# Algebraic complements in vector space of functions without the axiom of choice

The axiom of choice is equivalent to the statement that every subspace $U$ of every vector space $V$ has an algebraic complement, i.e. another subspace $W$ that has a trivial intersection with the former subspace ($U\cap W=\{0\}$) and their direct sum gives the whole vector space ($U\oplus W=V$). It is also equivalent to the statement that every vector space has a basis.

However, in the simple case of the vector space $V$ (over field $\mathbb R$) of all functions from a given set S to the real line $\mathbb R$, does every subspace have a direct complement? For example, it is easy to find a basis without the axiom of choice: $\{s\mapsto(\text{1 if$s=t$, otherwise 0})\;:\;t\in S\}$. Is it also easy to complement any subspace $U$ of $V$ without the axiom of choice? If not, what conditions on $U$ (e.g. finite dimensional subspace) make it possible to complement it without the use of the axiom of choice?

• That's not a basis: how do you express functions that are non-zero at infinitely many points? – Zhen Lin Sep 11 '13 at 8:36

If you take a look at the proof that the existence of a complement implies the axiom of choice then by a careful proof analysis one can construct a counterexample that is actually in a space which has a basis.

The proof of the algebraic principle goes through proving that the axiom of multiple choice holds, rather than the axiom of choice. The two are equivalent over $\sf ZF$, so it's fine. This version is as follows:

The axiom of multiple choice. Let $X$ be a family of pairwise disjoint non-empty sets, each having at least two elements. Then there exists a function $f$ whose domain is $X$ such that $f(x)$ is a proper finite subset of $x$, for all $x\in X$.

The proof of the direct complement begins by taking $X$ as above, and defining the following vector space: $$V=\bigoplus_{x\in X}F^{(x)}$$

Where $F^{(x)}$ is just the space of functions with finitely many non-zero values from $x$ into $F$. We also define $F^{(x)}_0$ to be the space $f\in F^{(x)}$ such that $\sum_{u\in x}f(u)=0$.

I will leave it to you to find a basis for this space (hint, $F^{(\bigcup X)}$). Now if we can find a complement to the subspace $W=\bigoplus_{x\in X}F^{(x)}_0$ then we can construct a function $f$ as above.

Begin now with a model where the axiom of choice fails, then the axiom of multiple choice fails, and we can find such $X$ where the vector space $V$ has a basis but $W$ doesn't have a direct complement.