Vector Space over an Infinite Field is not equal to the Union of Proper Subspaces
Theorem
Let $U_1, U_2, \dots, U_n$ be proper subspaces of a vector space $V$ over an infinite field $F$,
then $V$ is not equal to the union of $U_1, U_2, \dots, U_n$.
Proof
Aiming for a contradiction, suppose $V = V_1 \cup \cdots \cup V_n$.
We can assume that $n \geq 2$, and that $m$ is minimal $n$ with this property.
Choose and fix some $y \in V \setminus V_1$.
Let $x \in V_1$.
Since $F$ is infinite field, we can choose a subset $S \subset F$ of size $m+1$.
For each $\alpha \in S$, we can find some
- $\map i \alpha \in \set{1, \cdots, m}$
such that $x+\alpha y \in V_{\map i \alpha}$ because $V = V_1 \cup \cdots \cup V_m$.
The function $i: S \to \set{1, \cdots, m}$ cannot be injective.
So we can find $\alpha \neq \beta$ in $S$ such that $x + \alpha y$ and $x + \beta y$ both lie in $V_{\map i \alpha}$.
But then
- $\ds y = \frac{\paren{x + \alpha y} - \paren{x + \beta y}}{\alpha - \beta} \in V_{\map i \alpha}$.
Since $y \notin V_1$ by assumption, we conclude that $\map i \alpha > 1$.
Then $x = \paren{x + \alpha y} - \alpha y \in V_{\map i \alpha} \subseteq V_2 \cup \cdots \cup V_m$ for every $x \in V_1$.
Hence
- $V_1 \subseteq V_2 \cup \cdots \cup V_m$
This implies that $V = V_2 \cup \cdots \cup V_m$, which contradicts the minimality of $m$.
$\blacksquare$
Source
- 1987: D.J.H. Garling: A Course in Galois Theory: Exercise $1.17$