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 maximal element $\BB$.

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

Let:


 * $G = \map \span \BB$

Suppose that $G \ne E$.

Let $x \in E \setminus G$.

We argue that $\BB \cup \set x$ is linearly independent.

Take $x_1, \ldots, x_n \in \BB$ and take $\alpha_1, \ldots, \alpha_n, \alpha_{n + 1} \in K$ such that:


 * $\ds \alpha_{n + 1} x + \sum_{k \mathop = 1}^n \alpha_i x_i = 0$

If $\alpha_{n + 1} = 0$, then we have:


 * $\ds \sum_{k \mathop = 1}^n \alpha_i x_i = 0$

and so $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$ from the linearly independence of $\BB$.

Suppose that $\alpha_{n + 1} \ne 0$, then we would have:


 * $\ds x = -\alpha^{-1}_{n + 1} \sum_{k \mathop = 1}^n \alpha x_i$

This would mean that $x \in G$, (from Linear Span is Linear Subspace) contradicting that $x \not \in G$.

So we have $\alpha_{n + 1} = 0$, and hence $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$.

So $\BB \cup \set x$ is linearly independent, while:


 * $H \subseteq \BB \subseteq \BB \cup \set x$

So:


 * $\BB \cup \set x \in \SS$

Since $\BB$ is a maximal element of $\SS$, we have that:


 * $\BB \cup \set x = \BB$

That is, $x \in \BB$, so that $x \in G$.

This is contradicts that $x \in E \setminus G$.

So we must have $G = E$.

So $\BB$ is a linearly independent subset of $E$ that generates $E$.

So $\BB$ is a basis of $E$ that contains $H$.