Theorem 4, Section 4.2 of Hungerford’s Algebra

Question

Every vector space $V$ over a division ring $D$ has a basis and is therefore a free $D$-module. More generally every linearly independent subset of $V$ is contained in a basis of $V$.

Sketch of proof: The first statement is an immediate consequence of the second since the null set is a linearly independent subset of every vector space. Consequently, we assume that $X$ is any linearly independent subset of $V$ and let $\mathcal{S}$ be the set of all linearly independent subsets of $V$ that contain $X$. Since $X\in S$, $S\neq \emptyset$. Partially order $\mathcal{S}$ by set theoretic inclusion. If $\{C_i \mid i\in I\}$ is a chain in $\mathcal{S}$ verify that the set $C = \bigcup_{i\in I} C_i$ is linearly independent and hence an element of $\mathcal{S}$. Clearly $C$ is an upper bound for the chain $\{C_i\mid i\in I\}$. By Zorn's Lemma $\mathcal{S}$ contains a maximal element $B$ that contains $X$ and is necessarily a maximal linearly independent subset of $V$. By Lemma 2.3 $B$ is a basis of $V$.

I am trying to show $C=\bigcup_{i\in I}C_i$ is linearly independent. Let $\{x_1,…,x_n\}$ be finite subset of $C$. Then $x_j\in C_{i_j}$ for all $1\leq j\leq n$. We need to show $r_1x_1+…+r_nx_n=0$$\implies$$r_i=0$ for all $i$. Since $\{C_i\mid i\in I\}$ is a chain, we have $C_{i_1}\subseteq C_{i_2}$ or $C_{i_2}\subseteq C_{i_1}$. Similarly, $C_{i_2}\subseteq C_{i_3}$ or $C_{i_3}\subseteq C_{i_2}$, and so on. How to progress from here?

I have a intuition $\exists C_{i_p}$ ($1\leq p\leq n$) such that $C_{i_j}\subseteq C_{i_p}$ for all $j$. So $\{x_1,…,x_n\}\subseteq C_{i_p}$ and by linear independence of $C_{i_p}$ we have $r_i=0$ for all $i$.

Answer 1

Ahh, how I missed induction.

Claim: If $\{x_1,…,x_j\}\subseteq C$ with $x_q\in C_{i_q}$, then $\exists 1\leq p\leq j$ such that $C_{i_1},…,C_{i_j}\subseteq C_{i_p}$.

Proof: We use induction on $j$. Base case: $j=2$. Let $\{x_1,x_2\}\subseteq C$ with $x_1\in C_{i_1}$ and $x_2\in C_{i_2}$. Since $\{C_i\mid i\in I\}$ is a chain, $C_{i_1}\subseteq C_{i_2}$ or $C_{i_2}\subseteq C_{i_1}$. In either case, $C_{i_1},C_{i_2}\subseteq C_{i_p}$ for some $1\leq p\leq 2$. Inductive step: Suppose $\{x_1,…,x_j\}\subseteq C$ with $x_q\in C_{i_q}$$\implies$$\exists 1\leq p\leq j$ such that $C_{i_1},…,C_{i_j}\subseteq C_{i_p}$. Let $\{x_1,…,x_j,x_{j+1}\}\subseteq C$ with $x_q\in C_{i_q}$. By inductive hypothesis, $\exists 1\leq p\leq j$ such that $C_{i_1},…,C_{i_j}\subseteq C_{i_p}$. Since $\{C_i\mid i\in I\}$ is a chain, $C_{i_{j+1}}\subseteq C_{i_p}$ or $C_{i_p}\subseteq C_{i_{j+1}}$. In either case, $C_{i_1},…,C_{i_{j+1}}\subseteq C_{i_s}$ for some $1\leq s\leq j+1$. By mathematical induction, claim holds.

Put $j=n$. Thus $\exists 1\leq p\leq n$ such that $C_{i_1},…,C_{i_n}\subseteq C_{i_p}$. So $\{x_1,…,x_n\}\subseteq C_{i_p}$ and by linear independence of $C_{i_p}$ we have $r_i=0$ for all $i$.