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?

  • 3
    $\begingroup$ 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 :). $\endgroup$ Nov 12, 2011 at 21:37
  • 2
    $\begingroup$ 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. $\endgroup$ Nov 12, 2011 at 21:38
  • $\begingroup$ Francois, I am not sure that I understand the remark. $\endgroup$
    – Asaf Karagila
    Nov 13, 2011 at 7:10
  • $\begingroup$ @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 $\oplus_{v \in V} \mathbb{F} \to V$. $\endgroup$ Nov 13, 2011 at 12:08
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    $\begingroup$ 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. $\endgroup$
    – Goldstern
    Nov 14, 2011 at 1:31

1 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.

  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Nov 12, 2011 at 21:59
  • $\begingroup$ 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. $\endgroup$
    – Asaf Karagila
    Nov 13, 2011 at 5:02
  • $\begingroup$ @Asaf The induction is on the (well-ordered) set of finite subsets of $\omega$, not on $S$. $\endgroup$ Nov 13, 2011 at 11:30
  • $\begingroup$ @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. $\endgroup$
    – Asaf Karagila
    Nov 13, 2011 at 12:08
  • $\begingroup$ 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. $\endgroup$
    – Goldstern
    Nov 14, 2011 at 1:08

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