Let there be a set $A$, and lets define: $B=\{X \in A: X \notin X \}$. Show the Russel paradox doesn't occur.

How is this done? And what's this 'mysterious' $A$? I'm not sure I'm using $A$ properly...

$B \in B$, so lets take $A$ so that $B \in A$, therefore $B \notin B$. Seems like I did something wrong here...

Thanks in advance for any assistance.


Russell's paradox occurs when you consider the class $R=\big\{x\big\vert x\notin x\big\}$. The paradox arises when you ask whether $R\in R$. It turns out that $$R\in R \iff R\notin R.$$ This is the paradox. There is no truth value for $R\in R$ which resolves this sentence.

However, the paradox is resolved by localizing. It doesn't matter which set $A$ you use to localize. The mere act of localizing resolves the paradox:

Let's work through the details of whether $B\in B$. If $B\in B$, then $B\in A$ and $B\notin B$. So this is a contradiction. But if $B\notin B$, this means that either $B\notin A $ or $B\in B$. The possibility that $B\notin A$ is not problematic, and offers us a way out of the paradox. So, we conclude that $B\notin A$ and $B\notin B$.

The point here is to see that the localized version of the paradox (which is permitted under ZFC) is only a problem when you assume that there is a set of all sets. If such a set exists (call it $V$), then taking the Russell class, localized to $V$, gives you a paradox. We therefore conclude that there is no set of all sets.

More generally, as Dan Christensen has noted in the comments, Russell's paradox serves to illustrate why we have to be careful about which things we allow to be sets. At first blush, the axiom of unrestricted comprehension seems quite innocuous:

Given a formula $\phi(x)$, the collection $\{x \vert \phi(x)\}$ is a set.

However, this axiom is too liberal: it forces us to permit $R$ and $V$ to be sets, which in turn exposes us to the paradox. This observation provides a lot of the motivation for the various axiomatizations of set theory (such as ZFC): we can't afford to be too naive, and so have to put a lot of work into preventing things like $R$ from being considered sets (while still permitting all sorts of useful things, like $\mathbb{N}$ to be sets). In general, we have to we reject unrestricted comprehension and replace it with something more subtle. ZFC, for instance, includes the axiom of restricted comprehension, which only forces objects of the form $\{x\in A\vert \phi(x)\}$ to be sets, where $A$ is assumed to be known to be a set.

In other words, ZFC (and restricted comprehension) allows you to consider $B$ to be a set, but not $R$.

  • $\begingroup$ Your answer was helpful, but the I think Russell's paradox can also be resolved by simply proving by contradiction that $$\neg \exists R: \forall x: [x\in R \iff x\notin x]$$ and introducing an axiom of restricted comprehension as in ZFC, i.e. not assuming a set exists simply because you can describe some property of its elements . $\endgroup$ – Dan Christensen Apr 28 '14 at 17:16
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    $\begingroup$ I agree. I think that the main value of Russell's paradox is to show that unrestricted comprehension is not viable. My take on the original question was that dsfsf wanted to understand the details of how the use of restricted comprehension resolves the paradox. $\endgroup$ – Unwisdom Apr 28 '14 at 17:17
  • $\begingroup$ Yes, I found that part was helpful and up-voted your answer. I just wanted to comment on the resolution of the paradox itself. I thought the part about "truth value" was a bit vague and needed clarification. $\endgroup$ – Dan Christensen Apr 28 '14 at 17:25
  • $\begingroup$ @DanChristensen: Does this edit address your concerns? $\endgroup$ – Unwisdom Apr 28 '14 at 17:40
  • $\begingroup$ To say that ZF allows you to consider $B$ to be a set is a little bit strange. In ZF, it is easily shown that $B=A$. It is somewhat more interesting to consider this construction in other set theories where $B\ne A$ might hold. $\endgroup$ – MJD Apr 28 '14 at 18:52

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