6
$\begingroup$

let me say that I am not a set theorist, but I have to settle up some things in category theory and I need your help.

What I'd like to do is, in some way, use axiom of choice for proper classes. I say that axiom of choice holds for a set $X$ if there exist a function $f:X \to \bigcup X$ such that $f(x) \in x$.

My attempt is the following:

Assume the Tarski Grothendieck axiom, i.e. every set is contained in a Grothendieck Universe.

Fix a Universe $U$ and an enlarged universe $U^+$ that contains $U$. Now note that every subset of $U$, i.e. a $U$ proper class, is contained in $U^+$ by the subset axiom. Now I read from wikipedia that "axiom of choice holds" in Tarski Grothendieck framework. My question is:

Does the axiom of choice holds in TG for every element of some universe?

This would mean that for every subset of $U$, axiom of choice holds, being an element of $U^+$. To be honest, I would be ok with the following:

Does the axiom of choice holds for a Universe? Can you provide a reference?

$\endgroup$
8
$\begingroup$

If you have ZFC in the ambient theory, including the axiom of choice, then indeed the axiom of choice holds in every Grothendieck-Zermelo universe (also sometimes known as Grothendieck universes). A Grothendieck-Zermelo universe is a rank-initial segment $V_\kappa$ of the cumulative hierarchy, where $\kappa$ is an inaccessible cardinal. And every set in $V_\kappa$ has a well-ordering in $V$, by the axiom of choice in the ambient theory, and this order must also be in $V_\kappa$, since $V_\kappa$ is closed under subsets of its elements.

Basic lesson: if the axiom of choice holds in the background theory, then it also holds in every Grothendieck-Zermelo universe.

Meanwhile, it is not a consequence of the Universe axiom over ZF that AC must hold, since one can easily use forcing to make AC false, while preserving the truth of the universe axiom. So without AC in the ambient theory, you cannot conclude that AC holds in every Grothendieck-Zermelo universe.

$\endgroup$
  • 1
    $\begingroup$ Wow! Thank you very much. Another question: now let's suppose I just need choice for the class of finite sets, and the class $\{X: |X| = k\}$, where $k$ is some fixed cardinal (the cardinality of $\mathbb{R}$ is ok for me). Which axioms would you suggest me to assume? I suspect that TG is not necessary, even if it solves the issue. On the other side, it seems to me that there are " a lot" of finite sets, in the sense that the function $\{\}: U \to Finite Sets$ that send $X$ to $\{X\}$ is injective. So... what to do? $\endgroup$ – Andrea Marino Dec 10 '19 at 15:25
  • 1
    $\begingroup$ The axiom of choice for (families of) finite sets is not provable in ZF, but it is strictly weaker than AC, since it follows from the axiom that every set is linearly orderable, which is weaker than AC. And there are complicated relations between the principles, if asserted for specific finite sizes. I suggest asking as a separate question, and some of the AC experts will weigh in. $\endgroup$ – Joel David Hamkins Dec 10 '19 at 18:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.