Let the poset $P$ be the set of all LI subsets of a vector space. Why any union $\bigcup T$ of totally ordered LI subsets T is still LI? Let the poset $P$ involved be the set of all linearly independent (L.I.) subsets of a vector space, ordered by inclusion, i.e. $A\leq B$ iff $A \subseteq B$.
In proving "every vector space has a basis" using Zorn's lemma, one usual step is to claim that: If a set of linearly independent subsets is linearly ordered by inclusion, then their union is also linearly independent.
But I cannot see why this is true...
My intuitive example is as follows:
Let {$e_1$} $\subseteq$ {$e_1$,$e_2$} be the first totally ordered subset.
Let {$5e_2$} $\subseteq$ {$5e_2$,$e_3$} be the second totally ordered subset.
We can choose $e_1$,$e_2$,$e_3$ such that {$e_1$,$e_2$} and {5$e_2$,$e_3$} are both
linearly independent (L.I.) subsets, but their union
{$e_1$,$e_2$,$5e_2$,$e_3$} is not L.I.
This seems to contradict the fact that: Any union $\bigcup T$ of totally ordered L.I. subsets T is still L.I., and thus $\bigcup T \in P$.
Can anyone tell me where my thinking goes wrong? Thank you.
 A: Your attempt at an example fails because it is not linearly ordered: you have the set $\{e_1,e_2\}$, and the set $\{5e_2,e_3\}$. Neither set contains the other. So this collection is not linearly ordered.
Suppose you have a linearly ordered family $\{S_i\}_{i\in I}$; that means that:

*

*Each $S_i$ is linearly independent; and

*For each $i,j\in I$, either $S_i\subseteq S_j$ or $S_j\subseteq S_i$.

Let $S=\cup_{i\in I}S_i$ be the union. If the union were linearly dependent, then there would exist a finite collection of vectors $s_1,\ldots,s_n\in S$, and scalars $\alpha_1,\ldots,\alpha_n$,m all nonzero, such that
$$\alpha_1s_1+\cdots +\alpha_ns_n=0.$$
Because $S$ is a union, each $s_k$ must be in some $S_i$. That is, there exist $i_1,\ldots,i_n\in S$ such that $s_k\in S_{i_k}$, $1\leq k\leq n$.
It is now easy to see that the linear ordering of the family means that there is a $k_0$ such that $S_{i_k}\subseteq S_{i_{k_0}}$ for all $k$, $1\leq k\leq n$. But that means that all the vectors $s_1,\ldots,s_n$ are in $S_{i_{k_0}}$, and therefore that $S_{i_{k_0}}$ is linearly dependent (since the same linear combination still equals zero, but now we see that all the vectors are in $S_{i_{k_0}}$. This contradicts our assumption that all the $S_i$ were linearly independent and that the family was linearly ordered.
Where did the contradiction come from? It came from the unwarranted assumption that $S$ was linearly dependent. Thus, we conclude that in fact $S$ is linearly independent, as claimed.
This is used to verify the hypotheses of Zorn's Lemma: consider the collection $\mathcal{C}$ of all linearly independent subsets. This is a partially ordered set. Given a chain $\{S_i\}_{i\in I}$ in $\mathcal{C}$, $\cup S_i$ is also a linearly independent subset (as noted above), and thus lies in $\mathcal{C}$. In addition, it is an upper bound for the chain. Thus every chain in $\mathcal{C}$ has an upper bound in $\mathcal{C}$.
Note that none of this requires the Axiom of Choice. You don't use the Axiom of Choice until you use Zorn's Lemma to conclude there is a maximal such linearly independent subset.
