# General Steinitz exchange lemma

Where can I find a proof of the following general Steinitz exchange lemma:

Let $B$ be a basis of a vector space $V$, and $L\subset V$ be linearly independent. Then there is an injection $j:L\rightarrow B$ such that $L\cup(B\setminus j(L))$ is a disjoint union and a basis of the vector space $V$.

Or can someone suggest a proof of this result? Probably using Zorn's lemma.

Thank you!

• @Censi LI: The axiom of choice tag is for questions asking specifically about the role of the axiom of choice in a proof, or what sort of statement are consistently true without the axiom. This question does not seem to fit into that tag. Similarly, this is not a set theory question. – Asaf Karagila Jun 2 '15 at 21:36
• @AsafKaragila I'm so sorry... – Censi LI Jun 2 '15 at 21:39
• @Censi LI: No need to apologize. Just for future reference. – Asaf Karagila Jun 2 '15 at 21:40
• Here you can find a blogpost about a very general replacement theorem of Steinitz for modules. – Pedro Tamaroff Jul 1 '17 at 21:19

Let $W$ denote the subspace generated by $L$, $\mathcal B$ denote the set of subsets of $B$ whose images in $V/W$ are linearly independent. There is an obvious partial order on $\mathcal B$, namely the one induced by inclusion. Apply Zorn's Lemma to obtain a maximal element of $\mathcal B$, denoted by $C$, then the disjoint union $C\cup L$ forms a basis of $V$, just as $B=C\cup(B\setminus C)$ do. So there must be a bijection $j:L\xrightarrow{\sim}B\setminus C$. This $j$ will fulfill the requirement.
• It is not true that every chain in $\mathcal{L}$ has an upper bound--see my answer at math.stackexchange.com/questions/2341188/…. The problem is that at each step of the chain you are removing new elements of your spanning set since $j(L')$ is getting larger, and the limit may end up not spanning. – Eric Wofsey Jul 1 '17 at 17:06
Choose (by using Zorn's Lemma) a subset $B'$ of $B$ maximal with the property that $\langle L\rangle\cap \langle B'\rangle=0$. Then $\langle L\rangle+\langle B'\rangle=V$: if $b\in B-B'$ then $\langle L\rangle\cap \langle B'\cup\{b\}\rangle\ne0$ hence there are $l_i\in L$, and $b_i'\in B'$ such that $\sum\alpha_il_i=\sum_i\beta_ib_i'+\beta b$ with $\beta\ne0$, so $b\in \langle L\rangle+\langle B'\rangle$.
Now use that any two bases of a vector space are equipotent. Let $V'=\langle B'\rangle$. The quotient vector space $V/V'$ has the following bases: $L$ and $B\setminus B'$. Then there is a bijection $f:L\to B\setminus B'$ which composed by the inclusion $i:B\setminus B'\to B$ gives us the desired injection, that is, $j=i\circ f$.