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The following theory contains a comprehension axiom that is a naïve-like schema. This theory definitely looks inconsistent at first glance. However, I tried to find this inconsistency, but to no avail. The main idea pivotes around an acyclic membership relation $\in^*$, so an acyclic member of a set is an element of a set that doesn't contain that set in its transitive closure. Now, this theory allows free construction of sets after all formulas in the language $\operatorname{FOL}(=, \in^*)$.

Formal workup:

Language: the first order language of set theory.

Extensionality: $\forall x \, (x \in A \leftrightarrow x \in B) \to A=B$

Transitive Closures: $\forall x \exists t: t=\operatorname{TC}(x)$

$\DeclareMathOperator\TC{TC}\DeclareMathOperator\trs{trs}$Define: $t=\TC(x) \iff \trs(t) \land x \subseteq t \land \forall k (\trs(k) \land x \subseteq k \to t \subseteq k)$

Where "$\trs$" stands for "is transitive", that is: closure under relation $\in$.

Induction: if $\phi$ is a formula, then:

$$\forall y \in x \ (\phi(y)) \land \forall k \, (\phi(k) \to \forall l \in k (\phi(l))) \to \\ \forall m \in \TC(x)( \phi(m))$$

Define: $y \in^* x \iff y \in x \land \neg \, x \in \TC(y)$

Comprehension: $\exists x \forall y \, (y \in x \iff \phi^*)$

Where $\phi^*$ is a formula not using “$x$”, whose predicates are among $=$, $\in^*$ symbols.

Questions:

Is there a clear inconsistency with this theory?

If not, then can this theory prove Infinity?

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  • $\begingroup$ What's $x(\phi)$ in the Induction schema? $\endgroup$ Jan 9 at 18:26
  • $\begingroup$ @PeterGerdes it is $\forall y \in x \ (\phi)$, it means for all y in x such that formula $ \phi$ is true $\endgroup$ Jan 9 at 18:47
  • $\begingroup$ Ahh, thanks...and yah I misread the comprehension axiom when I gave my first answer let me think for a moment. $\endgroup$ Jan 9 at 19:20
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    $\begingroup$ In your various posts, you consistently write $``x''$, which looks awful (even in proper TeX, as opposed to MathJax)—for example, the two straight quotes in math mode appear as a double prime, not as a right double quotation mark. Please use instead “$x$”. I have edited accordingly. $\endgroup$
    – LSpice
    Jan 10 at 18:50
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    $\begingroup$ @LSpice, agreed! Thanks $\endgroup$ Jan 10 at 19:24

1 Answer 1

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I misread comprehension the first time. Now that I understand it correctly can't you construct the standard Russell set:

$\phi(y) = \lnot y \in^{*} y$

So let's ask if the resulting $x$ satisfies $x \in^{*} x$. If so then we have $x \not\in x$ which contradicts the assumption $x \in^{*} x$.

Now let's assume that $x \not\in^{*} x$. But now that implies $x \in x$ so we must have $x \not\in TC(x)$. But since $x \in x$ and the transitive closure is a superset of $x$ we must have $x \in TC(x)$. Contradiction.

I haven't checked this carefully so maybe I made a dumb error but it seems right.

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    $\begingroup$ $\forall y \, (y \not \in^* y )$ is a theorem of this theory! So your set is the universe. Note that $`` x∈x→x∉TC(x)"$ is not a theorem of this theory, neither is $ z \not \in^* x \to z \not \in TC(x)$. You can indeed have $z \not \in^* x \land z \in TC(x)$ like for example the universe $V$, where we have $V \in V \land V \not \in^* V$ $\endgroup$ Jan 10 at 10:16

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