I know that Zorn's lemma or its equivalent axiom of choice and so on can be applied to a set - but I am not sure if it can be applied to a class.

I think I've seen an usage of axiom of choice to a class, but I am not sure if I really did. So the question comes.


Let me try and be more accurate when you say "apply Zorn's lemma to a class":

Suppose that $(A,\leq)$ is a partially ordered class (that is $A$ is a class, and $\leq$ is a class, such that $\leq\subseteq A\times A$, and the axioms of partial orders hold for $\leq$), such that for every chain $B\subseteq A$ there exists some $a\in A$ such that $a$ is an upper bound of $B$. Then there is $a\in A$ which is a maximal element.

This would be Zorn's lemma for proper class, and it is equivalent to the axiom of choice for class, which is called "the axiom of global choice". To be more explicit about this:

Every class of non-empty sets admits a choice function (which may be a proper class itself).

Similarly to the case of the usual axiom of choice, this is equivalent to the well-ordering principle with a slight twist:

Every proper class can be put in bijection with the class of ordinals.

(It is enough to require this only for the entire universe, of course; and in fact the global choice thing too - the universe except the empty set).

All the usual proofs work out just fine.

Now it is important to add that these three equivalent principles are stronger than the axiom of choice/Zorn's lemma/well ordering principle for sets. Indeed it is consistent with $\sf ZFC$ that there is no well-ordering of the entire universe, and similarly there is no choice function from all non-empty sets, and there is a partially ordered class where every chain is bounded but there is no maximal element.

  • $\begingroup$ When you say "every chain is bounded" do you mean every set chain or every class chain? In the former case, of course, the ordinals themselves are an example. $\endgroup$ – Carl Mummert Jul 3 '13 at 14:32
  • $\begingroup$ @Carl: I mean a class chain, of course. And let me add, I mean a class of the universe. So it has to be first-order definable too! :-) $\endgroup$ – Asaf Karagila Jul 3 '13 at 14:33
  • $\begingroup$ Only if you think about things in terms of ZFC :) $\endgroup$ – Carl Mummert Jul 3 '13 at 14:36
  • $\begingroup$ Anyway, I just wanted to document that the chain could be a class (and the $\subseteq$ in the first quoted statement means subclass) to clarify that for people who read the answer later. $\endgroup$ – Carl Mummert Jul 3 '13 at 14:36
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    $\begingroup$ @Ric: In this context a function means a class (which may or may not be definable) of ordered pairs which satisfies the requirement of a function (i.e. $\langle x,y\rangle$ and $\langle x,z\rangle$ are both in this class means that $y=z$), and it has the property that $\langle x,y\rangle$ in the class, then $y\in x$. That is, it is a choice function. It's just not a set. $\endgroup$ – Asaf Karagila Mar 23 '14 at 21:03

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