If $V$ is a vector space with a basis. $W\subseteq V$ has to have a basis too?

Question

Suppose $V$ is a vector space, we say that $\mathcal B$ is a basis for $V$ if:

1. Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
2. If $\sum\alpha_i b_i = 0$, where $\alpha_i$ are scalars and $b_i\in\mathcal B$, then $\alpha_i=0$ for all $i$.

Assuming the axiom of choice, every vector space has a basis. In particular, every subspace have a basis.

However assuming the axiom of choice does not hold, there are spaces without a basis. Of course that if $V$ is a vector space without a basis it may have a subspace which has a basis, e.g. a span of a single vector.

It is simple to have a vector space which has a non-$\aleph$ basis as well, since in the absence of choice there are sets whose cardinals are not $\aleph$ numbers, let $A$ be such set and consider the functions from $A$ into $\mathbb F$ with finite support. That is:

$$V=\left\lbrace f\colon A\to\mathbb F\ \colon\ |A\setminus f^{-1}(0)|<\aleph_0\right\rbrace$$

Addition and multiplication by scalar defined pointwise make it pretty clear how this is a vector space over $\mathbb F$. Every such function can be defined as a linear combination of $\delta$ functions, that is functions which are $1$ at a single point only.

It is also pretty clear that $a\mapsto\delta_a$ is a bijection between $A$ and this basis, therefore we have a basis which is not well-orderable.

Question: $(\lnot AC)$ Suppose $V$ is a vector space, and $\mathcal B$ is a basis of $V$. Is it true that every subspace of $V$ has a basis? Or can we find a counterexample, namely a vector space spanned by a basis with a subspace which has no basis?

Does this depend on the definition of basis above?

Answer 1

The answer is no, I think. Here is a proof sketch. (An unclear point in a previous version has now been removed, by slightly modifying the construction of the sequence.)

Let $(S_n)_{n\in\omega}$ be a family of ``pairs of socks''; that is, each $S_n$ has 2 elements, the $S_n$ are disjoint, but there is no set which meets infinitely many $S_n$ in exactly one point. Let $S$ be the union of the $S_n$.

Let $V$ be a vector space with basis $S$ over the 3-element field. For each $v\in V$, each $s\in S$ let $c_s(v)$ be the $s$-coordinate of $v$. (In your notation: $v(s)$.)

Consider the subspace $W$ of all vectors $w$ with the following property: For all $n$, if $S_n = \{a,b\}$, then $c_a(w)+c_b(w)=0$. The set of all $n$ such that for both/any $a\in S_n$ we have $c_a(w) \neq0$ will be called the domain of $w$. Clearly, each domain is finite, and for each finite subset of $\omega$ of size $k$ there are $2^k$ vectors $w\in W$ with this domain.

[Revised version from here on.]

I will show

• From any basis $C$ of $W$ we can define a 1-1 sequence of elements of $W$.
• From any 1-1 sequence of elements of $W$ we can define a 1-1 sequence of elements of $S$. Together, this will show that there is no basis, as $S$ contains no countably infinite set.

For each set $D$ which appears as the domain of a basis vector, let $x_D$ be the sum of all basis vectors with this domain. So $x_D \neq 0$, and for $D\neq D'$ we get $x_D\neq x_{D'}$. From a well-order of the finite subsets of $\omega$ we thus obtain a well-ordered sequence of nonzero vectors. Since there must be infinitely many basis vectors, and only finitely many can share the same set $D$, we have obtained an infinite sequence of vectors in $W$.

We are now given an infinite sequence $(w_n)$ of distinct vectors of $W$. The union of their domains cannot be finite, so we may wlog assume that the sequence $k_n:= \max(dom(w_n))$ is strictly increasing. (Thin out, if necessary.)

Now let $a_n$ be the element of $S_{k_n}$ be such that $c_{a_n}(w_n)=1$. Then the set of those $a_n$ meets infinitely many of the $S_k$ in a singleton.