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Add a primitive partial unary function symbol $F$ to the first order language of set theory.

Working in Zermelo (Separation restricted to the language of set theory), add the following axioms:

$F$ function: $\exists X ( F: \mathcal P(X) \longrightarrow X, F \text { is injective} )$

Forward copying:$ \forall A \exists B: B = \{ \{ F(n)| n \in m \}| m \in A \} $

Backward copying: $ \forall A \exists B: B = \{ \{ n| F(n) \in m \}| m \in A \} $

Note: It needs to be emphasized that $F$ is an external function, and that it cannot be used in instances of separation. The proof that such an injective $F$ can exist can be seen from Boffa model construction for NFU. See here page 5, only restrict the automorphism $j$ to $V_{\alpha +1}$, and you get our $F$ where $V_\alpha$ would witness $X$ here.

Now if the graph of $F$ happens to be an isomorphism on $\in$ [between $dom(F)$ and $Range(F)$], then $F$ cannot be bijective! Since the parity of the set of all ordinals in $\mathcal P(X)$ is different from that in $X$. Hence the following questions:

Is there an example of $F$ whose graph is not an isomorphism on $\in$ between its domain and range?

If so, then is it the case that the qualifications of $F$ in the above system are still enough as to forbid $F$ from being a bijection?

The rationale beyond the above question is that if there is no clear argument against $F$ being bijective, then this might entail opening the door for a possible proof of $\sf Con(NF)$, as $F$ being bijective would easily interpret a finite axiomatization of $\sf NF$. The interpretation is over $\mathcal P^{-1} (dom(F))$ using a new membership relation $\in^F$ defined as: $$ y \in^F x \iff y \in F^{-1}(x)$$

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  • $\begingroup$ There is no such $F$ in any elementary topos, if I am not mistaken this then is also not possible in Zermelo, no? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 1, 2021 at 15:33
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    $\begingroup$ Here is an argument adapted from Johnstone's "Sketches of an Elephant", D.4.1, Proposition 4.1.8(i) (page 948). It is in fact the standard Russell paradox kind of argument. I believe it does not use anything beyond Zermelo. In fact it does not even use tertium non datur. Let $$R:=\{F(Y)\mid Y\subseteq X \& F(Y)\notin Y\},$$and suppose $F(R)\in R$. This means that $F(R)=F(Y)$ for some $Y\subseteq X$ with $F(Y)\notin Y$. But since $F$ is injective, $F(R)=F(Y)$ implies $R=Y$, so $F(R)\notin R$. This in turn implies that $F(R)\ne F(Y)$ for any $Y\subseteq X$ with $F(Y)\notin Y$, so $F(R)\ne F(R)$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jan 1, 2021 at 16:19
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    $\begingroup$ Now the map $G$ sends sets of subsets of $X$ to their $F$-set images (i.e. sets of all F images of their elements), while map $H$ sends sets of sets of subsets of $X$ to their $G$-set images. The forwards and backward axioms are about map $H$ and its converse. $\endgroup$
    – Zuhair Al-Johar
    Commented Jan 1, 2021 at 22:46
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    $\begingroup$ I think this is an interesting question, and I’m curious about the anonymous downvote — would the downvoter care to explain their reasoning? $\endgroup$
    – Alec Rhea
    Commented Jan 2, 2021 at 17:20
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    $\begingroup$ @მამუკაჯიბლაძე, you can have a model of Zermelo in which all von Neumann ordinals are (internally) countable definitely. But, even in those models there are well orderings that are (internally) uncountable, for example a well ordering on the set of all equivalence classes of well orderings over N under order isomorphism. $\endgroup$
    – Zuhair Al-Johar
    Commented Jan 3, 2021 at 9:53

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