Informally the axiom schema of accessibility states that for each unary function $F$ that is definable over the whole universe of discourse "in the language of set theory", like the powerset function $P$, or the singleton function $\iota$, or the set Union function $\bigcup$, etc.. for each such a function $F$ we can coin an accessibility notion [dependent on $F$] whereby a set $x$ is said to be reached (i.e. is accessible) from a set $\alpha$ if and only if there is a set $\beta$ that is hereditarily strictly subnumerous to $x$ such that $F(\beta)$ is supernumerous to $x$ and $\alpha \subseteq \beta$; or otherwise $x$ is hereditarily subnumerous to $\alpha$ . Now the scheme states that per each definable unary function $F$ for every set $x$ there exists a set of all sets that are hereditarily $F$ accessible from $x$. The real intention is to have a theory that can prove the existence of large cardinals. Infinity is provable since all hereditarily finite sets are seen as hereditarily power accessible from the set $\emptyset$. Now it can be proved that the set of all cardinals that are hereditarily Power (of Set Union) accessible from $\omega_0$ exists and this would be a regular limit of regular cardinals, i.e. the first inaccessible cardinal, and from that one can prove the existence of a set of all sets that are hereditarily strictly subnumerous to this first inaccessible, and this would serve as a domain of a model of $\text{ZF}$. To get to that, we only need to add this axiom scheme on top of axioms of Zermelo set theory minus infinity and add an axiom of Transitivity that asserts that every set is a subset of some transitive set, this way we can define transitive closures and define the hereditary notions. The issues are: what would be the consistency strength of this theory? how much this theory is stronger than the theory with the $F$ function fixed to be the Power of union set function, i.e. $\forall x,y \ [F(x)=y \leftrightarrow y=P(\bigcup(x))]$? what kinds of inaccessible cardinals can this theory prove? Would this theory prove Replacement?

Formal workup:

$\text{Axiom schema of Accessibility:}$ if $``F"$ is a symbol that denotes a unary function that is definable in the language of set theory over the whole universe of discourse, and if $\phi(Y)$ is a formula in which $X$ doesn't occur free and $Y$ occurs free and only free, then all closures of:

$$[\exists \alpha \forall Y (\phi(Y) \to Y \ ..ACC^F \ \alpha)] \to \exists X \forall Y (Y \in X \leftrightarrow \phi(Y))$$ are axioms.

Where $ACC^F$ is defined as:

$$Y \ ACC^F \ \alpha \iff Y\ ..\leq \ \alpha \lor \exists \beta \ [\beta \ ..< \ Y \wedge \alpha \subseteq \beta \wedge F(\beta) \geq Y] $$

Where generally $``..R"$ denotes "hereditarily $R$" relation defined as:

$$ X \ ..R \ \ Y \iff X \ R \ Y \wedge \forall m \in TC(X) [m \ R \ Y]$$

Where $TC(X)$ is defined in the customary manner as the minimal transitive superset of $X$.

Where: $ x < y \iff \exists f (f:x\to y \wedge f \text{ is an injection}) \wedge \not \exists g (g: y \to x \wedge g \text{ is an injection})$

and: $ x \geq y \iff \exists f (f: y \to x \wedge f \text { is an injection} )$;

and: $ x \leq y \iff y \geq x$

Now the question is:

What is the consistency strength of the theory whose axioms are the axioms of $[\text{Z} - \text{INF.}] + \text{Transitivity} + \text{Accessibility}?$

Where Axiom of Transitivity is the axiom stating that every set is a subset of some transitive set.

In particular would Replacement be provable in this theory?

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    $\begingroup$ @Zuhair You do not have the "set" of hereditarily finite sets. $\endgroup$ – Not Mike Mar 12 '18 at 17:24
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    $\begingroup$ In first order logic, when one proposes a theory in a language with a function symbol $F$, then the meaning is that every model provides one and only one interpretation of that function. Evidently you have some other convention, and so I no longer know what your theory is. Evidently you intend to have a scheme over the definable functions, and in this case, you don't really have or need a function symbol. $\endgroup$ – Joel David Hamkins Mar 12 '18 at 18:07
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    $\begingroup$ If you will rewrite the question, then let me suggest that you state your intended axiom mainly in plain language, first, or most semi-formal language, and especially that you try to explain or motivate the idea of it. The purpose of a formal statement is mainly to resolve ambiguities that might be present in the natural language or semi-formal account of the axiom. Since well-stated axioms or principles usually don't have such ambiguities, the formal assertions are rarely needed or desired. $\endgroup$ – Joel David Hamkins Mar 12 '18 at 18:50
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    $\begingroup$ Zuhair, in the question you write, "All ZF sets are seen as hereditarily Power accessible from $\omega_0$, and so a model of ZF is definable in this theory," but I don't find this to be a meaningful assertion. $\endgroup$ – Joel David Hamkins Mar 15 '18 at 17:05
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    $\begingroup$ All of what? I don't know what "ZF sets" are. I don't think this phrase has a meaning. $\endgroup$ – Joel David Hamkins Mar 15 '18 at 18:55


