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Let say we live in the category of vector spaces over $\mathbb{R}$ or $\mathbb{C}.$ Here are three sentences:

  1. Axiom of choice
  2. Every vector space has a base.
  3. For every vector space $V$ and its subspace $E\subset V$ there is a subspace $F\subset V$ such that $V=E\oplus F.$

I know how to prove that (1)->(2)->(3). How about the inverese? Do (2)->(1) and (3)->(2) hold?

If this is not the case, then is there some weaker version of AC which imply (3)?

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  • $\begingroup$ By the way, how do you prove (2) implies (3)? The thing you'd want to use is "every linearly independent set is contained in a basis", but that's not the same as "every vector space has a basis", at least not as far as "naive assumptions go". $\endgroup$ – Asaf Karagila Jun 16 '18 at 17:35
  • $\begingroup$ @AsafKaragila I can't recall what the proof I had in mind then. I recall however that the produced space $F$ should be isomorphic to $V/E$. Now I think about following construction of $F$: Given base $v_\alpha$ of $V$ any $v\in V$ has a form $v=\sum_\alpha a_\alpha v_\alpha$ and thus gives rise to the $i_v\in V^*$ in such a way that $i_v(w) = \sum_{\alpha,\beta} a_\alpha b_\beta \delta_{\alpha}^{\beta}$ where $w = \sum_\beta b_\beta v_\beta$. We put $F$ as $\bigcap_{z\in E}\ker(i_z)$. However, I am not sure if $F\cong V/E$. Do you think that (2)->(3) is a false statement? $\endgroup$ – Fallen Apart Jun 16 '18 at 19:24
  • $\begingroup$ No, both statements are equivalent to choice. I just don't know how to prove (2) implies (3) without proving Zorn's Lemma first. :) $\endgroup$ – Asaf Karagila Jun 16 '18 at 19:35
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No, there is no weaker choice principle implying (3). It was shown that (3) implies the axiom of choice in $\sf ZF$.

The proof is via an equivalent of the axiom of choice called "The Axiom of Multiple Choice". You can find the details in Rubin & Rubin's "Equivalents of the Axiom of Choice II" as Theorem 6.35 (pp. 119-120 and 122).

The proof is due to Bleicher from 1964

M. N. Bleicher, Some theorems on vector spaces and the axiom of choice, Fund. Math. 54 (1964), 95--107.

It is interesting to note that in a more relaxed setting where there might be atoms (non-set objects) or that the axiom of regularity fails, it is not known whether or not (3) implies the axiom of choice.

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  • $\begingroup$ Perfect answer. Since I live in vector spaces I assume axiom of regulatiry. $\endgroup$ – Fallen Apart Mar 25 '16 at 21:30
  • $\begingroup$ @Fallen Apart: I'm not sure why, or how this is relevant. But as far as set theory goes, it apparently is relevant. $\endgroup$ – Asaf Karagila Mar 25 '16 at 21:36
  • $\begingroup$ You are right, but I am still little bit ignorant about what happen when I abandon some of ZF axioms. I was just playing with non-degenerated bilenar forms and over and over again I was using (3). So I asked above question. But do not undersand me wrong. I really like this "pathologies" that you offer and you did not write it in vain. $\endgroup$ – Fallen Apart Mar 25 '16 at 21:44

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