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?
Best Answer
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.