Let $\varphi$ be the formula $$\varphi(x) \iff [1 \subset \mathsf{trcl}(x)=x \wedge (\forall a, b\in x)(a \in b\text{, }a=b\text{, or }b\in a)]$$

(read "$x$ is an ordinal larger than $1$.") and define the function $F$ by recursion, as follows

  • $F(a) = \mathcal{P}(a \cup \bigcup \{ F(x) : x\in a\})$ with $F(\emptyset)=\omega_0.$

Proposition: Assuming replacement and choice, Either,

There exists an inaccessible cardinal.


For every ordinal $\delta \ge 1$, there exists some $A \subset \delta$, such that

  • $\vert A \vert < \delta$, and
  • $\delta \le F(A)$ (here we are using $\le$ in sense defined in the question.)

Proof: To begin, note that for every $\delta\ge 0$, we have $\delta+1 \le F(\{\delta\})$ (since $F(\{\delta\}) = \mathcal{P}(\{\delta\} \cup F(\delta))$ and $\delta \subset F(\delta)$). It follows that the second alternative holds for every successor ordinal $\xi = \delta+1 \ge 1$, as witnessed by $A =\{\delta\}$.

Now, assume $\delta$ is a limit ordinal and the least ordinal for which the second alternative fails. Then, $\delta > \omega = F(\emptyset)$. Moreover, $\delta$ is regular; to see this, note that for any unbounded $A\subset \delta$, we have

$$\delta \subset \cup \{ \xi: \xi \in A \} \subset \cup \{ F(\xi): \xi\in A\}$$ as such, our assumptions on $\delta$ entail $ot(A) = \delta$, and so $\delta$ is regular. Next, let $\gamma \in \delta$, then we have $\vert \gamma \vert < \delta $, so noting $\mathcal{P}(\gamma) \subset F(\gamma)$, we must have $\vert \mathcal{P}(\gamma) \vert \le \vert F(\gamma) \vert < \delta.$ $\square$.

Remark: It would seem that either the instance of your axiom for the function $F$ and formula $\varphi$ is not valid, or an inaccessible cardinal is lurking around which ensures the hypothesis of the axiom are not satisfied (at least assuming replacement anyway.) I'm not entirely sure how to parse the situation.

Edit: Cleaned up proof.

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  • $\begingroup$ what is $ot$? in your expression $ot(A)=\delta$ $\endgroup$ – Zuhair Al-Johar Mar 14 '18 at 8:50
  • $\begingroup$ first of all, we are not sure that Replacement is consistent with this theory? but if it is, then (although I didn't understand your proof) if your proof is correct, then this would only prove the existence of an F inaccessible set. $\endgroup$ – Zuhair Al-Johar Mar 14 '18 at 20:23
  • $\begingroup$ @Zuhair and it means order-type. $\endgroup$ – Not Mike Mar 15 '18 at 2:49
  • $\begingroup$ I have some problems in understanding the relevance of your answer to the hypothesis of the accessibility schema. For example your second alternative doesn't seem to be directly related to the accessibility question, I mean even if we have a subset $A$ of an ordinal $\delta$ where $ |A| < \delta$ that doesn't entail that $A$ is hereditarily strictly subnumerous to $\delta$, for example $|\{\omega_0\}| < \omega_0 +1$ however it is not the case that $\{\omega_0\} \ .. < \ \omega_0 +1$ , so I don't see it an important issue.A better argument can be phrased using "Successor cardinals"! $\endgroup$ – Zuhair Al-Johar Mar 15 '18 at 16:31
  • $\begingroup$ Let $\kappa$ be the largest cardinal less than $\delta$. Then, you can find $A \subset \kappa$ such that $\vert A \vert < \kappa \le \delta$ and $\kappa < F(A)$. Since $\kappa $ was the largest one below $\delta$ you get $\delta\le F(A)$ for free. I'll leave it as an exercise that every element of the transitive closure of $A$ has cardinality strictly less then $\kappa$ (and hence $\delta$) $\endgroup$ – Not Mike Mar 15 '18 at 17:11

A limited form of Replacement is provable in this theory! That of Replacing elements of a set of ordinals by ordinals, where an ordinal is defined in the usual manner after von Neumann.

