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:
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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 $n$.
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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.
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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?
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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 $u_1, u_2,\ldots, u_n \in B_n$.
Best Answer
If you quoted that accurately, it's a bit of a mess. The maximal principle actually says that there is a maximal chain $C$ in $F$, and that's the one that the author proceeds to consider. (There may of course be more than one, but any one of them will serve.) If $U\in F$ contains each member of $C$, then $C\cup\{U\}$ is a chain in $F$, and clearly $C\subseteq C\cup\{U\}$, so by the maximality of $C$ we must have $C=C\cup\{U\}$ and hence $U\in C$. Clearly, then, there cannot be any $M\in F$ such that $U\subsetneqq M$, since then $C\cup\{M\}$ would be a strictly larger chain than the maximal chain $C$. Thus, $U$ is a maximal element of $F$; in general it is not the maximal element of $F$, because in general $F$ has no unique maximal element.
The rest of the paragraph is better: it's simply a proof that if we set $U=\bigcup C$, then $U\in F$. Since obviously this $U$ contains each member of $C$, that establishes (by the argument above) that $U$ is a maximal element of $F$. Proving that $U\in F$ merely requires establishing that $U$ is linearly independent. To do this, we must show that every finite subset of $U$ is linearly independent: that's the definition of linear independence for infinite sets. The author's $\{u_1,\dots,u_n\}$ is simply an arbitrary finite subset of $U$. Then each $u_i$ belongs to some member $A_i$ of the chain $C$. Because $C$ is a chain, we know that these sets $A_1,\dots,A_n$ are nested, and there is no harm in assuming that we've numbered them so that $A_1\subseteq A_2\subseteq\ldots\subseteq A_n$. Then $\{u_1,\dots,u_n\}\subseteq A_n$, and $A_n$ is a linearly independent set, so $\{u_1,\dots,u_n\}$ is also linearly independent, as desired.