This question was inspired by the question on examples of classes that are not sets.

From the discussion in the comments there, it seems there does not exist a set of all sets of a given cardinality.

To me, it seems easy to see that this is true for any nonzero finite cardinality $n$. Given any set $x$, you can always make a set of size $n$ by adding $n-1$ other sets to $x$ to make a set of size $n$, which exists by repeated use of the pairing axiom.

But how would you do this for infinite cardinalities? For suppose $\kappa$ is some given infinite cardinality. To show that the set of all sets of cardinality $\kappa$ does not exist, it seems you would have to show that for any set $x$, there exists a set of that size with $x$ as an element, and then you could take the union of that set to find the set of all sets, and thus a contradiction. But it doesn't seem reasonable to simply say, for a given set of size $\kappa$ if $x$ is in the set, we have no problem. If not, just take an element out of the set and put $x$ in. Something about that seems like it would not be allowed.

So how would you do this for infinite cardinalities?


Suppose there is a set $x$ whose elements are precisely the sets of size $\kappa$. We will define a function $f$ with domain $x$ whose range is the universe $V$ of all sets. Since the latter is known not to be a set (and functions map sets to sets---this is the replacement axiom), it follows that $x$ itself cannot be a set either.

Fix a set $Y$ of size $\kappa$. Given a set $a\in x$, let $f(a)=0$, unless there is a set $z$ such that $a=\{(b,z)\mid b\in Y\}$, in which case we set $f(a)=z$. This $f$ works.

Let me add: The universe $V$ of sets can be seen as "constructed by stages." More precisely, $$V=\bigcup_{\alpha\in ORD}V_\alpha,$$ where the set $V_\alpha$ is obtained by iterating the power set operation $\alpha$ times, starting with the empty set, and $ORD$ is the (proper) class of all ordinals. The details do not matter much right now. The point is that something is a set if it appears at some stage $V_\alpha$ of the construction, and a set cannot appear until all its elements do (i.e., the stage $\alpha$ at which a set $t$ appears is larger than any stage $\beta$ at which an element of $t$ appears).

But something like "the collection of all sets of size $\kappa$" cannot possibly appear at any stage $V_\alpha$, since we can always build a set of size $\kappa$ with $V_\alpha$ as one of its elements, forcing the set to appear at a stage larger than $\alpha$. In general, one can verify whether a collection is a set or a proper class by checking whether it contains sets that appear at arbitrarily large stages (proper class) or, instead, all its elements appear by some stage $\alpha$ (set).

This suggests the heuristic that a proper class ought to be as large as the universe, or at least, as large as the class ORD of all ordinals. This, however, is independent of the usual axioms of set theory. (An additional problem appears here, since the standard presentation of set theory does not treat proper classes as objects, so one cannot even make the statement "all proper classes have the same size". But there are ways around this obstacle.)

  • $\begingroup$ @Hobbie: Given $z$, to say $a=\{(b,z)\mid b\in Y\}$ is the same as saying $a=Y\times\{z\}$. And for any set $z$, $Y\times\{z\}$ has the same size as $Y$, so it is in our collection of all sets of size $\kappa$. Of course, many sets of size $\kappa$ won't be of this form, so they are mapped to 0 by $f$. But, if $a$ happens to have the special form $Y\times\{z\}$, then $f(a)=z$. $\endgroup$ – Andrés E. Caicedo Feb 14 '11 at 8:08
  • $\begingroup$ Oh ok, so for a set $c$ which is not of the form $Y\times\{z\}$ for some $z$ will still get hit when considering the set $Y\times\{c\}$. And $0$ is just some image arbitrarily chosen so that $f$ is defined everywhere. $\endgroup$ – Dani Hobbes Feb 14 '11 at 8:16
  • $\begingroup$ Exactly. (There are of course many other ways to define a function $f$ that maps this collection of sets onto $V$, perhaps some are more natural that the way I chose.) $\endgroup$ – Andrés E. Caicedo Feb 14 '11 at 8:20

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