Give a proof that there is no universal set, using the Subset Axiom and a Russell’s-Paradox-type argument.

so that is the question that I am working on. My approach at the moment is to have if all $x$ in $U$, then define $T=\{x \in U: x \notin x\}$, then $(T \in T) \Rightarrow (T \notin T)$, and $(T \notin T) \Rightarrow (T \in T)$, so either case is a contradiction.

Does this satisfy the above prompt?

  • $\begingroup$ What is $U$ in this case? Also please try and use tex for the math. $\endgroup$ – UserB1234 Jan 22 '14 at 16:44
  • $\begingroup$ @DanulG The meaning of $U$ is specified by the introductory "if all $x$ in $U$, then ..." $\endgroup$ – Hagen von Eitzen Jan 22 '14 at 16:46
  • $\begingroup$ Ah, I see. Your idea is essentially correct. However you have to pay attention to two things: It is probably better to start of by saying: Assume that the collection $U$ of all sets is a set. Now in-order to get the final bit of your argument, you need the fact that $T$ itself is a set. You should probably quote the axiom which gives you that. $\endgroup$ – UserB1234 Jan 22 '14 at 16:51
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    $\begingroup$ I would not phrase it as "... either case is a contradiction.", as $A\implies \not A$ itself is no contradiction. The contradiction comes only when you combine both implications, then you get $A\iff\not A$. $\endgroup$ – Stefan Hamcke Jan 22 '14 at 17:14
  • $\begingroup$ Now when you say "subset axiom", what exactly do you mean? $\endgroup$ – Asaf Karagila Jan 22 '14 at 17:28

You have the right idea, but need to better structure your proof. Sketching the proof...

Start by supposing $\exists U: \forall a:a\in U$. Then use the subset axiom to prove the existence of $T$ such that $\forall a: [a\in T\iff a\in U \land a\notin a]$. Then obtain the contradiction $T\in T \land T\notin T$. Thus your original premise would have to be false.


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