# vector space over an infinite field is not a finite union of proper subspaces

###### Theorem 1.

A vector space^{} $V$ over an infinite field $\mathrm{F}$ cannot be
a finite union of proper subspaces^{} of itself.

###### Proof.

Let $V={V}_{1}\cup {V}_{2}\cup \mathrm{\dots}\cup {V}_{n}$ where each ${V}_{i}$ is a proper subspace of $V$ and $n>1$ is minimal. Because $n$ is minimal, ${V}_{n}\not\subset {V}_{1}\cup {V}_{2}\cup \mathrm{\dots}\cup {V}_{n-1}$.

Let $u\notin {V}_{n}$ and let $v\in {V}_{n}\setminus \left({V}_{1}\cup {V}_{2}\cup \mathrm{\dots}\cup {V}_{n-1}\right)$.

Define $S=\{v+tu:t\in \mathbb{F}\}$. Since $u\notin {V}_{n}$ is not the zero vector and the field $\mathbb{F}$ is infinite, $S$ must be infinite.

Since $S\subset V={V}_{1}\cup {V}_{2}\cup \mathrm{\dots}\cup {V}_{n}$ one of the ${V}_{i}$ must contain infinitely many vectors in $S$.

However, if ${V}_{n}$ were to contain a vector, other than $v$, from $S$ there would exist non-zero $t\in \mathbb{F}$ such that $v+tu\in {V}_{n}$. But then $tu=v+tu-v\in {V}_{n}$ and we would have $u\in {V}_{n}$ contrary to the choice of $u$. Thus ${V}_{n}$ cannot contain infinitely many elements in $S$.

If some $$ contained two distinct vectors in $S$, then there would exist distinct ${t}_{1},{t}_{2}\in \mathbb{F}$ such that $v+{t}_{1}u,v+{t}_{2}u\in {V}_{i}$. But then $\left({t}_{2}-{t}_{1}\right)v={t}_{2}\left(v+{t}_{1}u\right)-{t}_{1}\left(v+{t}_{2}u\right)\in {V}_{i}$ and we would have $v\in {V}_{i}$ contrary to the choice of $v$. Thus for $$ cannot contain infinitely many elements in $S$ either. ∎

Title | vector space over an infinite field is not a finite union of proper subspaces |
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Canonical name | VectorSpaceOverAnInfiniteFieldIsNotAFiniteUnionOfProperSubspaces |

Date of creation | 2013-03-22 17:29:43 |

Last modified on | 2013-03-22 17:29:43 |

Owner | loner (106) |

Last modified by | loner (106) |

Numerical id | 9 |

Author | loner (106) |

Entry type | Theorem |

Classification | msc 15A03 |