Existence of vector space complement and axiom of choice

Question

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)?

Answer 1

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.