3
$\begingroup$

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?

$\endgroup$
  • 3
    $\begingroup$ That's not a basis: how do you express functions that are non-zero at infinitely many points? $\endgroup$ – Zhen Lin Sep 11 '13 at 8:36
3
$\begingroup$

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.

$\endgroup$

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.