Generator of Vector Space Contains Basis/Infinite Dimensional Case

Theorem
Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $G$ be a generator of $X$.

Then:


 * $G$ contains a basis for $X$.

Proof
If $X = \set { {\mathbf 0}_X}$, then we must have $G \subseteq \set { {\mathbf 0}_X}$, while the only basis for $X$ is $\O$.

So in this case, we have the claim immediately.

Now take $X \ne \set { {\mathbf 0}_X}$.

Let:


 * $\SS = \set {L \subseteq G : L \text { is linearly independent} }$

Since $G$ generates $X$, it contains some $x \ne {\mathbf 0}_X$.

Then we have $\set x \in \SS$, so $\SS \ne \O$.

With view to apply Zorn's Lemma, we show that every non-empty $\subseteq$-chain in $\SS$ has an upper bound.

Let $\CC$ be a non-empty $\subseteq$-chain in $\SS$.

Let:


 * $\ds C = \bigcup \CC$

Let $x_1, x_2, \ldots, x_n \in C$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ such that:


 * $\ds \sum_{k \mathop = 1}^n \alpha_k x_k = 0$

For each $1 \le i \le n$, let $C_i \in \CC$ be such that $x_i \in C_i$.

Since $\CC$ is a $\subseteq$-chain, there exists $1 \le j \le n$ such that:


 * $C_i \subseteq C_j$ for all $1 \le i \le n$.

Then $x_i \in C_j$ for each $1 \le i \le n$.

Since $C_j$ is linearly independent, we have:


 * $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$

So $C$ is linearly independent with $C \subseteq G$.

So $C \in \SS$, while $F \subseteq C$ for all $F \in \CC$.

So $C$ is an upper bound for $\CC$.

So every non-empty $\subseteq$-chain in $\SS$ has an upper bound.

So by Zorn's Lemma, $\SS$ has a $\subseteq$-maximal element $\BB$.

We show that $\BB$ is a basis for $E$.

Let:


 * $V = \map \span \BB$

Suppose that $V \ne X$.

We then have that $V$ cannot contain $G$, otherwise we would have $V = X$, since $G$ generates $X$.

So there exists $y \in G$ such that $y \not \in V$.

From Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set, $\BB \cup \set y$ is linearly independent, contradicting the maximality of $\BB$.

So there exists no such $y$ and we have $V = X$.

Since $\BB$ is a linearly independent generator for $X$, it is a basis.

Since $\BB \in \SS$, we have $\BB \subseteq G$ as desired.