9
$\begingroup$

I'm not sure this question is more suitable for MO or for MSE, so feel free to move it to MSE if necessary.

I work here in ZF theory. Consider the following statements:

$(C)$ Axiom of choice: for any non empty family $(E_i)_{i\in I}$ of non empty subsets , there exists a function $f:I\to\displaystyle \bigcup_{i\in I}E_i$ such that $f(i)\in E_i$ for all $i\in I$

$(MC)$ Axiom of multiple choice: for any non empty family $(E_i)_{i\in I}$ of non empty subsets , there exists a function $f:I\to\displaystyle \mathscr{P}(\bigcup_{i\in I}E_i)$ such that $f(i)$ is a non empty finite subset of $E_i$ for all $i\in I$

$(B)$ for any field $K$, any $K$-vector space $E$ has a basis

$(G)$ For any field $K$ and any $K$-vector space $E$, any generating family of $E$ contains a basis.

In ZF, it is known that all these statements are equivalent.

I'm interested in writing a self-contained proof of $(B)\Longrightarrow (C)$.

I've found two proofs in the literature: one is proving $(B)\Longrightarrow (MC)$, the other one is proving $(G)\Longrightarrow (C)$.

So, if I now how to prove $(MC)\Longrightarrow (C)$ or $(B)\Longrightarrow (G)$, I'm done. Now, here comes my questions.

In all the references I found, the proofs of $(MC)\Longrightarrow (C)$ are drowned in the looong proof of the equivalence of 5 statements in ZF, but I couldn't find any direct proof. I'm not that familiar with non elementary set theory, and i'm not sure the effort I'll have to make to understand the proof of the equivalence of all these statements will worth it. So:

Question 1. Do you know a direct proof of $(MC)\Longrightarrow (C)$ in ZF ?

Moreover, I'm a bit puzzled by the fact that I can't find a direct proof of $(B)\Longrightarrow (G)$ in ZF. I thought it would be straightforward, but I can't find any proof for the moment. It's possible I'm missing an easy argument though. So :

Question 2. Do you know a direct proof of $(B)\Longrightarrow (G)$ in ZF ? Thanks in advance for your help !

Edit. After reading Asaf comment, I think I should say that I'm happy to include the axiom of regularity in ZF.

I reformulate my question: using ZF+regularity, can we find a direct proof of (MC) implies (C), or at least a proof which is more direct thatn the ones I've read in books (which are a chain of 4 implications, see my comment to Asaf answer).

Also, using ZF+regularity, can we find a proof of (B) implies (G) ?

Of course, I anticipate the the answers will be probably NO, but I prefer to make sure, before I choose to use a black box in my proof.

$\endgroup$
  • $\begingroup$ Would you mind giving a reference of $G\implies C$? $\endgroup$ – Wojowu Sep 10 '17 at 10:23
  • 1
    $\begingroup$ In $MC$, presumably you mean $f(i)$ to be non-empty? $\endgroup$ – Andrej Bauer Sep 10 '17 at 10:28
  • $\begingroup$ Yes, of course. I edit my post right away. $\endgroup$ – GreginGre Sep 10 '17 at 10:30
  • $\begingroup$ Regarding Question 2, $(G)\Rightarrow(C)$ is much easier than $(B)\Rightarrow(C)$ (and I don't think it uses regularity, although IANAST). So I wouldn't expect a very simple direct proof of $(B)\Rightarrow(G)$. $\endgroup$ – Jeremy Rickard Sep 10 '17 at 11:19
  • $\begingroup$ Have you read my reply to your comment before making this edit? Regularity is an axiom of ZF. I just meant to imply that there is no naively simple proof. Which is what a lot of people would mean by "elementary". $\endgroup$ – Asaf Karagila Sep 11 '17 at 11:58
11
$\begingroup$

There is no fully elementary proof that you are looking for. The reason is that the axiom of regularity is needed in these proofs. Multiple Choice does not imply Choice without it, and the only proofs we know about vector spaces go through Multiple Choice. While regularity is not a "difficult axiom" it does mean the proof makes some non-trivial use of the structure of the universe of set theory.

However, you can look in Jech's The Axiom of Choice, in the 9th chapter he covers the implication of $\sf MC\to AC$, and more.

$\endgroup$
  • $\begingroup$ How do you know it's not possible to have a fully elementary proof which would avoid multiple choice? $\endgroup$ – Wojowu Sep 10 '17 at 10:27
  • $\begingroup$ Have I said something remotely like that? $\endgroup$ – Asaf Karagila Sep 10 '17 at 10:31
  • 1
    $\begingroup$ This is how I understood the first sentence, as there being no elementary proof of AC from existence of bases. $\endgroup$ – Wojowu Sep 10 '17 at 10:32
  • 1
    $\begingroup$ And of course, there was quite some research into it, and so far no regularity free proof of bases to choice exists either. $\endgroup$ – Asaf Karagila Sep 10 '17 at 10:37
  • 1
    $\begingroup$ I think you can somewhat simplify Jech’s chain of implications by proving directly that (MC) implies that the power set of a well ordered set can be well ordered (using much the same argument, just not spelling out the intermediate steps). $\endgroup$ – Emil Jeřábek supports Monica Sep 11 '17 at 11:47

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.