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

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?

• Let me just remark that $V$ is the direct sum of copies of $\mathbb{F}$ indexed by $A$ and that a vector space has a basis iff it arises as such a direct sum; these standard arguments don't use AC. So the question really is: Can we conclude that in (ZF) every subspace of a direct sum $\oplus_{i \in I} \mathbb{F}$ of copies of $\mathbb{F}$ is again a direct sum of copies of $\mathbb{F}$? The answer is yes if there is a well-ordering on $I$, but probably no in general; somebody here will force you a counterexample :). Nov 12, 2011 at 21:37
• Minor remark : the corresponding problem for surjections admits easy counterexamples, namely for every vector space $W$ there is a surjection $V \to W$ with $V$ admitting a basis (one can take $V=k^{(W)}$). One can then ask whether the kernel of this surjection admits a basis. Nov 12, 2011 at 21:38
• Francois, I am not sure that I understand the remark. Nov 13, 2011 at 7:10
• @Asaf : I just meant that if $V$ is a vector space that doesn't admit a basis (such a vector space exists under $\lnot AC$), then you can cover it by the vector space $\oplus_{v \in V} \mathbb{F}$, which admits a basis. One could then try to produce a counterexample to your question by looking at the kernel of the canonical map $\oplus_{v \in V} \mathbb{F} \to V$. Nov 13, 2011 at 12:08
• One could try, but it would not work, or at least contradict my claim below. The set $W$ below is the kernel of the linear map from $V$ to $3^{<\omega}$, defined by $f(a) = (0,0,..., 1)\in 3^n$ for both $a\in S_n$. So in this case, both $V$ and $V/W$ have a basis, but $W$ does not. Nov 14, 2011 at 1:31

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.

• Thank you for the construction. I will have to look at it more carefully tomorrow morning, though. Unless of course, someone will point a mistake. In the meantime, you are missing a couple \ in the LaTeX code which you may want to correct. Nov 12, 2011 at 21:59
• I think there is a problem with the induction. $S$ is amorphous, so DC does not hold. There an induction might not prove the existence of an infinite set as you like. This is due to the strange cardinality of the vector space. Nov 13, 2011 at 5:02
• @Asaf The induction is on the (well-ordered) set of finite subsets of $\omega$, not on $S$. Nov 13, 2011 at 11:30
• @Guillaume: Yes, I know that. However countable unions of finite sets need not be countable in our model. What we want to try and do is transfer the problem from finite subsets of $S$ (or even $V$) to finite subsets of $\omega$, which I am not 100% certain that we can do nicely. Remember that $S$ itself is amorphous, anything acting on infinitely many members of $S$ is acting the same way on almost all of them. I am currently working on this very model to see whether or not the example is good, but it is not yet clear. Any further insights will be most welcomed. Nov 13, 2011 at 12:08
• Note 1: $S$ is not amorphous; for example, the union of all $S_{2k}$ is infinite co-infinite. Note 2: It is correct that a construction of a 1-1 subsequence of $S$ is not possible - neither by induction not otherwise. But this is an indirect proof. Nov 14, 2011 at 1:08