Editing Cardinality (section)

=== Real paradoxes ===

==== Cantor's paradox ====
{{Main|Cantor's paradox}}
[[Cantor's theorem]] state's that, for any set <math>A,</math> possibly infinite, its [[powerset]] <math>\mathcal{P}(A)</math> has a strictly greater cardinality. For example, this means there is no bijection from <math>\N</math> to <math>\mathcal{P}(\N) \sim \R.</math> [[Cantor's paradox]] is a paradox in [[naive set theory]], which shows that there cannot exist a "set of all sets" or "[[Universe (mathematics)|universe set]]". It starts by assuming there is some set of all sets, <math>U := \{x \; | \; x \,\text{ is a set} \},</math> then it must be that <math>U</math> is strictly smaller than <math>\mathcal{P}(U),</math> thus <math>|U| \leq |\mathcal{P}(U)| .</math> But since <math>U</math> contains all sets, we must have that <math>\mathcal{P}(U) \subseteq U,</math> and thus <math>|\mathcal{P}(U)| \leq |U|.</math> Therefore <math>|\mathcal{P}(U)| = |U|,</math> contradicting Cantor's theorem. This was one of the original paradoxes that added to the need for a formalized set theory to avoid these paradoxes. This paradox is usually resolved in formal set theories by disallowing [[unrestricted comprehension]] and the existence of a universe set.

==== Set of all cardinal numbers ====
Similar to Cantor's paradox, the paradox of the set of all cardinal numbers is a result due to unrestricted comprehension. It often uses the definition of cardinal numbers as ordinal numbers for representatives. It is related to the [[Burali-Forti paradox]]. It begins by assuming there is some set <math>S := \{ X \, | X \text{ is a cardinal number}\}.</math> Then, if there is some largest element <math>\aleph \in S ,</math> then the powerset <math>\mathcal{P}(\aleph)</math> is strictly greater, and thus not in <math>S.</math> Conversly, if there is no largest element, then the [[Union (set theory)#Arbitrary union|union]] <math>\bigcup S</math> contains the elements of all elements of <math>S,</math> and is therefore greater than or equal to each element. Since there is no largest element in <math>S,</math> for any element <math>x \in S,</math> there is another element <math>y \in S</math> such that <math>|x| < |y|</math> and <math>|y| \leq \Bigl| \bigcup S \Bigr|.</math> Thus, for any <math>x \in S,</math> <math>|x| < \Bigl| \bigcup S \Bigr|,</math> and so <math>\Bigl| \bigcup S \Bigr| \notin S.</math>