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

###### Proof.

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

Let $u\not\in V_{n}$ and let $v\in V_{n}\setminus\left(V_{1}\cup V_{2}\cup\ldots\cup V_{n-1}\right)$.

Define $S=\left\{v+tu:t\in\mathbbmss{F}\right\}$. Since $u\not\in V_{n}$ is not the zero vector and the field $\mathbbmss{F}$ is infinite, $S$ must be infinite.

Since $S\subset V=V_{1}\cup V_{2}\cup\ldots\cup V_{n}$ 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 $t\in\mathbbmss{F}$ 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 $V_{i},1\leq i contained two distinct vectors in $S$, then there would exist distinct $t_{1},t_{2}\in\mathbbmss{F}$ 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 $1\leq i cannot contain infinitely many elements in $S$ either. ∎

Title vector space over an infinite field is not a finite union of proper subspaces VectorSpaceOverAnInfiniteFieldIsNotAFiniteUnionOfProperSubspaces 2013-03-22 17:29:43 2013-03-22 17:29:43 loner (106) loner (106) 9 loner (106) Theorem msc 15A03