Given an arbitrary set $X$, can one always find an element outside $X$? I think this question boils down to whether there exists a universal set that contains everything, including itself. But wiki says that assuming the existence of a universal set leads to Russell's paradox. So given a set $X$, can one explicitly construct an element that does not belong to $X$? For example, is $\{X\}$ such an element?
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5Yes, in the usual axiomatizations of set theory ${X}$ is such an element: this follows from the axiom of regularity. – Brian M. Scott Sep 30 '20 at 14:55
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@Calclulix The exact grounds on which you will be able to establish the existence of $a \notin X$ given an arbitrary $X$ will of course depend on the technicalities of the axiomatic system/formal language you are employing, but this is indeed possible in general. This will also depend on what you mean by ''explicit construction'', but the axiomatic system developed by Bourbaki in their "Théorie des Ensembles" can be argued to provide a method for exhibiting a very syntactically explicit element outside any prior given set $X$ (it will no longer be ${X}$, though). – ΑΘΩ Sep 30 '20 at 15:07
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Russell's paradox itself gives a way to find such an element. Namely, let $$Y=\{x\in X:x\not\in x\}.$$ If $Y$ were an element of $X$, then Russell's paradox would give a contradiction: we would have $Y\in Y$ iff $Y\not\in Y$. Thus $Y$ is not an element of $X$.
(This construction works in any axiomatization of set theory that includes the axiom schema of separation, so that we can be sure there really does exist a set $Y$ whose elements are those $x\in X$ such that $x\not\in x$. As mentioned in Noah's answer, the usual ZFC axioms for set theory imply that actually no set is an element of itself, so $Y$ would just be $X$.)
Alex Ortiz
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Eric Wofsey
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It depends what set theory we're using. All set theories have to find a way around Russell's paradox, but there are different possible responses.
The usual foundational system is $\mathsf{ZF(C)}$. Here we have the axiom of regularity (or foundation), which implies that we don't have any "$\in$-loops" - e.g. we don't have a pair of sets $a,b$ with $a\in b$ and $b\in a$. Since $X\in\{X\}$, this rules out $\{X\}\in X$.
In other systems things play out differently: systems like $\mathsf{NF}$ and $\mathsf{GPK_\infty^+}$ actually have a universal set, and systems like $\mathsf{ZFC-Foundation+Antifoundation}$ lack a universal set but do allow $\{X\}\in X$. These latter theories however do still, to the best of my knowledge, all allow an explicit construction of a $Y\not\in X$ given a set $X$: specifically, by the Burali-Forti paradox we argue that no set contains every ordinal, so "the least ordinal not in $X$" provides a non-element as desired.
I'm not aware of any "natural" set theory in which there is no universal set but also no way of explicitly building non-elements of given sets. However, my background in alternative set theories isn't too strong, so I could be missing something.
Noah Schweber
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