定理（我想是康托尔）。 设 X 是一个集合，2^X 是其子集的集合。 那么卡(X) < 卡（2^X）。
证明。 当然，card(X) <= card(2^X)，因为 X 和 2^X 中的单例之间存在微不足道的双射。 我们必须证明card(X)!=card(2^X)。
假设 X 和 2^X 之间存在双射。 然后X中的每个xk被映射到2^X中的集合Ak。
x1 ---> A1
×2---> A2
...
xk ---> Ak
...
对于每个 xk 的机会是：xk 属于 Ak，或者不属于。 令 M 为所有不属于其相应集合 Ak 的 xk 的集合。 M是X的子集，因此必须存在X的元素m通过双射映射到M。
m 属于 M 吗？ 如果是，那么就不是，因为 M 是那些不属于它们映射到的集合的 x 的集合。 如果不是，那么它就是，因为 M 包含所有这样的 x。 这个矛盾源于双射存在的假设。 因此双射不存在，两个基数不同，定理得证。
The most elegant proof I've ever seen resembles Russell's paradox closely.
Theorem (Cantor, I suppose). Let X be a set, and 2^X the set of its subsets. Then card(X) < card(2^X).
Proof. Surely card(X) <= card(2^X), since there is a trivial bijection between X and the singletons in 2^X. We must prove that card(X) != card(2^X).
Suppose there is a bijection between X and 2^X. Then each xk in X is mapped to a set Ak in 2^X.
x1 ---> A1
x2 ---> A2
...
xk ---> Ak
...
For each xk the chances are: either xk belongs to Ak, or it does not. Let M be the set of all those xk that do not belong to their corresponding set Ak. M is a subset of X, thus there must exist an element m of X which is mapped to M by the bijection.
Does m belong to M? If it does, then it does not, for M is the set of those x that do not belong to the set they're mapped to. If it does not, then it does, for M contains all such x's. This contradiction stems from the assumption that a bijection exists. Thus a bijection cannot exist, the two cardinalities are different, and the theorem is proved.
The question is ill-posed in the standard ZFC (Zermelo-Fraenkel + axiom of Choice) set theory because the object thus defined is not a set.
Since (again, assuming standard ZFC) your class {x : x\not\in x} is not a set, the answer becomes no, it's not an element of itself (even as a class) since only sets can be elements of classes or sets.
By the way, as soon as you agree to the axiom of foundation, no set can be an element of itself.
Of course the nice thing about math is you can choose whichever axioms you want :) but believing in paradoxes is just weird.
在冯·诺依曼-伯奈斯-哥德尔集合论 (NBG) 中，类 R = { x : x > 是一个集合，并且 x ∉ x } 是可定义的。 然后我们问是否R ∈ R； 如果是这样，那么也 R ∉ R，产生矛盾。 因此我们必须有R ∉ R。 但这里并不矛盾，因为对于任何给定的类 A，A ∉ R 意味着 A ∈ < em>A 或 A 是一个真类。 由于 R ∉ R，我们必须简单地认为 R 是一个真类。
当然，如果没有限制，类 R = { x : x ∉ x } 根本就不是可在NBG中定义。
另外值得注意的是，上述过程在NBG中可以形式化地构造为证明，而在ZFC中则必须诉诸元推理。
In ZFC, either the axiom of foundation [as mentioned] or the axiom (scheme) of comprehension will prohibit this. The first, for obvious reasons; the second, since it basically says that for given z and first-order property P, you can construct { x ∈ z : P(x) }, but to generate the Russell set, you would need z = V (the class of all sets), which is not a set (i.e. cannot be generated from any of the given axioms).
In New Foundations (NF), "x ∉ x" is not a stratified formula, and so again we cannot define the Russell set. Somewhat amusingly, however, Vis a set in NF.
In von Neumann--Bernays--Gödel set theory (NBG), the class R = { x : x is a set and x ∉ x } is definable. We then ask whether R ∈ R; if so, then also R ∉ R, giving a contradiction. Thus we must have R ∉ R. But there is no contradiction here, since for any given class A, A ∉ R implies either A ∈ A or A is a proper class. Since R ∉ R, we must simply have that R is a proper class.
Of course, the class R = { x : x ∉ x }, without the restriction, is simply not definable in NBG.
Also of note is that the above procedure is formally constructable as a proof in NBG, whereas in ZFC one has to resort to meta-reasoning.
