2
$\begingroup$

Language (first order logic with equality "$=$" and membership "$\in$", and constant symbol "$j$")

Axiom: ID axioms +

There exists a set $A$, such that:

  1. Field: $\forall x \in j \ \exists a \in A \ \exists b \subseteq A (x=\langle a,b \rangle)$

$$\text{Define: } j(a)=b \iff \exists x \in j \ (x=\langle a,b \rangle)$$

  1. Extensionality: $\forall a,b \in A \ [\forall x (x \in j(a) \leftrightarrow x \in j(b)) \rightarrow j(a)=j(b)]$

  2. Functionality: $\forall a \in A \ \exists ! b \ (j(a)=b)$

  3. Functional singletons: $\forall a \in A \ \exists b \ (b=\{a\} \wedge \exists k \in A (j(k)=b))$

  4. Functional complements: $\forall a \in A \ \exists a' \ (a'=\{x \in A| x \not \in j(a)\} \wedge \exists k \in A (j(k)=a'))$

  5. Functional binary union: $\forall a,b \in A \ \exists u \ (u=j(a) \cup j(b) \wedge \exists k \in A (j(k) = u))$

  6. Functional set Union: $\forall x \in A \ \exists y \ ( y=\{z| \exists m \in j(x) (z \in j(m))\} \wedge \exists k \in A (j(k)=y) )$

  7. Functional unordered relative products: $$\forall x,y \in A \ \exists z \\ (z=\{m| \exists k,l,a,b,c \ ( k \in j(x) \wedge l \in j(y) \wedge j(k)=\{a,c\} \wedge j(l)=\{c,b\} \wedge j(m)=\{a,b\}) \} \wedge \exists k \in A (j(k)=z))$$

  8. Functional unordered intersection relation set: $$\exists x (x=\{y| \exists a,b,c,a^*,b^* (c \in a \wedge c \in b \wedge j(a^*)=a \wedge j(b^*)=b \wedge j(y)=\{a^*,b^*\})\}\wedge \exists k \in A (j(k)=x))$$

In English: this theory speaks of existence of a function $j$ that sends elements of a set $A$ to subsets of $A$ such that all singleton subsets of $A$ are in the range of $j$ and such that the range of $j$ is closed under Boolean union and complements relative to $A$, functional set union refer to existence of a set of all elements of subsets of $A$ that are $j$ coded by elements of a subset of $A$ present in the range of $j$, and that this set is $j$ coded by an element of $A$, functional intersectional relation set is the set of all $j$ codes of pairs of $j$ codes of intersecting subsets of $A$ that are present in the range of $j$, and that this set is in the range of $j$, while functional relative products is better be understood formally as written.

/Theory definition finished.

Then we get to interpret $NFU$, by defining a new membership relation $\epsilon$ as:

$y \ \epsilon \ x \leftrightarrow y \in j(x)$

over domain $A$.

If we additionally add that $j$ is an injection, i.e.:

$\forall a,b \ [j(a)=j(b) \rightarrow a=b]$

Then we get to interpret $NF$.

I don't see any clear argument that prevents $j$ from being an injective relation!

Question: is there any clear argument against $j$ being an injective function?

The idea behind this question is that if we use the usual Boffa model construction for NFU [click [here] (Page 5)], then take the downward rank moving automorphism $J$ (written as $j$ in the referred article, changed here to $J$ to avoid confusion with the above mentioned function $j$ of this posting), then take its converse from $V_{J(\alpha)+1}$ to $V_{\alpha+1}$ which would be an isomorphism, now take the Boolean union of it with the set $\{\langle x,\emptyset \rangle| x \in V_{\alpha} \setminus V_{J(\alpha)+1}\}$, then this union would be a witness of the above-mentioned function $j$, and this clearly interprets $NFU$ through relation $\epsilon$ defined above over domain $V_{\alpha}$. However this modified Boffa approach cannot be further modified here by letting $J$ be a bijection between $V_{\alpha +1}$ and $V_{\alpha}$ as to get $NF$, since $J$ is an isomorphism and sets are never isomorphic to their powers. Yet the point is that $j$ as presented in the posting here need not be an isomorphism!!! so this obstacle is removed, and hence the question posed here about whether $j$ can be an injective function while still fulfilling the above axioms.

$\endgroup$
1
$\begingroup$

Once you have such a $j$ you can make another one that is not injective: take a bijection $b:A\to A\times A$ and the projection $\pi:A\times A\to A$ onto the first coordinate. Then the composition $j\circ\pi\circ b$ is far from injective and it also satisfies your conditions.

On the other hand: by means of a well-order $\prec$ of $A$ in type $|A|$ you can turn any non-injective $j$ into an injective one. From $\prec$ define $B=\{x\in A:(\forall y\in A)(j(y)=j(x)\Rightarrow x\preceq y)\}$, now the restriction of $j$ to $B$ is injective and $j[B]=j[A]$; let $b:A\to B$ be the unique order-preserving bijection defined by $b(x)=\min B\setminus\{b(y):y\prec x\}$. Then the composition $j \circ b$ would be an injection that satisfy all properties 1--9.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't know what is the relevance of the first remark? about the second remark, there is a problem, we know that $NFU + infinity + choice$ is consistent, we do have such a non-injective function $j$ (I already mentioned that explicitly), but you cannot use the well order to get an injective $j$ that satisfy the above conditions, since if you do that then you get $NF$ and this is known to kill choice. You need to elaborate a bit on your second remark. Actually if you can demonstrate what you said in the last line without violating choice? then you would have proved the consistency of $NF$ $\endgroup$ – Zuhair Al-Johar Oct 9 '18 at 11:41
  • 3
    $\begingroup$ The relevance of my remark is that given a function $j$ with your list of properties there is another function with those same properties that is not injective, that is, properties 1--9 do not guarantee injectivity. That is what I intended to show. Likewise the second remark shows that an injective $j$ with properties 1--9 is possible if there is one at all. That's it. I'm looking from the outside at your potential interpretation of NF(U) and observe, in ZFC, that your conditions neither imply nor prevent injectivity. $\endgroup$ – KP Hart Oct 9 '18 at 11:55
  • 2
    $\begingroup$ I was busy yesterday and didn't have time to react but it seems we are working in different frameworks. All my definitions are in 'normal' set theory (ZFC) and from the outside; the well-order of $A$ is something I imposed, then $B$ and $b$ are definable from that well-order. I felt justified in doing that because it looked like you were using 'normal' set theory as well to talk about a possible $A$ and $j$ and only later use these to talk about NF(U). I do not know enough of NF(U) to be able to say something about the existence of $B$ and $b$ within that context. $\endgroup$ – KP Hart Oct 11 '18 at 13:38
  • 1
    $\begingroup$ Indeed, I did not study 8 and 9 carfully enough. One can meet demands 1-7 by having a surjection from $A$ onto the finite-cofinite algebra on $A$ but 8 and 9 are quite different.Alas $\endgroup$ – KP Hart Oct 13 '18 at 20:44
  • 1
    $\begingroup$ Clause 4 implies that $\{\{a\}:a\in A\}$ is in the range of $j$, so $B$ has a subset that maps onto $A$; this implies that $|B|=|A|$. $\endgroup$ – KP Hart Oct 15 '18 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.