Lemma: $\forall ordinals \ \alpha, \beta \ (\alpha \ ..\leq \ \beta \lor \beta \ ..< \ \alpha)$

The proof of this lemma is obvious, and it is the main tool that would enable the following proof of ordinal replacement.

Let $S$ be an infinite set of ordinals, then $\cup(S)$ is a ordinal such that $\forall m \in S [ m \ ..\leq \cup(S)]$ Let $F$ by any function that sends ordinals to ordinals. Now because of schema (see prior answer) we have the set $$A=\{x|\exists m \in S (x=F(m)) \wedge x \ .. \leq \cup(S)\}$$ Now what remains are the $F$-images of elements of $S$, that are strictly bigger than $\cup(S)$, call those the "outside of $A$ ordinals", and thus $\cup(S)$ is hereditarily strictly smaller than each outside of $A$ ordinal, now we take the subset $S_1$ of $S$ that contains all elements of $S$ that are sent by $F$ to those outside of $A$ ordinals, i.e. $$S_1 = \{x \in S| \cup(S) \ ..< F(x)\}$$

Take the set $D$ of all ordinals that are equinumerous with $\cup(S)$ [schema in prior answer], take any injective function $g$ from $S_1$ to $D$, now take the converse function $g^{-1}:rng(g) \to S_1$, let $\kappa= min(rng(g))$, now define a new relation $h(x)=F(g^{-1}(x))$, now clearly we have:

$$\forall y (\exists x \in rng(g) [y=h(x)] \to y \ ..ACC^{h} \ \kappa)$$

The reason is because every such $y$ is an outside of $A$ ordinal so each element $x$ in $rng(g)$ would be hereditarily strictly subnumerous to $y$, and since $h(x)=y$ and $\kappa$ is a subset of each element of $rng(g)$ then by definition we have $y \ ACC^h \ \kappa$, now each element in $y$ [thus in $TC(y)$] is either strictly supernumerous to $x$ and thus $ACC^h$ from $\kappa$ [through y], or is subnumerous to $\kappa$ thus hereditarily subnumerous to $\kappa$ thus $ACC^h$ from $\kappa$!

Then from axiom of Accessibility we get the set $B=\{y| \exists x \in S_1 (y=F(x))\}$ exists. Now $A \cup B$ is our Replacement set. The case when $S$ is a finite set of ordinals can in a similar manner be proved, we just take $D$ to be any set of infinite countable ordinals, such that $S$ equinumerous to $D$, then take $g$ to be a bijective function that witnesses this equinumerousity, and then carry the rest of the arguemnt as above. QED

Now I'm not sure if that can lead to an argument that can prove full Replacement in this theory, but I guess so. I think if Choice is assumed, and the definition of $ACC^F$ is changed to: $$ X \ ACC^F \alpha \iff X \ ..\leq \alpha \lor \exists \beta (\beta \ ..< TC(X) \wedge \alpha \subseteq \beta \wedge F(\beta) \geq X)$$

Then I think by a modified form of the above argument full Replacement can be proven. Instead of $\cup(S)$ we take $TC(S)$, and use the same argument, but let $h(x)=TC(F(g^{-1}(x)))$.

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This is a partial answer to the theory I've presented here. This would answer to as whether $\text{ZF}$ is interpretable in this theory, it actually answers to doubts raised in comments, however, these were not among the original questions, so it is an aside answer.

The answer is to the positive. It will be shown that the axiom of Infinity is provable, and the schema of Replacement being interpretable over sets that are hereditarily strictly subnumerous to the first inaccessible cardinal, the first inaccessible is provable to exist in this theory and so is the existence of the set of all sets hereditarily strictly subnumerous to it.

