Let $V$ be a vector space. In order to show that any linearly independent set in $V$ can be extended to form a basis, or equivalently, any basis of a subspace of $V$ can be extended to a basis of $V$, is it required to use choice principles (i.e. Axiom of Choice or weaker forms of Choice). If so, what is the weakest choice principle that can prove this statement?
Is it required to use choice principles to prove that any linearly independent set can be extended to a basis
axiom-of-choicelinear algebra
Related Solutions
Yes. The axiom of choice is needed in order to show that every vector space which is not finitely generated contains an infinite linearly independent subset.
The original consistency proof due to Lauchli (1962) was to construct a model in which there is a vector space that is not spanned by any finite set, but every proper subspace is finite.
Namely every collection of linearly independent vectors is finite.
If you have the axiom of dependent choice then you can actually perform the induction which you suggest and have a countably infinite set of independent vectors. But this is quite far from the full axiom of choice.
If one tries real hard, one can get away with just countable choice (which is strictly weaker than dependent choice). The argument is as follows:
Let $\cal A_n$ be the collection of all sets of linearly independent of size $n$, since the space is not finitely generated $\cal A_n$ is non-empty for all $n\neq 0$. Let $A_n\in\cal A_n$ be some chosen set. Again using countable choice let $A$ be the union of the $A_n$'s, and $A$ is countable so we can write it as $\{a_n\mid n\in\omega\}$.
Pick $v_0=a_0$, and proceed by induction to define $v_{n+1}$ to be $a_k$ whose index is the least $k$ such that $a_k$ not in the span of $\{v_0,\ldots, v_n\}$. This $a_k$ exists because $A_{n+1}$ spans a vector space of dimension $n+1$ so it cannot be a subset of $\operatorname{span}\{v_0,\ldots,v_n\}$.
The set $\{v_n\mid n\in\omega\}$ is linearly independent, which follows from the choice of the $v_n$'s.
Interestingly, Lauchli's example was of a space that every endomorphism is a scalar multiplication and as the above indicates this construction also contradicted $\mathsf{DC}$. In my masters thesis I showed that you can have $\mathsf{DC}_\kappa$ and still have a vector space which has no endomorphisms except scalar multiplication - even if it has relatively large subspaces.
The completion is certainly not unique. Multiplying any of the new vectors by a nonzero constant will not affect the span or the linear independence, but will change the basis.
To prove that you can extend any linearly independent set $S$ to a basis, you proceed by an iterative argument. If $S$ spans you are done. Otherwise, the span of $S$ does not include some vector $v$. You claim that $S \cup \{v\}$ is linearly independent. Write down the condition for linear independence and observe for yourself that if linear independence fails, you can deduce that $v$ was in the span of $S$ (solve the non-zero equation for $v$ in terms of elements in $S$). Now you ask if $S \cup \{v\}$ spans, and if so, you are done. If not, take some vector $w$ not in the span and consider, $S \cup \{v\} \cup \{w\}$. Iterate this argument.To prove in general that this iteration eventually terminates in a spanning set, you actually need to use Zorn's Lemma. However, if you have a finite spanning set $B$ then you can pick the elements $v,w, \cdots$ from $B$ and the most number of steps our iteration might take is to exhaust all the elements of $B$.
Best Answer
Just the existence of a basis for every vector space is enough to conclude the axiom of choice, as one can extend the empty set (or any singleton, except $\{0\}$).