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Let $X$ be an infinite set.

Under ZFC, it is true that $X\notin X$.

So there is at least one set which is not an element of $X$.

Is there another element distinct from $X$ which is not a member of $X$?

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The power set of $X$ has greater cardinality than $X$, therefore most of its elements are not elements of $X$.

For another answer, since you're assuming the full ZFC and thus the axiom of foundation, $\{X, a\}$ is never an element of $X$.

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Of course. $\{X\}$, and $\mathcal P(X)$ and $\{\{X\}\}$. And many many more.

Note that $\sf ZFC$ proves there is a rank function defined by $\operatorname{rank}(X)=\sup\{\operatorname{rank}(Y)+1\mid Y\in X\}$. So every $Y$ such that $\operatorname{rank}(Y)>\operatorname{rank}(X)$ must be different from $X$ itself, and it cannot be a member of $X$.

And $\sf ZFC$ proves that for every ordinal $\alpha$, there is a set of rank $\alpha$. So "most" sets are not elements of $X$.

But even more naively than that. Even without the axiom of regularity. Since we can prove that the collection of all sets is not a set, the collection $\{Y\mid Y\notin X\}$ is not a set (since the union of two sets is a set).

So the axiom of regularity is not even needed to show that most sets are not elements of $X$.

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I'm a bit confused by the question, because it seems to be shifting levels of analysis:

You could be asking a very general question, in which case, given any particular infinite set (e.g., integers, all possible combination of letters, etc.), it would be obvious that one could construct sets that were not contained in that particular infinite set (e.g., the set of integers does not contain the set of presidential candidates from the last election cycle).

Or you might be asking a very specific question, about whether or not set X lacks elements that one would intuitively think should be in. If set X is defined as containing all elements X1, X2, etc., then there is no element of that type that it does not contain, but it lacks all sets with more than one element.

We can nit pick about whether X1 (a value) differs from [X1] (the set containing only one element, of that value)... but I suspect that is a distraction here, because if those are different types, then X contains no sets, and your question is again pretty obvious.

Probably what you are doing is defining your starting set as also containing all possible combinations of those elements, then the set contains those combinations as well. We might introduce a formality to specify that the set does not contain itself, but that is just part of setting up our initial game rules.

That formality was introduced to avoid a very specific problem, and so it contains no other exceptions.

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Sets have mathematical objects. Presidential candidates are not mathematical objects. Therefore the "set of presidential candidates from the last election cycle" does not exist at all. – Asaf Karagila Jul 24 '14 at 18:04
Also, in modern set theory everything is a set. Including integers. There are no types, these are superimposed on sets as needed. If $X$ is a non-empty set, then all its elements are sets as well. – Asaf Karagila Jul 24 '14 at 18:05
Well, yeah, formally you would say something like: What if I had set infinite set X, which contained X1, X2, X3, etc.? And I would say something like: Does your universe contain infinite set Y, which contains Y1, Y2, Y3, etc.? If so, then unless it is also the case that all members of Y are also members of X, then there are lots of sets not in X. – Eric Charles Jul 24 '14 at 19:45
I literally don't understand your comment. – Asaf Karagila Jul 24 '14 at 20:36
Sorry about that, trying to clarfiy: The original question was "Is there another element distinct from X which is not a member of X?" and the question was in the context of ZFC, which invokes the rule "X∉X", because it needs that to avoid Russel's paradox. So, either the answer to the original question is "No, there is only that one exception, which was created to avoid a somewhat obscure problem." or "Yes, because there are things other than elements of X in the wider system." It seems to me this is more an issue of starting axioms and definitions than a deep question for proof. – Eric Charles Jul 25 '14 at 15:15

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