  1. Proof of Infinity: I'll present 2 proofs:

Proof A: Let $x$ be a non empty hereditarily finite $``HF"$ set, let $n$ be the natural number (a finite von Neumann ordinal) that is equinumerous with $x$, now $n−1 ..< x$, since $P(n−1) \geq x$ and $ \emptyset \subseteq n-1$, then by definition of accessibility $x$ is Power accessible from $\emptyset$, since we have $\emptyset.. \leq \emptyset$. So every $HF$ set is power accessible from $\emptyset$, since every element in the transitive closure of a $HF$ set is a $HF$ set, then every $HF$ set is hereditarily power accessible from $\emptyset$, and so by the axiom of accessibility, we get the set of all $HF$ sets.

Proof B: Let $F$ be the fixed function $F(x)=\emptyset$, for all $x$. Now the second condition in the definition of $..ACC^F$ would be unfulfillable, so it would be reduced to the first condition, this would lead to proving the following scheme:

If $\phi(y)$ is a formula in which x is not free, then:$$\forall \alpha \exists X \forall y ( y \in X \leftrightarrow y\ ..\leq \alpha \wedge \phi(y))$$ is an axiom.

Now from that schema we get the set $\{\emptyset\}$, then from the above schema let $\alpha= \{\emptyset\}$ and we get the set of all sets that are hereditarily subnumerous to $\{\emptyset\}$, which is the set $N^{Zermelo}$of all Zermelo naturals. To get the usual Von Neumann ordinal $\omega_0$, we simply let $\alpha= N^{Zermelo}$ and we get the set of all hereditarily countable sets, now we apply the above schema on that set letting $\phi$ being "is a finite von Neumann ordinal" and we get $\omega_0$.

  1. Proof of Interpreting Replacement: Let the function $PU$ be defined as: $$\forall x,y [ PU(x)=y \iff y=P(\cup(x))],$$ Now it is easy to see that for each cardinal $\aleph_i$ we have: $$\aleph_i \ ..ACC^{PU} \ \omega_0 \to \aleph_{i+1} \ ..ACC^{PU} \ \omega_0$$. Now we show that for any singular uncountable limit cardinal $\aleph_\alpha$ we have: $$\forall \delta \in \aleph_\alpha (\delta \ ..ACC^{PU} \ \omega_0) \to \aleph_\alpha \ ..ACC^{PU} \ \omega_0$$ ,this is easy, since $\aleph_\alpha$ is singular, therefore, there is a subset $s$ of $\aleph_\alpha$ such that $s < \aleph_\alpha$ [thus $s ..< \aleph_\alpha$] and $P(\cup(s)) \geq \aleph_\alpha$. Now $\omega_0 \cup s$ would also possess the same qualities of $s$. This mean that $\aleph_\alpha \ ACC^{PU} \ \omega_0$, now since $\forall \delta \in \aleph_\alpha (\delta \ ..ACC^{PU} \ \omega_0)$ then $\aleph_\alpha \ ..ACC^{PU} \ \omega_0$. Now apply axiom of accessibility for the predicate $(cardinal (Y) \wedge \ Y..ACC^{PU} \ \omega_0)$, and we get the set $D$ of all those cardinals, now let $\aleph_{\kappa} = \bigcup (D)$, then $\aleph_{\kappa}$ cannot be a successor cardinal nor it can be a singular cardinal, so it must be a regular limit uncountable cardinal , i.e. the first inaccessible cardinal, now let $M=\{x| x \ ..< \aleph_{\kappa}\}$ (above scheme) then we get $M$ being the domain of a model of $\text{ZF}$, using power and separation we can easily get the membership relation $\in^M$ set on that domain, and the pair $\langle M, \in^M \rangle$ would be a model of $\text{ZF}$. QED

What is more interesting is that if it can be shown that all axioms of Replacement are provable in this theory, i.e. $\text{ZF}$ is a proper sub-theory of this theory?

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    $\begingroup$ I am not the downvote, but if this is your own theory and you are able to work out the answers to your own problems in ~3 days, why are you bringing them to MO so quickly? It is my understanding that proper etiquette for this site dicatates that we should spend a signifigant amount of our own time trying to work things out before coming here, not come here immedately and begin a public exposition on each problem we encounter. $\endgroup$ – Alec Rhea Mar 15 '18 at 22:38
  • $\begingroup$ In reality, my answer is just a partial answer to questions raised within the discussion, it is not the answer to the question that I've raised, It appears that some had doubts about infinity, proving inaccessibles, etc.. so I've presented it, but the questions that I've raised I didn't answer yet, I still don't know if replacement is PROVABLE in this theory? what is the exact consistency strength of this theory? if there is a clear inconsistency with this theory? $\endgroup$ – Zuhair Al-Johar Mar 15 '18 at 22:42

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