I've been trying to find a proof of the Steinitz exchange lemma (the general one, not the one for finite bases).
The closest I've found is the proof given here:
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 $V$.
with solution:
Let $\mathcal L$ denote the set of all pairs $(L',j)$ such that $L'\subseteq L$ and $j:L'\to B$ an injection satisfying that $L'\cup(B\setminus j(L'))$ is a disjoint union and a basis of the vector space $V$. Define a partial order $\preceq$ on $\mathcal L$ by $(L_1,j_1)\preceq(L_2,j_2)$ iff $L_1\subseteq L_2$ and $j_{2\mid L_1}=j_1$. Note that a subset $S$ of $V$ is linearly independent iff every finite subset of $S$ is linearly independent, and generating set iff every element of $V$ is generated by a finite subset of $S$. Therefore it's it's easy to see that every sub-chain of $\mathcal L$ admits a maximal element in $\mathcal L$. Then applying Zorn's Lemma we reach the conclusion.
I can just about follow it, it looks like for a given chain a max element for a particular chain could be made by taking the union $M$ of all the first elements of each member of the chain and the (unique) $m$ function that agrees with all the second elements of each member of the chain. I can prove this max $(M,m)$ is such that $M\cup B\setminus m(M)$ is a disjoint union and linearly independent but can't seem to prove it spans $V$, and is thus in $\mathcal L$, so I can't yet apply zorns' lemma and finish the proof.
can someone please provide a explanation as to why or if I have gone wrong an explanation as to where.
Note: new question and not comment on original question because I lack the points to ask the question as a comment.
Best Answer
You are right to be stuck: this proof is incorrect. For instance, suppose $B=\{b_0,b_1,b_2,\dots\}$. You can then one-by-one replace elements of this basis: first replace $b_0$ by $b_0-b_1$, then replace $b_1$ by $b_1-b_2$, then replace $b_2$ by $b_2-b_3$, and so on. At each finite stage of this, you still have a basis, but in the limit you get the set $L=\{b_0-b_1,b_1-b_2,b_2-b_3,\dots\}$ which does not span the entire space. This then gives a chain in $\mathcal{L}$ that has no upper bound.
Here is a correct proof. Let $C\subseteq B$ be a maximal subset of $B$ with the property that $C\cup L$ is linearly independent (such a $C$ exists by Zorn's lemma). By maximality, the span of $C\cup L$ must contain every element of $B$ and this is all of $V$, so $C\cup L$ is a basis. To finish the proof, we just need to know that $|B\setminus C|=|L|$ so we can choose an injection $j:L\to B$ with image $B\setminus C$. To prove this, note that $|B\setminus C|$ and $|L|$ are both the dimension of the vector space $V/\operatorname{span}(C)$, since the images of $B\setminus C$ and $L$ are each bases for this vector space.
[Note that the fact that any two bases of an infinite-dimensional vector space have the same cardinality can be proven quite easily without the Steinitz exchange lemma. If $B$ and $B'$ are both infinite bases, then each element of $B$ is in the span of finitely many elements of $B'$, so you need at most $|B|\cdot \aleph_0=|B|$ elements of $B'$ to span all of $B$. Since $B'$ is a basis, this means $|B'|\leq |B|$. Similarly, $|B|\leq|B'|$ as well, so $|B|=|B'|$.]