**homotopy theory, (∞,1)-category theory, homotopy type theory**

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**Theorems**

The **basis theorem** for vector spaces states that every vector space $V$ admits a basis, or in other words is a free module over its ground field of scalars. It is a famous classical consequence of the axiom of choice (and is equivalent to it by Blass (1984).

If $V$ is a vector space over any field (or, say, a left vector space over a skewfield) $K$, then $V$ has a basis.

We apply Zorn's lemma as follows: consider the poset consisting of the linearly independent subsets of $V$, ordered by inclusion (so $S \leq S'$ if and only if $S \subseteq S'$). If $(S_\alpha)$ is a chain in the poset, then

$S = \bigcup_\alpha S_\alpha$

is an upper bound, for each $v$ in the span of $S$ belongs to some $S_\alpha$ and is uniquely expressible as a finite linear combination of elements in $S_\alpha$ and in any $S_\beta$ containing $S_\alpha$, hence uniquely expressible as a finite linear combination of elements in $S$. Thus the hypothesis of Zorn’s lemma obtains for this poset; therefore this poset has a maximal element, say $B$.

Let $W$ be the span of $B$. If $W$ were a proper subspace of $V$, then for any $v$ in the set-theoretic complement of $W$, $B \cup \{v\}$ is a linearly independent set (for if

$a_0 v + a_1 w_1 + \ldots + a_n w_n = 0 \qquad (w_1, \ldots, w_n \in B)$

we must have $a_0 = 0$ – else we can multiply by $1/a_0$ and express $v$ as a linear combination of the $w_i$, contradicting $\neg(v \in W)$ – and then the remaining $a_i$ are 0 since the $w_i$ are linearly independent). This contradicts the maximality of $B$. We therefore conclude that $W = V$, and $B$ is a basis for $V$.

I’ll write out a proof of the converse, that the axiom of choice follows from the basis theorem, as soon as I’ve digested it – Todd.

Any two bases of a vector space are of the same cardinality.

Given any linearly independent set $A$ and spanning set $C$, if $A \subseteq C$, then there is a basis $B$ with $A \subseteq B \subseteq C$; the theorem above is the special case where $A = \empty$ and $C = V$. The proof of this more general theorem is a straightforward generalisation of the proof above.

The original proof of the existence of Hamel bases (after which the concept was named) was for the case of the real numbers regarded as a rational vector space and used (not directly Zorn's lemma but) the well-ordering theorem:

- Georg Hamel, pp. 460 in:
*Eine Basis aller Zahlen und die unstetigen Lösungen der Funktionalgleichung: $f(x + y) = f(x) + f(y)$*, Mathematische Annalen**60**(1905) 459–462 [doi:10.1007/BF01457624]

Lecture notes:

- Christoph Schweigert,
*Basis und Dimension*, §2.4 in:*Lineare Algebra*, lecture notes, Hamburg (2022) [pdf]

See also:

- Andreas Blass,
*Existence of bases implies the axiom of choice*, Contemporary Mathematics**31**(1984) 31-33 [doi:10.1090/conm/031 pdf]

Last revised on November 11, 2023 at 19:41:19. See the history of this page for a list of all contributions to it.