5

The usual argument I see for proving that, assuming S is a set, $S\not\in S$ (in ZFC) is the following:

Take $S$. Assume $S$ is a set in ZFC. Define $T=\{S\}$ ; by axiom of pairing, $T$ is a set. Thus it satisfies the axiom of regularity, i.e. $$\exists X\in T((\forall x\in X(x\not\in T))\land (\forall t\in T(t\not\in X))$$ Because the only element of $T$ is $S$, it must be that $$\forall s\in S(s\not\in T)\land S \not\in S $$

But, is it the case that $\{S\}\not\in S$ ?

I ideally wish to show using only axiom of regularity and axiom of pairing (as was done above) that if $S$ is a set, $\{S\}\not\in S$. (If this is not possible using just these 2 axioms, using full ZFC would be ok too.)

Because I am suspending axiom schema of specification, I can form the set $S=\{S\}$ or $S=\{\{S\}\}$ (etc) and try to find a contradiction. I have tried applying axiom of regularity to various sets of this form but I can not reach my desired result.

Any ideas on how I could do this?

edit: Asaf's answer is a good method. Maybe I am just not "on that level" yet but I had to work each step out explicitly, so I will post this here for anyone who also needs a step-by-step.

$X=\{S,\{S\}\}$ is a set by axiom of pairing.

By axiom of regularity, $\exists x\in X((\forall y\in x(y\not\in X))\land(\forall z\in X(z\not\in x))$

Because there are only 2 elements of $X$, it is satisfied for either $S$ or $\{S\}$. Thus at least one of the following is true:
(1) $\forall y\in S(y\not\in X)\land\forall z\in X(z\not\in S)$
or (2) $\forall y\in\{S\}(y\not\in X)\land\forall z\in X(z\not\in\{S\})$

$S\in X,S\in\{S\}$, so (2) is false, thus (1) must be true. Thus $\forall z\in X(z\not\in S)$

$\{S\}\in X\implies \{S\}\not\in S$. QED

This also can be extended by taking $X=\{S,\{S\},\{\{S\}\},\ldots\}$ to show that $S\not\in S, \{S\}\not\in S,\{\{S\}\}\not\in S,\ldots$

FD_bfa
  • 4,757
JustAskin
  • 802
  • 6
  • 14
  • More answers: http://math.stackexchange.com/questions/14697/how-to-prove-t-notin-t and http://math.stackexchange.com/questions/366136/how-to-show-x-neq-x – Chris Culter Dec 12 '14 at 09:43

1 Answers1

6

In order to prove that $\{S\}\notin S$ you can use the axiom of pairing to obtain the set $X=\{S,\{S\}\}$, and since $X$ must have an $\in$-minimal element and it cannot be $\{S\}$, it means that $S\cap X=\varnothing$ so in particular $\{S\}\notin S$.

Asaf Karagila
  • 405,794