# Can this naïve-like set theory using acyclic membership be consistent?

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?

• What's $x(\phi)$ in the Induction schema? Jan 9 at 18:26
• @PeterGerdes it is $\forall y \in x \ (\phi)$, it means for all y in x such that formula $\phi$ is true Jan 9 at 18:47
• Ahh, thanks...and yah I misread the comprehension axiom when I gave my first answer let me think for a moment. Jan 9 at 19:20
• 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. Jan 10 at 18:50
• @LSpice, agreed! Thanks Jan 10 at 19:24

$$\phi(y) = \lnot y \in^{*} y$$
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.
• $\forall y \, (y \not \in^* y )$ 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 $V \in V \land V \not \in^* V$ Jan 10 at 10:16