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

Theorem

Let $K$ be a field.

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

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

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:

$\ds C = \bigcup \CC$

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 $\CC$.

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

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 $\BB = \BB'$.

$\blacksquare$

This theorem depends on the Axiom of Choice, by way of Zorn's Lemma.

Because of some of its bewilderingly paradoxical implications, the Axiom of Choice is considered in some mathematical circles to be controversial.

Most mathematicians are convinced of its truth and insist that it should nowadays be generally accepted.

However, others consider its implications so counter-intuitive and nonsensical that they adopt the philosophical position that it cannot be true.