Russell's Paradox in *Naive Set Theory* by Paul Halmos

Question

In Naive Set Theory, Paul Halmos introduces an arbitrary set $A$ and another set $B = \lbrace{x \in A: x \notin x \rbrace}$. He then asserts that $B \notin A$ because $B \in A$ implies $B \in B$ or $B \notin B$, both of which lead to a contradiction.

My question is how does $B \notin A$ resolve the contradiction? It seems to me that no matter what the condition is, a set is either in another set or not: if $B \notin A$, it still follows that either $B \in B$ or $B \notin B$.

Note: the fact that,

$(*)$ for all $y$, $y \in B$ if and only if ($y \in A$ and $y \notin y$)

is also given as a consequence of the definition of $B$.

Answer 1

If $B\in B$ then $B\in A$ and $B\notin B$, which is a contradiction.

If $B\notin B$ then $B\notin A$ or $B\in B$, which is not a contradiction, seeing as $B\notin A$.

Answer 2

1. Suppose $\forall x:[x\in B \iff x\in A \land x\notin x]$ (the definition of $B$)

2. Specifying $x=B$ in (1), we obtain $B\in B \iff B\in A \land B\notin B$

3. Suppose $B\in A$

4. Suppose $B\in B$

5. From (2) and (4), we obtain $B\notin B$, and the contradiction $B\in B \land B\notin B$

6. Thus (4) is false and we have $B\notin B$

7. From (2), (4) and (6), we obtain $B\in B$, and the contradiction $B\in B \land B\notin B$

8. Thus (3) is false and we have $B\notin A$

Note that we cannot now obtain $B\in A$, as in the resolution of Russell's Paradox, to get yet another contradiction.

Answer 3

This approach makes a mistake that many people make about logic.

You can't resolve a paradox by introducing more axioms. Ever.

This question mentions one of the many ways that formal logical has tried to reconcile with Russell's paradoxical definition: the introduction of a universal set. In this case the author is using a subset of the universal set, $A$.

Either way, it just begs the question: "does the universal set contain Russell's self contradictory object?"

Maybe the author himself addresses this issue, I don't have the text. But $B \not \in A$ doesn't "resolve a [paradoxical] contradiction", it is simply a statement that follows from the definition of $B$. While a definition may lend insight into ways of resolving RP, adding a definition can never eliminate it; you have to take a different approach if your goal is to resolve RP.