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This question already has an answer here:

Is there a set that contains all sets?

This question was answered in our Set Theory course by providing Russell's Paradox. I understand the logic behind Russell's Paradox and that there exists no set whose condition is not being a member of itself.

However, how is this directly relevant to the question of whether there is a set that contains all sets?

How does it contradict the existence of such a universal set?

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marked as duplicate by Asaf Karagila set-theory Mar 9 '17 at 12:18

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  • $\begingroup$ If there's a set that contains all sets, then it would have a subset containing the sets that don't contain themselves. $\endgroup$ – Gerry Myerson Mar 9 '17 at 11:51
  • $\begingroup$ By Russell's Paradox that set which contains sets that don't contain themselves doesn't exist in the first place. Given that it's not a set to begin with, it cannot/shouldn't be in the set that contains all sets. $\endgroup$ – M.Ezzat Mar 9 '17 at 11:53
  • $\begingroup$ I think the trick is to prove by contradiction. Suppose $U$ is a set containing all sets, then $U \in U$ since U is a set by assumption! Contradicting no set is an element of itself. $\endgroup$ – Alex Vong Mar 9 '17 at 11:53
  • $\begingroup$ Because if it is a set then as a set it has to belong to itself and that is Russel paradox. Thre exist something which is the collection of all sets, but it is not a set, the terminology is class if I recall correctly. en.wikipedia.org/wiki/Class_(set_theory) $\endgroup$ – zwim Mar 9 '17 at 11:53
  • $\begingroup$ Read carefully: IF there's a set $X$ that contains all sets, then among the sets that $X$ contains there are some that don't contain themselves. All of those sets are elements of $X$, so the collection of all of them is a subset of $X$. But Russell says no such set can exist. Therefore, $X$ can't exist. $\endgroup$ – Gerry Myerson Mar 9 '17 at 11:57
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Russell's paradox is based on the assumption that if $A$ is a set and $P$ is a predicate then $\{ x \in A : P(x) \}$ is a set. It tells us that if $A$ is allowed to be a set of all sets, then $P$ can't be an arbitrary predicate; and if $P$ is allowed to be any predicate, then $A$ can't be allowed to be a set of all sets. Either one of these approaches can form an ostensibly consistent set theory. ZF set theory happens to choose the second one, but NF set theory chooses the first one instead.

In other words, there is no inherent contradiction in defining a universal set, but if you do, then you can't use comprehension by arbitrary predicates anymore.

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  • $\begingroup$ So the actual problem is a set being an element of itself, right? (No set can ever be an element of itself) $\endgroup$ – M.Ezzat Mar 9 '17 at 12:08
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    $\begingroup$ @M.Ezzat No, that is not the problem. Non-well-founded set theories are a thing, indeed NF that I mentioned is one of them. The problem is having both a universal set and comprehension over sets by arbitrary predicates. So if you take comprehension over sets by arbitrary predicates as given, then you have to forfeit a universal set. $\endgroup$ – Ian Mar 9 '17 at 12:15
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Well if you defined a "collection" $U$ like this, $U = \{ u \in U \ iff\ u\ a\ set\}$. Your statement is about $U$. In the definition I put, I defined the elements of $U$ but I did not qualify it as a set. Your statement asserts that the statement which asserts that $U$ defined above can not be a set.

The problem isnt that a set is in itself, it's that there is no bound or maximal set.

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  • $\begingroup$ I think it's probably complicated more than simple discussion can handle. It probably relates to complexity of the set or logical statements which qualify it. $\endgroup$ – marshal craft Mar 9 '17 at 12:21

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