-2
$\begingroup$

Language: mono-sorted ${\sf FOL}(=,\in,S)$, where $S$ is a unary predicate standing for ".. is a stage".

Axioms:

  1. Extensionality: $\forall z \, (z \in x \leftrightarrow z \in y) \to x=y$

  2. Positive Separation: $[\exists y \in A: \phi, \to \exists x \forall y (y \in x \leftrightarrow y \in A: \phi)]$; if $x$ doesn't occur in formula $\phi$.

  3. No Empty set: $\forall x \exists y: y \in x$

  4. Acyclicity: $n=1,2,3, \ldots; \\ \neg \, \exists x_0, \ldots, \exists x_n: x_0 \in \cdots \in x_n \land x_0 = x_n$

  5. Existence: $\exists x: S(x)$

  6. Stages: $\forall x: S(x) \to \exists y: S(y) \land x=\mathcal P(y)$

  7. Staging: $\forall x \exists y: S(y) \land x \in y$

  8. Multiplicity: $\exists x \exists y \exists z : S(x) \land y \in x \land z \in x \land y \neq z$

Where $\mathcal P$ is the powerset operator.

Is this theory consistent?

Does this theory prove infinity?

Is it consistent with Choice?

Improve this question
$\endgroup$
4
  • 1
    $\begingroup$ Is a bottomless hierarchy a lowerarchy? $\endgroup$ Commented Jul 22 at 20:42
  • $\begingroup$ @bof, I don't know what a lowerarchy means but if it means emphasis on lower stages then no. here lower stages which are the standard ones are washed out. If you mean a descending kind of construction, then also no, this goes both up and down. If one is to use a new terminology then possibly "birarchy" would be better. $\endgroup$ Commented Jul 22 at 20:54
  • $\begingroup$ I confess that I have long been a little confused by your manner of using colons and commas in formal expressions, since it is different from what I am used to in first-order logic or in type theory. For example, how am I to read the meaning of "$\exists y\in A:\phi\to,$"? And how are we to read "$y\in x\leftrightarrow y\in A:\phi$" exactly? To be honest, I have often been confused by your questions in this regard. $\endgroup$ Commented Jul 23 at 2:02
  • $\begingroup$ Ah, sorry, you had the comma before $\to$ not after, namely, $\exists y\in A:\phi,\to$. $\endgroup$ Commented Jul 23 at 2:22

1 Answer 1

Reset to default
1
$\begingroup$

I think we can find a model for this theory. Work in $\sf ZFA$ where we have an infinite set $A$ of all atoms (i.e., distinct empty objects) minus Regularity + there is a model $M$ of that theory that is countable, transitive, and not $\omega$-model. So, we have $\in^M$ being $\in \restriction M$, and there are elements $n \in M$ which are not naturals externally speaking, but seen inside $M$ as naturals.

Build an iterative hierarchy inside $M$ over $A^M$, as:

$W^M_0= A^M \\ W^M_{{n+1}^M}= \mathcal P^M (W^M_{n^M}) \setminus \{\emptyset^M\}$

Where $n^M$ is an $M$-natural number.

Where $\emptyset^M$ is the empty set inside $M$; $A^M$ is the set of all atoms in $M$; and $\mathcal P^M(x)$ is the set in $M$ of all subsets of $x$ that are elements of $M$.

This way for each positive $n$ we have all elements of $W^M_n$ being nonempty, and for each $n,m \in M$ if $n \neq m$ then $W^M_n;W^M_m$ are disjoint!

Now, let $W^M=\bigcup_{n \in \omega^M} W^M_n$. Now working form outside of $M$ we can see which $M$-naturals are standard and which are not. So, externally we define the set $\mathbb W = \bigcup_{n \in \mathbb N} W^M_n$, then cut it form $W^M$. That is, define $\mathbb W' = W^M \setminus \mathbb W$, and $\langle \mathbb W', \in \restriction \mathbb W'\rangle$ would be a model of this theory.

I don't think infinity is provable in this theory, where infinity is phrased as there exists a nonempty set $x$ and a relation $R$ on $x$ that is areflexive, and transitive, and such that for every $a \in x$ there is $b \in x$ such that $a \ R \ b$. Since we can have a similar construction but starting with $A$ having only two distinct elements. Thereby the theory would be consistent with Choice.

Improve this answer
$\endgroup$
18
  • 1
    $\begingroup$ What does it mean to "trim" a set? What are the stages? Why should axiom 3 hold? $\endgroup$ Commented Jul 22 at 4:07
  • $\begingroup$ OK, I've edited. trim means cut it out. The stages are interpreted in $\mathbb W'$ as $W^M_n$ where $n$ is nonstandard natural. Axiom 3 holds because there is no empty set in all of these stages, and positive stages only contain sets. $\endgroup$ Commented Jul 22 at 16:31
  • 1
    $\begingroup$ Why does axiom 2 hold in 𝕎′? $\endgroup$ Commented Jul 22 at 17:20
  • $\begingroup$ @GregKirmayer, yes this is the weak point, actually I myself has been doubting it, but I think if we make $\phi$ a bounded formula by stages then it would go through. Anyhow I didn't mange to produce a clear counterexample. $\endgroup$ Commented Jul 22 at 20:49
  • $\begingroup$ @GregKirmayer, Since we we have: $\forall A_1 \in W^M_{i_1}, \cdots , \forall A_n \in W^M_{i_n} : \\ [\exists y \in W^M_{i_{n-1}} : y \in A_n \land \phi^{W^M}] \to \\\exists x \in W^M_{i_n} \forall y \in W^M_{i_{n-1}} (y \in x \leftrightarrow y \in A_n \land \phi^{W^M})$, And $\mathbb W'$ is transitive. Then we get $\forall A_1 \in \mathbb W', \cdots , \forall A_n \in \mathbb W' : \\ [\exists y \in \mathbb W' : y \in A_n \land \phi^{\mathbb W'}] \to \\\exists x \in \mathbb W' \forall y \in \mathbb W' (y \in x \leftrightarrow y \in A_n \land \phi^{\mathbb W'})$. to be continued.. $\endgroup$ Commented Jul 23 at 11:38

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.