# Zorn’s lemma and bases for vector spaces

In this entry, we illustrate how Zorn’s lemma can be applied in proving the existence of a basis for a vector space^{}. Let $V$ be a vector space over a field $k$.

###### Proposition 1.

Every linearly independent subset of $V$ can be extended to a basis for $V$.

This has already been proved in this entry (http://planetmath.org/EveryVectorSpaceHasABasis). We reprove it here for completion.

###### Proof.

Let $A$ be a linearly independent subset of $V$. Let $\mathcal{S}$ be the collection^{} of all linearly independent^{} supersets^{} of $A$. First, $\mathcal{S}$ is non-empty since $A\in \mathcal{S}$. In addition^{}, if ${A}_{1}\subseteq {A}_{2}\subseteq \mathrm{\cdots}$ is a chain of linearly independent supersets of $A$, then their union is again a linearly independent superset of $A$ (for a proof of this, see here (http://planetmath.org/PropertiesOfLinearIndependence)). So by Zorn’s Lemma, $\mathcal{S}$ has a maximal element^{} $B$. Let $W=\mathrm{span}(B)$. If $W\ne V$, pick $b\in V-W$. If $0=rb+{r}_{1}{b}_{1}+\mathrm{\cdots}+{r}_{n}{b}_{n}$, where ${b}_{i}\in B$, then $-rb={r}_{1}{b}_{1}+\mathrm{\cdots}+{r}_{n}{b}_{n}$, so that $-rb\in \mathrm{span}(B)=W$. But $b\notin W$, so $b\ne 0$, which implies $r=0$. Consequently ${r}_{1}=\mathrm{\cdots}={r}_{n}=0$ since $B$ is linearly independent. As a result, $B\cup \{b\}$ is a linearly independent superset of $B$ in $\mathcal{S}$, contradicting the maximality of $B$ in $\mathcal{S}$.
∎

###### Proposition 2.

Every spanning set of $V$ has a subset that is a basis for $V$.

###### Proof.

Let $A$ be a spanning set of $V$. Let $\mathcal{S}$ be the collection of all linearly independent subsets of $A$. $\mathcal{S}$ is non-empty as $\mathrm{\varnothing}\in \mathcal{S}$. Let ${A}_{1}\subseteq {A}_{2}\subseteq \mathrm{\cdots}$ be a chain of linearly independent subsets of $A$. Then the union of these sets is again a linearly independent subset of $A$. Therefore, by Zorn’s lemma, $\mathcal{S}$ has a maximal element $B$. In other words, $B$ is a linearly independent subset $A$. Let $W=\mathrm{span}(B)$. Suppose $W\ne V$. Since $A$ spans $V$, there is an element $b\in A$ not in $W$ (for otherwise the span of $A$ must lie in $W$, which would imply $W=V$). Then, using the same argument as in the previous proposition^{}, $B\cup \{b\}$ is linearly independent, which contradicts the maximality of $B$ in $\mathcal{S}$. Therefore, $B$ spans $V$ and thus a basis for $V$.
∎

###### Corollary 1.

Every vector space has a basis.

###### Proof.

Either take $\mathrm{\varnothing}$ to be the linearly independent subset of $V$ and apply proposition 1, or take $V$ to be the spanning subset of $V$ and apply proposition 2. ∎

Remark. The two propositions above can be combined into one: If $A\subseteq C$ are two subsets of a vector space $V$ such that $A$ is linearly independent and $C$ spans $V$, then there exists a basis $B$ for $V$, with $A\subseteq B\subseteq C$. The proof again relies on Zorn’s Lemma and is left to the reader to try.

Title | Zorn’s lemma and bases for vector spaces |
---|---|

Canonical name | ZornsLemmaAndBasesForVectorSpaces |

Date of creation | 2013-03-22 18:06:49 |

Last modified on | 2013-03-22 18:06:49 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 9 |

Author | CWoo (3771) |

Entry type | Result |

Classification | msc 16D40 |

Classification | msc 13C05 |

Classification | msc 15A03 |