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

- Every $v\in V$ can be written as a linear combination of elements of $\mathcal B$;
- 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?

noin general; somebody here will force you a counterexample :). $\endgroup$1more comment