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Please see this question for background on this question: $\{S\} \not\in S$ in ZFC?

In that question, I have proven using only the axiom of pairing and axiom of regularity that given a set $S$: $S\not\in\ S$ and $\{S\}\not\in S$. Now I wish to prove that $\{\{S\}\}\not\in S$, however after much scribbling I have formed a hypothesis that using only these two axioms, it is not possible to show such. (edit: apparently I also am using extensionality here.)

My reasoning is: Assume $S$ is a set. Because the only axiom allowing new sets to be formed from $S$ is pairing, the only sets that can exist under these axioms are (first) pairs of $S$ with itself, i.e. $\{S,S\}=\{S\}$, and (second) pairs of $S$ and $\{S\}$ (e.g. $\{S,\{S\}\},\{\{S\}\}$), and (third) pairs of the above. Every set has either one or two elements. (this is very useful for doing casewise proofs.)

To show $S\not\in S$, I took $T=\{S\}$ and applied regularity to find such as a consequence. Specifically, since T is a singleton, $S$ must be the element that satisfies regularity (the "regular element") and from such it follows that $S$ cannot be in $S$.

To show $\{S\}\not\in S$, I took $T=\{S,\{S\}\}$ and applied regularity to find such as a consequence. Specifically, $\{S\}$ cannot be the regular element because $S\in\{S\}\land S\in T$, so $S$ must be the regular element meaning that both $S\not\in S$ and $\{S\}\not\in S$. (I could have used this example to show the first one, too, I suppose.)

Aiming now to show $\{\{S\}\}\not\in S$, I first took $T=\{S,\{\{S\}\}\}$, but now $\{\{S\}\}$ can in fact be the regular element (because $\{S\}\in\{\{S\}\}$ is only in $T$ if either $S=\{S\}$ (which I have shown to be impossible) or if $\{\{S\}\}=\{S\}$ which because they are both singletons implies $\{S\}=S$ which once again is impossible), so we can conclude that $\{\{S\}\}$ is a regular element, and thus we can not assume that $S$ is a regular element (as we could in the previous example).

So now I tried finding a contradiction, from assuming that $\{\{S\}\}\in S$. Because our only set building axiom is pairing, $S$ is necessarily either a singleton or a pair or some other set built by pairing from $S$ (call this $T$).

Case singleton: $S=\{\{\{S\}\}\}$
The regular element of $S$ must be $\{\{S\}\},\implies\{S\}\not\in S$ which (because $\{S\}\neq\{\{S\}\}$) is true. So this case does in fact satisfy regularity.

Case pair: $S=\{\{\{S\}\},T\}$ The regular element of $S$ is either $\{\{S\}\}$ or $T$.
-Case $\{\{S\}\}\implies\{S\}\not\in S$, which is true under the restriction that $T\neq\{S\}$.
-Case $T\implies\forall t\in T(t\not\in S)\implies T\not\in T\land\{\{S\}\}\not\in T$ , one of which is already proven and the other of which simply puts a restriction on $T$.

From this I wish to conclude that in this axiom system consisting of only pairing and regularity, $\{\{S\}\}\not\in S$ is not a provable statement and thus $\{\{S\}\}\in S$ is at least "possible" (or "not provably impossible"). Can I at this point conclude this, or must I somehow go further?

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  • I'd like to note that ${{S}}\not\in S$ , ${{{S}}}\not\in S$, etc. is easily provable if you add the axiom of union, and create the set $T={S,{S},{{S}},\ldots}$ and apply regularity. Each of the elements contains the one-set-lower-than-it element e.g. ${{{S}}}\in{{{{S}}}}$ so it cannot be the regular element, leaving just $S$ to be the regular element, with the restriction that $\forall t\in T(t\not\in S)\implies S,{S},{{S}},\ldots\not\in S$ – JustAskin Mar 11 '14 at 18:56
  • It's hard to imagine what a set theory based only on pairing and foundation would look like. But I think that you are on the wrong track here. In order to assume that every set is constructed from a pair, you are assuming some sort of recursive structure on the universe of sets. But the usual recursive structure (the cumulative hierarchy) only makes sense within the context of the power set axiom. I think that you may also be implicitly assuming extensionality or separation somewhere in your argument too, since they are needed to make set builder notation work. – Unwisdom Mar 11 '14 at 19:01
  • I don't believe I am using separation, because all the sets I formed were by pairing. But I am indeed using extensionality (without even realizing it). Since it doesn't appreciably change the results, I'll note intro I suppose. – JustAskin Mar 11 '14 at 19:08

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I think you're right; you can't prove that $\{\{S\}\}\not\in S$ from pairing and regularity alone. To see this, we can build a counter model. Let $x_0 E_0 x_1 E_0 x_2 E_0 x_0$. Then, recursively, let $E_{n+1}\supseteq E_n$ be such that it codes (unique) pairs for any (one or) two objects in the field of $E_n$. A simple induction establishes that everything in the field of $E_\omega = \cup_{n\in\omega}E_n$ is a pair and that there are no $E_\omega$ loops from elements not among $x_0,x_1,x_2$ to themselves.

Now, suppose there's a pair $\{x, y\}$ in our model for which regularity fails. Then either $x\in x$, or $y\in y$, or $x\in y$ and $y\in x$. In all cases, we'd have a $E_\omega$ loop of at most two elements. But this is impossible, because there are no such loops for elements other than $x_0,x_1,x_2$ and by construction of $E_0$ there are no such loops for $x_0,x_1,x_2$.

(Note that our model also satisfies separation and extensionality)

  • Is there a way I could understand this counter model with no knowledge of fields? I am reading that $E_0$ is a binary relation (analogous to $\in$ I assume) but that's as far as I get. – JustAskin Mar 11 '14 at 20:34
  • The field of a relation $R$ is just the set of things it relates. In other words, the field of $R$ is ${x:\exists y(xRy \vee yRx)}$. The $E_n$'s are membership relations, analogous to $\in$ (as you point out). –  Mar 11 '14 at 21:13