Independent sets in the span of a countable set are countable

I am self-studying Artin's Algebra. Following is Exercise 6.1 from Chapter 3:

Let $V$ be a vector space that is spanned by a countably infinite set. Prove that every independent subset of $V$ is finite or countably infinite.

Though there's an answer on this site here, I want to know if the following proof idea of mine also works.

My proof idea:

Let $S = (u_1, u_2, \ldots)$ be a countably infinite spanning set and $L$ be an infinite independent set. For each $v\in L$, choose$^1$ a coordinate vector $X = (x_1, x_2, \ldots)$ with respect to $S$, so that $X$ is nonzero in only finitely many slots. This assignment is injective since $L$ is independent, and hence we can focus on the set of chosen coordinate vectors. Now, since each of them has coordinates eventually equal to zero, it is countable.

$^1$We can avoid AC here by ordering the possibly many coordinate vectors of a given $v$ by an analogue of lexicographic order, and taking the minimal of those. (The field of scalars need not be totally ordered for this.)

Solution 1:

Your argument is a sketch of a correct proof, though I feel there are some important steps you have glossed over. First, you ought to mention that because $L$ is linearly independent, so is the set of coordinate vectors you have chosen to represent its elements (because any nontrivial linear relation between those coordinate vectors would give a nontrivial linear relation between the corresponding elements of $L$).

So, you have a linearly independent subset of the space $k^\infty$ of sequences of scalars that are eventually $0$. Assuming that you have shown that there is no uncountable linearly independent subset of $k^\infty$, you are now done. That fact is nontrivial enough though (I would say it is basically as hard as the statement you are trying to prove) that you should probably mention it explicitly. (One way to prove that fact is to observe, for instance, that a given linearly independent subset of $k^\infty$ can contain at most $n$ elements of $k^n$ where you identify $k^n$ with the set of sequences that are $0$ after their $n$th term, and use this to write your linearly independent subset as a countably union of finite sets.)

In more abstract terms, here is what is going on in your argument. If $V$ is spanned by some set $S$, then this gives a surjective linear map $f:F(S)\to V$ where $F(S)$ is the free vector space on $S$ (the set of formal linear combinations of elements of $S$). Given any linearly independent family of elements of $V$, you can lift it along $f$ to get a family of linearly independent elements of $F(S)$. So, it suffices to show that there is no uncountable family of linearly independent elements of $F(S)$.

It is not clear to me how you are avoiding any use of the axiom of choice here, though. Defining a lexicographic order on a set of tuples requires an ordering on each coordinate set, so you need some sort of ordering on the field $k$. In fact, your argument would appear to require not just a total ordering on $k$ but a well-ordering, since you need the set of ways of representing a given element $v\in L$ to have a least element with respect to your ordering (and the set of such ways is typically infinite). Moreover, the final step of the argument which you glossed over (proving that $k^\infty$ has no uncountable linearly independent subset) appears to require some use of the axiom of choice (in the argument I suggested, to conclude the countable union of finite sets is countable), though I don't know whether this can be avoided somehow.