# every vector space has a basis

This result, trivial in the finite case, is in fact rather surprising
when one thinks of infinite^{} dimensionial vector spaces^{}, and the
definition of a basis: just try to imagine a basis of the vector space
of all continuous mappings $f:\mathbb{R}\to \mathbb{R}$. The theorem is
equivalent^{} to the axiom of choice^{} family of axioms and theorems. Here
we will only prove that Zorn’s lemma implies that every vector space
has a basis.

###### Theorem.

Let $X$ be any vector space over any field $F$ and assume Zorn’s lemma. Then if $L$ is a linearly independent subset of $X$, there exists a basis of $X$ containing $L$. In particular, $X$ does have a basis at all.

###### Proof.

Let $\mathcal{A}$ be the set of linearly independent subsets of $X$
containing $L$ (in particular, $\mathcal{A}$ is not empty), then $\mathcal{A}$
is partially ordered by inclusion. For each chain $C\subseteq \mathcal{A}$,
define $\widehat{C}=\bigcup C$. Clearly, $\widehat{C}$ is an upper bound of $C$. Next we
show that $\widehat{C}\in \mathcal{A}$. Let $V:=\{{v}_{1},\mathrm{\dots},{v}_{n}\}\subseteq \widehat{C}$
be a finite collection^{} of vectors. Then there exist sets ${C}_{1},\mathrm{\dots},{C}_{n}\in C$ such that ${v}_{i}\in {C}_{i}$ for all $1\le i\le n$. Since $C$ is
a chain, there is a number $k$ with $1\le k\le n$ such that
${C}_{k}=\bigcup _{i=1}^{n}{C}_{i}$ and thus $V\subseteq {C}_{k}$, that is $V$ is
linearly independent^{}. Therefore, $\widehat{C}$ is an element of $\mathcal{A}$.

According to Zorn’s lemma $\mathcal{A}$ has a maximal element^{}, $M$, which
is linearly independent. We show now that $M$ is a basis. Let $\u27e8M\u27e9$
be the span of $M$. Assume there exists an $x\in X\setminus \u27e8M\u27e9$. Let
$\{{x}_{1},\mathrm{\dots},{x}_{n}\}\subseteq M$ be a finite collection of vectors and
${a}_{1},\mathrm{\dots},{a}_{n+1}\in F$ elements such that

$${a}_{1}{x}_{1}+\mathrm{\cdots}+{a}_{n}{x}_{n}-{a}_{n+1}x=0.$$ |

If ${a}_{n+1}$ was necessarily zero, so would be the other ${a}_{i}$, $1\le i\le n$,
making $\{x\}\cup M$ linearly independent in contradiction^{} to the
maximality of $M$. If ${a}_{n+1}\ne 0$, we would have

$$x=\frac{{a}_{1}}{{a}_{n+1}}{x}_{1}+\mathrm{\cdots}+\frac{{a}_{n}}{{a}_{n+1}}{x}_{n},$$ |

contradicting $x\notin \u27e8M\u27e9$. Thus such an $x$ does not exist and $X=\u27e8M\u27e9$, so $M$ is a generating set and hence a basis.

Taking $L=\mathrm{\varnothing}$, we see that $X$ does have a basis at all. ∎

Title | every vector space has a basis |

Canonical name | EveryVectorSpaceHasABasis |

Date of creation | 2013-03-22 13:04:48 |

Last modified on | 2013-03-22 13:04:48 |

Owner | GrafZahl (9234) |

Last modified by | GrafZahl (9234) |

Numerical id | 14 |

Author | GrafZahl (9234) |

Entry type | Theorem |

Classification | msc 15A03 |

Synonym | every vector space has a Hamel basis^{} |

Related topic | ZornsLemma |

Related topic | AxiomOfChoice |

Related topic | ZermelosWellOrderingTheorem |

Related topic | HaudorffsMaximumPrinciple |

Related topic | KuratowskisLemma |