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 $U_1, U_2, \dots, U_n$.

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 $m$.

$\blacksquare$