Representing a vector space as a sum of subspaces

Question

In a linear algebra book I'm reading now, there was the following exercise:

Let $W\subseteq V$ be a subspace of vector space $V$. Do there always exist two subspaces $W_1,W_2\subseteq V$ such that:

1. $W_1+W_2=V$

2. $W_1\cap W_2=W$

3. $W_1\neq V,W_2\neq V$

The answer is clearly no if we allow $W=V$, but even without it we can find counterexamples, e.g. $W=\Bbb R\times\{0\},V=\Bbb R^2$.

Critical property of the example above is that there are no "intermediate" spaces, i.e. if $W\subseteq W'\subseteq V$, then $W'=W$ or $W'=V$. I started wondering whether this is an equivalent condition to failure of condition in problem, and I found out it is the case. Below I present a proof of this fact, which however makes heavy use of the axiom of choice (in existence of bases).

My question now is:

Can the equivalence which I state and prove below be shown without any appeal to axiom of choice?

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For $W$ subspace of $V$ the following are equivalent:

1. There exist subspaces $W_1,W_2$ which satisfy 1-3 above

2. There exists a proper subspace of $V$ properly containing $W$.

Proof: 1 $\Rightarrow$ 2: I claim $W_1$ is such a proper subspace. Clearly $W\subseteq W_1\subsetneq V$. If $W_1=W$, then $V=W_1+W_2=W+W_2=W_2$ as $W\subseteq W_2$, but this is a contradiction.

2 $\Rightarrow$ 1: Let $W\subsetneq W_1\subsetneq V$. Let $B_1$ be any basis of $W$, $B_2$ any basis of $W_1$ containing $B_1$ and $B_3$ any basis of $V$ containing $B_2$. Define $W_2=\text{span}((B_3\setminus B_2)\cup B_1)$. It's straightforward to see that $W_1,W_2$ satisfy the properties we want them to.

Answer 1

According to this answer on MO, it is consistent with ZF for there to exist a vector space $V$ which is not $\leq 1$-dimensional and has no nontrivial direct sum decomposition. Taking $W=0$, we get a counterexample to your question. (Note that more generally, your question is equivalent to asking whether the quotient $V/W$ admits a nontrivial direct sum decomposition.)

Answer 2

For every $\kappa$ it is consistent with $\sf DC_\kappa$ that there is a vector space $V$ (over a fixed, but any field) such that:

1. If $W\subseteq V$ is a proper subspace if and only if $W$ has a basis of size $<\kappa$.

2. $V$ is not the span of any $\kappa$ vectors.

This extends the result by Lauchli that Eric Wofsey appeals to. This is a result from my masters thesis.

In such situation any proper subspace of $V$ has an extension by one more vector which is therefore a proper subspace of $V$ still, but there is no decomposition of the space into any two non-zero subspaces. Interestingly, such a vector space will have no nonzero functionals and it cannot be topologized in a nontrivial way (and certainly cannot be made normed).