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Is the isomorphism class of a fixed cardinality a set(not a proper class)? Or a fixed ordinality for that matter? By "isomorphism" I mean just bijection for cardinals and order preserving bijection for ordinals, in the category of sets.

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Only if the cardinalty in question is zero. Each set is an element of a set of given nonzero cardinality, so the union of all sets of a given nonzero cardinality is the universe. –  Robin Chapman Aug 1 '10 at 10:53
    
See also Class of sets of a given infinite cardinality at MSE. –  Martin Sleziak Sep 20 '13 at 12:37
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2 Answers 2

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The isomorphism classes of the set {} and the ordinal 0 are sets, each with a single element.

let $\kappa$ be an non-zero ordinal and $\alpha$ # $\beta$ be the natural sum of the ordinals $\alpha$ and $\beta$. Since the natural sum is the restriction of the addition of surreal numbers to ordinals, and the surreal numbers are an ordered field, the natural sum is cancellative and strictly increasing. Therefore, for any ordinal $\gamma$, the set {$\gamma$#0, $\gamma$#1, ... $\gamma$#$\kappa$} of ordinals that are the natural sum of $\gamma$ and an ordinal less than $\kappa$ is order-isomorphic to $\kappa$. These sets are different for different values $\gamma$, so this provides a bijection between the ordinals and a subclass of the isomorphism class of $\kappa$. If the isomorphism class of $\kappa$ was a set, this subclass would be a set and therefore the ordinals would also form a set. The ordinals cannot form a set, therefore the isomorphism class of $\kappa$ does not form a set.

Assuming the axiom of choice, every set is well-orderable. Given a well-ordering of a set there is a bijection between the order-isomorphism class of that set and a subclass of it's isomorphism(bijection) class. if the set is non-empty, then since it's order-isomorphism class is not a set it's isomorphism class is not a set either.

This can also be proved without the axiom of choice. Suppose that a set x is bijectible with {y}$\sqcup$ z, where $\sqcup$ means disjoint union, for two sets y and z. Then for any set w, x is bijectible with {w}$\sqcup$z , and since these sets are different for different values of w, this gives a bijection of the class of all sets with a subclass of the isomorphism class of x. Since the class all sets cannot be a set, the isomorphism class of x cannot be a set. It remains to be shown that any non-empty set x is bijectible with {y}$\sqcup$z for two sets y and z. to see this, simply choose one element v from x (this does not require the axiom of choice, because we are only making one choice of one element from one set) and write x as the union of {v} and the set of other elements of x. these two sets are disjoint, so the obvious surjection from their disjoint union to their union is a bijection. The union is x, so v and the set of other elements of x are the y and z we were looking for, respectively.

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This sound very complicated to me. Why mention surreal numbers? Given a nonempty set $X$, consider the class of all sets of the form $X_y=\{(x,y):x\in X\}$, where $y$ is any set. Without AC there is a bijection between $X$ and $X_y$, by mapping each $x\in X$ to $(x,y)$. Clearly the collection of all $X_y$ is a proper class. –  Stefan Geschke Aug 1 '10 at 14:31
    
Still, the mention to surreal numbers makes it more illuminating... –  David FernandezBreton Jun 27 '11 at 18:42
    
@DavidFernandezBreton, lol that made me laugh, bitime. (You're are being sarcastic, right?) –  goblin Mar 29 at 8:44
    
@user18921: Ok., maybe "illuminating" was not the word I should have used, but nevertheless I like the argument (even if unnecessarily complicated) because I like surreal numbers :) –  David FernandezBreton May 5 at 18:00
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If I understand your question correctly, the first part can be rephrased as: Is the system of all sets of a given cardinality a set or a proper class. It is a proper class already for singletons: Just notice that $x\mapsto\{x\}$ is a bijection between all sets and all singletons. Or, for a given cardinal $\varkappa$, you can use the bijection $x\mapsto \{x\}\times\varkappa$ between the class of all sets and a class consisting only of sets of cardinality $\varkappa$ (although not all of them). If singletons (sets of cardinality $\varkappa$) formed a set, then (by axiom schema of replacement) the class of all sets would be a set too, a contradiction.

(By the way, this is pretty much along the same lines as what Robin Chapman and Stefan Geschke wrote in their comments.)

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This comment might be slightly off-topic, but it explains why this question interested me. A natural possibility to define a cardinal number would be the class of all sets of the same cardinality. The problem is that this is a proper class. This problem can be resolved if we accept AC for classes en.wikipedia.org/wiki/Axiom_of_global_choice or, as I learned while trying to find a reference for the original question, using Scott's trick en.wikipedia.org/wiki/Scott%27s_trick Forster's book referenced at wikipedia mentions answer to the original question in section 8.6.1. –  Martin Sleziak Aug 9 '10 at 9:56
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