Existence of extension of lin. indep. set to a basis A theorem from Greub 1975, Linear Algebra, pg 12-13:


I'm interested in the sentence at 2/3 of the proof:

Then the vectors of $T\cup x$ are linearly dependent because otherwise it would follow that $x\cup T\in \mathscr{A}(R,S)$ which contradict the maximality of $T$.

My observations / questions:


*

*$T\cup x$ is not necessarily linearly dependent, for instance if $x\in T\subset E$ we get an linearly independent union.

*The elements $X$ of $\mathscr{A}(R,S)$ are subsets of $S$, not a set of all linearly independent sets in $E$. What I mean is that being lin. indep. is not enough to be element of $\mathscr{A}(R,S)$. I do not get the argument here.
 A: The objective when picking an arbitrary $x$ is to show that it can be expressed as a linear combination of elements of $T$. If $x \in T$, then doing so is trivial. So, the only case with which we really need to deal is when $x \not\in T$.
Suppose $x \not\in T$ and $T \cup \{x\}$ is linearly independent. Then we have $R \subset T \subset T \cup \{x\}$ and since $T \cup \{x\}$ is LI it follows that $T \cup \{x\} \in \mathscr{A}(R,S)$. But, $T \subsetneq T \cup \{x\}$ which contradicts the maximality of $T$.
Edit: Expanding on my comment below, if $T \cup \{x\}$ is LI, we can write $x = \sum c_j x_j$ for $x_j \in S$. This will allow us to find some $x_k \in S$ so that $T \cup \{x_k\}$ is LI. Thus, reaching a contradiction. So, we may as well assume $x \in S$. 
A: You're right that there is an error. Greub should be letting $x\in S-T$ if $T\ne S$, in which case the rest of the argument then shows that the subspace generated by $T$ contains $S$, and hence contains all of $E$.
This isn't the only error in the proof. Note he's not actually showing that $A$ is a maximal element, as stated, rather that it's an upper bound, which is what's needed for Zorn's lemma to establish the existence of the maximal element $T$.
