Every Vector Space has a Base

# Every Vector Space has a Base

Recall that if $E$ is a vector space, then a base for $E$ is a subset $A$ of $E$ that is linearly independent and spans $A$. We will now show that every vector space possesses a base.

 Definition: Let $E$ be a vector space. A Chain for $E$, is a collection of subsets of $E$, call it $\mathcal C$, such that if $E_1, E_2 \in \mathcal C$ then either $E_1 \subseteq E_2$ or $E_1 \supseteq E_2$.
 The Maximal Axiom: Let $E$ be a vector space. If $\mathcal S$ is a collection of subsets of $E$ and if $\mathcal C$ is a chain for $E$, then there exists a maximal chain, $\mathcal M$, with $\mathcal C \subseteq \mathcal M \subseteq \mathcal S$.

By "maximal" chain, we mean that if $\mathcal M'$ is any other chain contained in $\mathcal S$, then $\mathcal M' \subseteq \mathcal M$.

It should also be remarked that the maximal axiom defined above, is an axiom that does not depend on $E$ being a vector space. We have only stated that way because in our applications, $E$ will be a vector space.

 Theorem 1: Let $E$ be a vector space. If $L \subset E$ is a linearly independent subset of vectors, and $S \subseteq E$ is a spanning set of vectors, then there exists a base $B$ of $E$ with $L \subset B \subset S$.
• Proof: Let $\mathcal L$ be the collection of all linearly independent sets in $E$, and consider the (trivial) chain $\{ L \} \subset \mathcal L$. By the maximal axiom, there exists a maximal chain, $\mathcal M$, with:
(1)
\begin{align} \{ L \} \subseteq \mathcal M \subseteq \mathcal L \end{align}
• Let $B = \bigcup_{M \in \mathcal M} M$. Then, since $\{ L \} \subseteq \mathcal M$, we have that $L \subseteq B$. Furthermore, since each $M$ is linearly independent and since $S$ spans $E$, we have that $M \subseteq S$ for each $M \in \mathcal M$, and consequently, $B \subseteq S$.
• We now show that $B$ is a base for $E$.
• First, observe that $B$ spans $E$. If instead, there exists $x \in E \setminus \mathrm{span}(B)$, then consider the set $B \cup \{ x \}$ is linearly independent, and so $B \cup \{ x \} \in \mathcal M$. But this is impossible unless $x \in B$, by the maximality of $\mathcal M$.
• Now observe that $B$ is linearly independent. For each $n \in \mathbb{N}$, and $x_1, x_2, ..., x_n \in B$, consider the equation:
(2)
\begin{align} \lambda_1x_1 + \lambda_2x_2 + ... + \lambda_nx_n = o \end{align}
• For each $1 \leq i \leq n$, there exists an $M_i \in \mathcal M$ with $x_i \in M_i$. So that $\lambda_1x_1 + \lambda_2x_2 + ... + \lambda_nx_n \in \bigcup_{i=1}^{n} M_i$. But since $\mathcal M$ is a chain, there exists a $1 \leq k \leq n$]] such that $\bigcup_{i=1}^{n} M_i = M_k$. But since $M_k$ is linearly independent, we have that $\lambda_1 = \lambda_2 = ... = \lambda_n = 0$.
• Thus $B$ is a base for $E$. $\blacksquare$
 Corollary 2: Let $E$ be a vector space. Then $E$ has a base.
• Proof: The subset $L := \emptyset$ is (trivially) linearly independent in $E$, and $S := E$ (trivially) spans $E$. So by Theorem 1, there exists a set $B$ with $\emptyset \subset B \subset E$ that is a base for $E$. $\blacksquare$