# Vector Space has Basis between Linearly Independent Set and Spanning Set

This article needs proofreading.Please check it for mathematical errors.If you believe there are none, please remove `{{Proofread}}` from the code.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Proofread}}` from the code. |

## Theorem

Let $V$ be a vector space over a field $F$.

Let $L$ be a linearly independent subset of $V$.

Let $S$ be a set that spans $V$.

Suppose that $L \subseteq S$.

Then $V$ has a basis $B$ such that $L \subseteq B \subseteq S$.

### Corollary

Let $K$ be a division ring.

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

Then $V$ has a basis.

## Outline of Proof

We use Zorn's Lemma to construct a maximal linearly independent subset.

## Proof

Let $\mathscr I$ be the set of linearly independent subsets of $S$ that contain $L$, ordered by inclusion.

Note that $L \in \mathscr I$, so $\mathscr I \ne \O$.

Let $\mathscr C$ be a nest in $\mathscr I$.

Let $C = \bigcup \mathscr C$.

Aiming for a contradiction, suppose that $C$ is linearly dependent.

Then there exist $v_1, v_2, \ldots, v_n \in C$ and $r_1, r_2, \ldots, r_n \in F$ such that $r_1 \ne 0$:

- $\ds \sum_{k \mathop = 1}^n r_k v_k = 0$

Then there are $C_1, C_2, \ldots, C_n \in \mathscr C$ such that $v_k \in C_k$ for each $k \in \set {1, 2, \ldots, n}$.

Since $\mathscr C$ is a nest, $C_1 \cup C_2 \cup \cdots \cup C_n$ must equal $C_k$ for some $k \in \set {1, 2, \ldots, n}$.

But then $C_k \in \mathscr C$ and $C_k$ is linearly dependent, which is a contradiction.

Thus $C$ is linearly independent.

By Zorn's Lemma, $\mathscr I$ has a maximal element $M$ (one that is not contained in any other element).

Since $M \in \mathscr I$, $M$ is linearly independent.

All that remains is to show that $M$ spans $V$.

Aiming for a contradiction, suppose there exists a $v \in V \setminus \map \span M$.

Then, since $S$ spans $V$, there must be an element $s$ of $S$ such that $s \notin \map \span M$.

Then $M \cup \set s$ is linearly independent.

Thus $M \cup \set s \supsetneq M$, contradicting the maximality of $M$.

Thus $M$ is a linearly independent subset of $V$ that spans $V$.

Therefore, by definition, $M$ is a basis for $V$.

$\blacksquare$

## Also see

- Existence of Vector Space Bases implies Axiom of Choice
- Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, for the finitely generated case

## Axiom of Choice

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.