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我见过的最优雅的证明与罗素悖论非常相似。
定理（我想是康托尔）。
设 X 是一个集合，2^X 是其子集的集合。 那么卡(X) < 卡（2^X）。
证明。 当然，card(X) <= card(2^X)，因为 X 和 2^X 中的单例之间存在微不足道的双射。 我们必须证明card(X)!=card(2^X)。
假设 X 和 2^X 之间存在双射。 然后X中的每个xk被映射到2^X中的集合Ak。
对于每个 xk 的机会是：xk 属于 Ak，或者不属于。 令 M 为所有不属于其相应集合 Ak 的 xk 的集合。 M是X的子集，因此必须存在X的元素m通过双射映射到M。
m 属于 M 吗？ 如果是，那么就不是，因为 M 是那些不属于它们映射到的集合的 x 的集合。 如果不是，那么它就是，因为 M 包含所有这样的 x。 这个矛盾源于双射存在的假设。 因此双射不存在，两个基数不同，定理得证。
The most elegant proof I've ever seen resembles Russell's paradox closely.
Theorem (Cantor, I suppose).
Let X be a set, and 2^X the set of its subsets. Then card(X) < card(2^X).
Proof. Surely card(X) <= card(2^X), since there is a trivial bijection between X and the singletons in 2^X. We must prove that card(X) != card(2^X).
Suppose there is a bijection between X and 2^X. Then each xk in X is mapped to a set Ak in 2^X.
For each xk the chances are: either xk belongs to Ak, or it does not. Let M be the set of all those xk that do not belong to their corresponding set Ak. M is a subset of X, thus there must exist an element m of X which is mapped to M by the bijection.
Does m belong to M? If it does, then it does not, for M is the set of those x that do not belong to the set they're mapped to. If it does not, then it does, for M contains all such x's. This contradiction stems from the assumption that a bijection exists. Thus a bijection cannot exist, the two cardinalities are different, and the theorem is proved.
这个问题在标准 ZFC 中是不恰当的（Zermelo-Fraenkel + 选择公理）集合论，因为这样定义的对象不是集合。
由于（再次假设标准 ZFC）您的 class {x : x\not\in x} 不是一个集合，所以答案变为“否”，它不是其自身的元素（即使作为一个类），因为只有集合可以是类或集合的元素。
顺便说一句，一旦您同意基础公理，任何集合都不能成为本身的元素。
当然，数学的好处是你可以选择你想要的任何公理:)，但相信悖论是很奇怪的。
The question is ill-posed in the standard ZFC (Zermelo-Fraenkel + axiom of Choice) set theory because the object thus defined is not a set.
Since (again, assuming standard ZFC) your class {x : x\not\in x} is not a set, the answer becomes no, it's not an element of itself (even as a class) since only sets can be elements of classes or sets.
By the way, as soon as you agree to the axiom of foundation, no set can be an element of itself.
Of course the nice thing about math is you can choose whichever axioms you want :) but believing in paradoxes is just weird.
在ZFC中，基础公理[如上所述]或理解公理（方案）将禁止这样做。 第一个，原因显而易见； 第二个，因为它基本上表示对于给定的 z 和一阶属性 P，您可以构造 { x ∈ z< /em> : P(x) }，但要生成罗素集，您需要 z = V（所有集合的类），它不是一个集合（即不能从任何给定的公理生成）。
在新基础（NF）中，“x ∉ x”不是分层公式，因此我们再次无法定义罗素集。 然而，有点有趣的是，V是NF中的一个集合。
在冯·诺依曼-伯奈斯-哥德尔集合论 (NBG) 中，类 R = { x : x > 是一个集合，并且 x ∉ x } 是可定义的。 然后我们问是否R ∈ R； 如果是这样，那么也 R ∉ R，产生矛盾。 因此我们必须有R ∉ R。 但这里并不矛盾，因为对于任何给定的类 A，A ∉ R 意味着 A ∈ < em>A 或 A 是一个真类。 由于 R ∉ R，我们必须简单地认为 R 是一个真类。
当然，如果没有限制，类 R = { x : x ∉ x } 根本就不是可在NBG中定义。
另外值得注意的是，上述过程在NBG中可以形式化地构造为证明，而在ZFC中则必须诉诸元推理。
In ZFC, either the axiom of foundation [as mentioned] or the axiom (scheme) of comprehension will prohibit this. The first, for obvious reasons; the second, since it basically says that for given z and first-order property P, you can construct { x ∈ z : P(x) }, but to generate the Russell set, you would need z = V (the class of all sets), which is not a set (i.e. cannot be generated from any of the given axioms).
In New Foundations (NF), "x ∉ x" is not a stratified formula, and so again we cannot define the Russell set. Somewhat amusingly, however, V is a set in NF.
In von Neumann--Bernays--Gödel set theory (NBG), the class R = { x : x is a set and x ∉ x } is definable. We then ask whether R ∈ R; if so, then also R ∉ R, giving a contradiction. Thus we must have R ∉ R. But there is no contradiction here, since for any given class A, A ∉ R implies either A ∈ A or A is a proper class. Since R ∉ R, we must simply have that R is a proper class.
Of course, the class R = { x : x ∉ x }, without the restriction, is simply not definable in NBG.
Also of note is that the above procedure is formally constructable as a proof in NBG, whereas in ZFC one has to resort to meta-reasoning.