1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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