) \in a\}$

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

There is no such model. Suppose that $M$ is a countable transitive model like that. Since $M$ is countable, it has only countably many subsets of $\omega$. Let $f:\omega\to\omega$ be a bijective function so that $f[e]$ is a subset of $\omega$ that $M$ lacks, where $e$ is the set of even numbers, and $f$ is defined on the odd numbers so as to be bijective. But $e\in M$, and so this contradicts your property, since $f[e]$ is not in $M$.

If you drop the countability requirement, then $\langle V_{\omega+\omega},\in\rangle$ is a model of Zermelo set theory with that feature, because it is supertransitive, and so the function $f$ itself will be a member.

Improve this answer
$\endgroup$
4
  • $\begingroup$ What's the proof that such a function exists in every countable transitive model of $\sf Z$? Because your argument begins with an assumption that clearly violates the needed condition. $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 21 at 11:44
  • $\begingroup$ Externally, the even numbers are bijective with any infinite subset of $\omega$, so we can define such an $f$ by mapping $2n$ to the $n$th element of the missing set. So $f[e]$ will be that missing set. And then we define $f(2n+1)$ so as to make it a bijection $f:\omega\to\omega$. (Note: the missing subsets will be infinite/coinfinite.) $\endgroup$
    – Joel David Hamkins
    Commented Jul 21 at 11:58
  • 2
    $\begingroup$ So, we take any missing subset $S$ of $\omega$, then we externally define a functin $f^1$ from $e$ to $S$ that is bijective, then we define another bijective function $f^2$ from the odds to the complementary set of $S$ relative to $\omega$. Then the union of these two functions would be a function from $\omega \to \omega$ that will contradict the stipulated condition. I think, that's what you are saying. Nice! $\endgroup$
    – Zuhair Al-Johar
    Commented Jul 21 at 12:03
  • 1
    $\begingroup$ Yes, that is exactly what I had in mind. $\endgroup$
    – Joel David Hamkins
    Commented Jul 21 at 12:16

You must log in to answer this question.

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