Linearly Independent Set is Contained in some Basis/Infinite Dimensional Case

Theorem
Let $K$ be a field.

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

Let $H$ be a linearly independent subset of $E$.

There exists a basis $B$ for $E$ such that $H \subseteq B$.

Proof
Let:


 * $\SS = \set {L \supseteq H : L \subseteq E \text { is linearly independent} }$

We have $H \in \SS$, so certainly $\SS \ne \O$.

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 $H \subseteq \CC$.

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

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 $E$.

From definition 2 of a basis, it suffices to argue that $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$.

Suppose that $\BB'$ is a linearly independent subset of $G$ such that $\BB \subseteq \BB'$.

Then since $H \subseteq \BB$, we have $H \subseteq \BB'$.

So $\BB' \in \SS$.

So by the maximality of $\BB$ in $\SS$, we have $\BB = \BB'$.

So $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$, and hence is a basis.