TITLE: Prove that Every Vector Space Has a Basis
QUESTION [18 upvotes]: My textbook extended the following proof to show that every vector space, including the infinite-dimensional case, has a basis.

Condition:  $S$ is a linearly independent subset of a vector space $V$. 
  Theorem:  There is a maximal linearly independent subset of $V$ that contains $S$.

Proof.  Let $F$ be the family of all linearly independent subsets of $V$ that contains $S$.  If $C$ is a chain in $F$ and there exists a member $U \in F$ that contains each member of $C$, by the maximal principle, $U$ is the maximal element of $F$, the family of all linearly independent subsets of $V$.  As a result, $U$ is the maximal linearly independent subset of $V$.  So $U$ is the basis of $V$. 
Let $U$ be the union of the elements of $C$.  Clearly $U$ contains each element of $C$.  To show that $U$ is a linearly independent subset of $V$, first note that $S \subset U$.  Let $u_1, u_2, \ldots, u_n$ be vectors in $U$ and $c_1, c_2 ... c_n$ be scalars such that $0 = \sum_{i=1}^{n} {c_i}{u_i}$.  Because $u_i \in U$ for all $i$, there exist sets $A_i$ in $C$ such that $u_i \in A_i$.  Since $C$ is a chain, there is one set, say $A_k$, that contains all the others.  So $u_1, u_2,\ldots, u_n \in A_k$.  But since $A_k$ is linearly independent, $c_i = 0$ for all $i$.  Therefore, $U$ is linearly independent.  By the maximal principle, $U$ is the maximal element of $F$. $\square$

My questions are as follows:

Is the author arguing that since each vector space has a basis, the infinite-dimensional vector space also has a basis?  This is similar to saying that $\lim_{n \rightarrow \infty} a_n = 0$ if $a_n = 0$ for all $n$.
How come the author is checking one chain $C \in F$ only?  I thought that the maximal principle requires that the maximal element contains all members of each chains.
I am still not sure of how $u_1, u_2, \ldots, u_n$ are picked out.  The greatest number of vectors in a linearly independent subset cannot exceed $\dim(V)$, but that is assuming that $V$ has a basis.  So, how does the author know what the finite number $n$ is?
When the author is assigning $u_i$ to $A_i$, the set $\{A_i\}$ is not yet a chain.  But the $A_i$ can be rearranged to form a chain.  For example, $u_1 \in B_1$, $u_1, u_2 \in B_2,\ldots,$ and $u_1, u_2,\ldots, u_n \in B_n$.

REPLY [4 votes]: Here are some answers to your questions:

This is not like saying that if the limit is $0$ then all the elements of the sequences are zero. This is like saying that if all the natural numbers are rational numbers then in particular $0$ is a rational number.
That is, we say that if all vector spaces have a basis, then infinitely dimensional vector spaces (which are in fact vector spaces) have a basis.

The author doesn't check for just one $C\subseteq F$ (not $\in$, by the way). We don't specify which $C$ we took; $C$ is arbitrary. Therefore this, if true, holds for all the chains in $F$.
In the spirit of previous analogues, if we pick an arbitrary natural number $n$, and we show that $n$ is rational, then we may conclude that every natural number is rational.

No. We don't use a basis. We use the fact that if a set of vector is not linearly independent, then there is a finite subset which exhibits that. This is because by their definition linear combinations are finite combinations, so if there is a non-trivial linear combination which gives out $0$ it only involves a finite number of vectors.

Now, since we only have a finite number of vectors, and they all appear somewhere in our chain $C$, there is some $A_i\in C$ which contains them all. This is because $C$ is a chain, and $U$ is the union of all the sets in that chain. Therefore if $u_i\in U$ there is some $A_i\in C$ such that $u_i\in A_i$. Since we only have a finite number of relevant vectors, we can pick $A_i$ for each $u_i$.
The collection $\{A_i\mid i=1,\ldots,n\}\subseteq C$. You should prove, in case you don't see it already, that a subset of a chain is a chain; and that a finite chain has a maximal element. Therefore we have some $A\in C$ such that $u_1,\ldots,u_n\in A$.