I am trying to understand the proof of the following theorem:
Let $S$ be a linearly independent subset of a vector space $V$, then there exists a maximal linearly independent subset of $V$ that contains $S$.
We consider a collection say $G$ which contains all the linearly independent sets of $V$ that contain $S$.
So, $G$ could be represented as $\{S_1,S_2,S_3......\}$, where $S_i$ is a linearly independent subset of V containing $S$.
Now we consider a 'chain' $X$ in $G$ which itself is a collection of sets such that for any $A,B \in X$ either $A \subseteq B$ or $B \subseteq A$.
Now, we claim that there exists some $U \in G$ where $U$ is the union of all the elements in $X$.
So, $U$ is the union of all the elements of $X$, whose elements contain $S$, so certainly, $S \subseteq U$.
Now we need to prove that $U$ is linearly independent.
$(*)$ Say, $\{u_1,u_2,.....,u_n\}$ be elements in $U$, for some scalars consider $\sum_{i=1}^{n}(a_iu_i)=0$.
Now, $u_i \in U$ which is the union of all the elements of $X$, so, $u_i \in A$ where $A$ is some set in $X$.
Since, $X$ is a chain, there exists a bigger set $B$ such that $u_i \in B$, in fact all $u_i$'s are in $B$.
Now, $B \in X\subseteq G$ and hence $B$ is linearly independent so the scalars are zero and hence according to the maximal principle, we have a maximal linearly independent subset containing $S$.
The problem is, where are we assuming that there are only a finite number of elements in $U$ in the $(*)$ statement above ?
We are proving this kind of theorems to be able to extend the results done earlier with finite dimensional vector spaces to infinite dimensional vector spaces, so we cannot assume $V$ to be finite dimensional, can we ?
Also, if we assume that there are only a finite number of elements in $U$, that would mean finite elements in the chain $X$ and that would imply finite elements in the set $S$, which could be false for an infinite dimensional vector space ?
Is that even a valid question ?
Please help !
You should't use enumerations such as $G=\{S_1,S_2,S_3,\dotsc\}$, because the set $G$ could be much more than countable.
Anyway, this is not the main point. A set of vectors is linearly independent if and only if every finite subset thereof is linearly independent.
Recall that a linear combination of vectors in a possibly infinite sets is one where all but a finite number of coefficients is $0$.
Thus you are reduced to prove that any finite subset of $U$ is linearly independent, which is true, because the finite subset is contained in a single element of the chain and each element of the chain is linearly independent by construction.
There's no assumption on the space to be finite-dimensional.