# Logical equivalence - Russell's Paradox

In 'How to Prove it' Velleman creates the following set: $R = \{A\in U| A \notin A \}$. This is, according to Velleman, equivalent to $\forall A \in U (A \notin A \iff A\in R)$. That is clear. However, based on what I learned from Velleman it appears that the set R is also equivalent to the logical statement: $\exists A (A \in U \land A \notin A \land A \in R )$. Is that correct? Is there a short way to show that the two logical statements are equivalent? Thanks in advance.

• I don't think the two are equivalent. Can you explain your intuition/reasoning maybe? Jul 23, 2015 at 17:31
• The statement $\exists A (A \in U \land A \notin A \land A \in R )$ says "There is some set in $U$, not a member of itself, that's in $R$". This doesn't force $R$ to contain all sets that don't contain themselves. It only forces there to be one. Jul 23, 2015 at 17:36
• What do you mean by "the set R is equivalent to the logical statement ..."? How can a set be equivalent to a statement? They're two entirely different sorts of things. Jul 23, 2015 at 17:39
• Your question is a bit vague (See Andreas Blass's comment). It may help to point out that $\forall A \in U (A \notin A \iff A\in R)$ is not equivalent to $\exists A (A \in U \land A \notin A \land A \in R )$ if that is what you are asking about. From the first statement we can infer that $R\notin U$. We cannot infer this from the second statement. Aug 7, 2015 at 11:04

The statement $(*)$ "$\exists A(A\in U\wedge A\not\in A\wedge A\in R)$" - that is, "$R$ is not empty" - is a consequence of one half of the paradox. It follows from the straightforward observation that $R\not\in R$ (since, if $R\in R$, then by definition of $R$ we know $R\not\in R$). Of course, the same reasoning, reversed, shows that $R\in R$. So $R\in R$ and $R\not\in R$, a contradiction.
The statement $(*)$ doesn't come up at all, specifically; I don't know why it's (equivalent to anything which is) relevant at all. This could be a typo; maybe the statement is supposed to read $$(**)\quad\exists A(A\in U\wedge A\in A\wedge A\in R),$$ which would be a contradiction with the definition of $R$? This would be another way to phrase the punchline of Russell's paradox. Note that $(**)$ is still not equivalent to anything in particular, the point is that it's a consequence of the existence of $R$: if $R\not\in R$, then $R\in R$ by definition of $R$, so by contradiction we know $R\in R$, and this implies $(**)$.