0
$\begingroup$

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?

$\endgroup$
5
  • $\begingroup$ What's $x(\phi)$ in the Induction schema? $\endgroup$ Jan 9, 2022 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, 2022 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, 2022 at 19:20
  • 2
    $\begingroup$ 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, 2022 at 18:50
  • 1
    $\begingroup$ @LSpice, agreed! Thanks $\endgroup$ Jan 10, 2022 at 19:24

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ $\forall y \, (y \not \in^* y )$ 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, 2022 at 10:16
  • $\begingroup$ As I wrote in the above comment, your statement "so we must have $x \not \in TC(x)$" is unjustified. This is the error. $\endgroup$ Jan 13 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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