# Abstract Nonsense

## Every Vector Space Has a Basis

Point of post: In this post we will give a careful proof that every vector space does, in fact, have a basis. This is mainly so that when I reference this fact in further posts, I can link back to this one, for the interested reader.

Theorem: Let $\mathscr{V}$ be a vector space (or more generally and $R$-module over a division ring)  and $L\subseteq\mathscr{V}$ any linearly independent subset of $\mathscr{V}$. Then, there exists a basis for $\mathscr{V}$ containing $L$.

Proof: Let

$\text{ }$

$\mathscr{P}=\left\{S\subseteq\mathscr{V}:L\subseteq S\text{ and }S\text{ is linearly independent}\right\}$

$\text{ }$

Then, $\left(\mathscr{P},\subseteq\right)$ is a partially ordered set. So, let $\mathcal{C}$ be a chain in $\left(\mathscr{P},\subseteq\right)$. Clearly, if $\displaystyle \bigcup_{C\in\mathcal{C}}C\in\mathcal{P}$ we have that $\displaystyle \bigcup_{C\in\mathcal{C}}C$ is an upper bound for $\mathcal{C}$. So, to see that $\displaystyle \bigcup_{C\in\mathcal{C}}C$ is, in fact, in $\mathscr{P}$ it suffices to prove that it is linearly independent. So, let $\displaystyle x_1,\cdots,x_n\in\bigcup_{C\in\mathcal{C}}C$. Then, there exists $C_1,\cdots,C_n\in\mathcal{C}$ such that $x_j\in C_j,\text{ }j=1,\cdots,n$. But, since $\mathcal{C}$ is a chain we may conclude that $C_1\cup\cdots\cup C_n=C_{s}$ for some $s\in\{1,\cdots,n\}$. But, this means that $x_1,\cdots,x_n\in C_s$ and thus $\{x_1,\cdots,x_n\}$ is a subset of a linearly independent set (namely $C_s$) and so it itself is linearly independent. It follows that any finite subset of $\displaystyle \bigcup_{C\in\mathcal{C}}C$ is linearly independent, and thus by definition $\displaystyle \bigcup_{C\in\mathcal{C}}C$ is, itself, linearly independent.

$\text{ }$

From this we see that every chain in $\left(\mathscr{P},\subseteq\right)$ has an upper bound, and so by Zorn’s Lemma, we are able to conclude that $\left(\mathscr{P},\subseteq\right)$ has some maximal element, call it $\mathfrak{M}$. Now, since $\mathfrak{M}$ is linearly independent (since it’ in $\mathscr{P}$) to prove it’s a basis it suffices to prove that $\text{span }\mathfrak{M}=\mathscr{V}$. To do this, suppose to the contrary that there is some $v\in\mathscr{V}-\text{span }\mathfrak{M}$. We claim that $\mathfrak{M}\cup\{v\}$ is in $\mathscr{P}$. Since $L\subseteq\mathfrak{M}$ it suffices to prove that $\mathfrak{M}\cup\{v\}$ is linearly independent. To see this let $\{x_1,\cdots,x_n\}$ be any finite subset of $\mathfrak{M}\cup\{v\}$. If $v\notin\{x_1,\cdots,x_n\}$ then linear independence of $\{x_1,\cdots,x_n\}$ follows immediately. If not, then $v=x_j$ for some $j\in\{1,\cdots,n\}$, assume without loss of generality that $v=x_1$. Then, if

$\text{ }$

$\alpha_1 v+\alpha_2x_2+\cdots+\alpha_nx_n=\bold{0}$

$\text{ }$

we must have that $\alpha_1=0$ otherwise

$\text{ }$

$\displaystyle v=-\alpha^{-1}\alpha_2x_1+\cdots+-\alpha_1^{-1}\alpha_nx_n$

$\text{ }$

contradicting that $v\notin\text{span }\mathscr{M}$. Thus, $\alpha_1=0$ and so the above equation may be rewritten as

$\text{ }$

$\alpha_2x_2+\cdots+\alpha_nx_n=\bold{0}$

$\text{ }$

but since $\{x_2,\cdots,x_n\}\subseteq\mathfrak{M}$ is linearly independent it follows that $\alpha_1=\cdots=\alpha_n=0$. Thus, since the finite subset was arbitrary it follows that $\mathfrak{M}\cup\{v\}$ is linearly independent, and so $\mathfrak{M}\cup\{v\}\in\mathscr{P}$. But, this contradicts the maximality of $\mathfrak{M}$ since $\mathfrak{M}\cup\{v\}\not\subseteq\mathfrak{M}$. Thus, the assumption that $v\in\mathscr{V}-\text{span }\mathfrak{M}$ must have been incorrect, in other words $\mathscr{V}-\text{span }\mathfrak{M}=\varnothing$. But, this is only true if $\mathscr{V}\subseteq\text{span }\mathfrak{M}$, but since $\text{span }\mathfrak{M}\subseteq\mathscr{V}$ we may conclude that $\mathscr{V}=\text{span }\mathfrak{M}$. Thus, $\mathfrak{M}$ is a basis. $\blacksquare$

October 19, 2010 -

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