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Prove that there is no set $\Omega$ such that $(\forall A)(A \subseteq \Omega)$
(1) I rewrite this proposistion as: $$(\forall x)(\forall A)(x\in A \Rightarrow x\in \Omega)$$ (2) Since this relation is supposed to hold for all possible sets $A$, let's define $A$ as: $$A = \{x : x\notin \Omega \}$$ (3) Let's assume that $x\in A$, therefore $x \notin \Omega$ and so there exists a set $A$ such that $(\forall x)(x\in A \Rightarrow \neg(x\in \Omega))$, therefore the theorem is true.
Is my proof valid, clear and understandable?

Aemilius
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  • If you simply define $A=\emptyset$ then you've got a set $A$ such that $(\forall x)(x\in A\implies\neg(x\in\Omega)).$ Now, what is supposed to follow from that? – bof Nov 03 '17 at 23:43
  • @bof I am quite ashamed that I haven't thought about the empty set in the first place.. Nevertheless, does my example work as well? – Aemilius Nov 03 '17 at 23:44
  • I'm not sure what you mean by "therefore the theorem is true". What theorem are you referring to? And why does it follow from what you've said? – bof Nov 03 '17 at 23:44
  • For every $A$? For every $A$ from what universe? There is indeed such an $\Omega$ such that for all $A$ in the family ${{1},{2}}$ you have $A\subseteq \Omega$. For example $\Omega={1,2}$ would work. If you are taking it to be for all possible sets then $\Omega$ would be the "set" of all possible elements... (though you should be wary since contradictions arise if you treat it as an actual set, it is unfortunately not) – JMoravitz Nov 03 '17 at 23:45
  • @bof Theorem: There is no set $\Omega$ such that $(\forall A)(A \in \Omega)$ This follows, because we have found an example of $A$ such that $x \in A$ and $x \notin \Omega$ – Aemilius Nov 03 '17 at 23:46
  • In your "proof" you tacitly assumed that $A$ was non-empty when you said "Let's assume that $x\in A$." This assumption is not valid. If $\Omega$ really was "the set containing all possible elements" then defining $A$ as the set of elements not in $\Omega$ would intuitively imply that $A$ is in fact empty. The true contradiction lies elsewhere. – JMoravitz Nov 03 '17 at 23:47
  • @JMoravitz The problem is, this is not clearly stated in the exercise what universum we are referring to – Aemilius Nov 03 '17 at 23:48
  • Well, I can see two scenarios here: (1) $\Omega$ is the set of all sets and so it does not exist (Russel) (2) $\Omega$ is just an arbitrary set and so my example applies to it. – Aemilius Nov 03 '17 at 23:50
  • No. You've found a set $A$ (the empty set) such that $\forall x(x\in A\implies x\in\Omega).$ For the same $A$ we *also* have $\forall x(x\in A\implies x\in\Omega).$ Of course, the empty set is the *only* set for which both statements are true. The upshot is that you haven't got a contradiction yet. What your argument shows is that *your* set $A$ must be the empty set. In other words, ${x:x\notin\Omega}$ is the empty set. In other words, $(\forall x)x\in\Omega.$ – bof Nov 03 '17 at 23:51
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    Actually, the existence of the set ${x:x\notin\Omega}$ is problematic. You don't need it. Given any $x,$ consider the set $A={x}.$ From $A\subseteq\Omega$ it follows that $x\in\Omega.$ In other words, the statement $(\forall A)A\subseteq\Omega$ is just a (very lightly) disguised way of saying that $\Omega$ is the universal set. – bof Nov 03 '17 at 23:54
  • I originally didn't think of this as a duplicate. Since this is a subtly different question, and more of a proof verification. But seeing the answer you accepted is this one, I figured that it might as well be a duplicate after all. – Asaf Karagila Nov 04 '17 at 00:34

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Suppose that such a set $\Omega$ existed. Then $\{\Omega\}$ is also a set.

According to our hypotheses then $\{\Omega\}\subseteq \Omega$. However this is a contradiction as no set is allowed to be an element of itself.

JMoravitz
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  • Is there a way to solve this that does not rely on the axiom of regularity? This hasn't been introduced yet during the course. – Aemilius Nov 03 '17 at 23:54
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    @Aemilius the existence of such a set would contradict Cantor's theorem that the power set of a set should be of strictly greater cardinality than the original set, but the power set of a universal set would contain elements, each of which would be elements of the universal set itself so the cardinality of the universal set would be greater than or equal to the cardinality of its own power set, a contradiction. – JMoravitz Nov 04 '17 at 00:02
  • How do you reconcile this with Quine's New Foundations? Or Positive set theory? ;) – Asaf Karagila Nov 04 '17 at 00:39