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How is Russell’s Paradox, which is defined as the collection $$ \{ x \mid x \notin x \}, $$ equal to the universal set $ \{ x \mid x = x \} $? Do we assume that if $ x = x $, then $ x \notin x $? My question is specifically what is written at the end of Page 21 of this document.

Thanks for any help!

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You seem confused about the relationship between Russell's Paradox and the non-existence of the so-called universal set. I will try to clarify without referring to the notion of a class.

If you assume the existence of a universal set $U$ such that $\forall x:x\in U$ and your set theory allows for arbitrary subsets and does not disallow $x\in x$, then you can define set $R$ such that $\forall x:[x\in R\iff x\in U \land x\notin x]$. Note the similarities to the standard presentation of Russell's Paradox.

Applying the definition of $R$ to itself, we have $R\in R \iff R\in U \land R\notin R$. Since, by definition, we must have $R\in U$, this is a contradiction. Thus, the existence of $U$, as defined above, results in a contradiction. Thus, $U$ cannot exist.

Thus, the non-existence of the universal set can be proven (in the set theories described here) by using the same kind of contradiction that arises in Russell's Paradox

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  • $\begingroup$ Thanks Dan. I was actually asking for how they get this equality, the one that I mention in my question in these notes. Its on page 21 of the notes math.toronto.edu/weiss/set_theory.pdf $\endgroup$ – Abdul Rahman Jun 21 '14 at 21:36
  • $\begingroup$ @AbdulRahman I can see why you might have been confused. $\endgroup$ – Dan Christensen Jun 22 '14 at 4:01
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There are a few problems with your statements.

  1. If there is no universal set, then $\{x\mid x=x\}$ is not a set. This is because this collection is the collection of all sets, since every set is equals to itself.

    In some set theories like $\sf NF$ there is a universal set, but in others like $\sf ZFC$ there is none.

  2. $\{x\mid x\notin x\}$ need not be equal to the entire class of sets. It is possible to arrange a situation where we have $x\in x$, but still $x=x$. In some set theories this is outright provable, for example in $\sf NF$ where you have a universal set $U$ then we necessarily have $U\in U$. In such situation, as I wrote, the two classes are not equal.

  3. Russell's paradox is not that $\{x\mid x\notin x\}=\{x\mid x=x\}$. It is a paradox that shows that the class $\{x\mid x\notin x\}$ is not a set. It begins by assuming otherwise, then it is a set $A$. If $A\in A$, then by definition $A\notin A$; and if $A\notin A$ then by definition $A\in A$. So we have a contradiction, therefore $A$ is not a set.

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  • $\begingroup$ Thanks Asaf. I was actually asking for how they get this equality, the one that I mention in my question in these notes. Its on page 21 of the notes math.toronto.edu/weiss/set_theory.pdf $\endgroup$ – Abdul Rahman Jun 21 '14 at 21:36
  • $\begingroup$ Are you sure the link is correct? It doesn't really respond very well. My guess is that they assume $\sf ZF$ or some extension thereof. The axiom of regularity (also known as the axiom of foundation sometimes) allows us to prove that $\forall x(x\notin x)$. Therefore the two classes are equal. Neither is a set, though. $\endgroup$ – Asaf Karagila Jun 21 '14 at 21:39
  • $\begingroup$ www.math.toronto.edu/weiss/set_theory.pdf $\endgroup$ – Abdul Rahman Jun 21 '14 at 21:45
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    $\begingroup$ The file says the following thing, "Upon reflection, you might say that in fact, nothing is an element of itself". This reflection is informal, and later on when you are introduced to the axiom of foundation (p. 23), this reflection becomes formal. $\endgroup$ – Asaf Karagila Jun 21 '14 at 21:59
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    $\begingroup$ Thanks Asaf, editted my question to include the link now. $\endgroup$ – Abdul Rahman Jun 21 '14 at 22:14

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