# Axiom(s) of choice and bases of vector spaces

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.

• Would you mind giving a reference of $G\implies C$? – Wojowu Sep 10 '17 at 10:23
• In $MC$, presumably you mean $f(i)$ to be non-empty? – Andrej Bauer Sep 10 '17 at 10:28
• Yes, of course. I edit my post right away. – GreginGre Sep 10 '17 at 10:30
• 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 $(B)\Rightarrow(G)$. – Jeremy Rickard Sep 10 '17 at 11:19
• 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". – Asaf Karagila Sep 11 '17 at 11:58

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.