Finite Unions of Proper Subspaces Over an Infinite Field

A vector space over an infinite field cannot be the union of finitely many proper subspaces.

Proof. Assume for the sake of contradiction that $V$ is a vector space over an infinite field $\mathbb{F}$ and $V = \bigcup_{i=1}^n W_i$ , where $n$ is the smallest positive integer satisfying this property and the $W_i$ are proper subspaces of $V$ .
Since $V$ cannot be a proper subspace of itself, $n$ must be greater than or equal to $2$ . Since $n$ is minimal, $W_n$ is not contained in the union of the remaining subspaces. Thus there exists a vector $u \in W_n$ such that $u \notin W_i$ for all $i < n$ . On the other hand, since $W_n$ is a proper subspace of $V$ , there exists a vector $v \notin W_n$ .
Consider the infinite set of vectors of the form $v + c u$ with $c \in \mathbb{F}$ . Since $V$ is the union of $n$ subspaces and the field $\mathbb{F}$ has infinitely many elements, by the pigeonhole principle there exists a subspace $W_k$ containing at least two distinct vectors from the considered set. Suppose $v + c_1 u \in W_k$ and $v + c_2 u \in W_k$ with $c_1, c_2 \in \mathbb{F}$ and $c_1 \neq c_2$ . Then the difference of these two vectors is $(c_1 - c_2) u \in W_k$ . Since $c_1 \neq c_2$ , the scalar $c_1 - c_2$ is non-zero, hence we have $u \in W_k$ . If $k = n$ , then from $v + c_1 u \in W_n$ and $u \in W_n$ we deduce that $v \in W_n$ , which contradicts the choice of $v$ . If $k < n$ , then $u \in W_k$ , which contradicts the choice of $u$ . All cases lead to a contradiction. Therefore, a vector space over an infinite field cannot be the union of finitely many proper subspaces. $\Box$