TITLE: A question on vector space over an infinite field
QUESTION [11 upvotes]: Can a vector space over an infinite field be a finite union of proper subspaces ?

REPLY [3 votes]: Denote by $K$ the field over which $V$ is a vector space.
Suppose the answer is true and consider write $V=\bigcup\limits_{i=1}^{n} V_i$, where $V_i$ are proper subspaces of $V$ and $n \in \mathbb{N}$ is minimal.
Because $n$ is minimal, there exists an element $v_n \in V_n \setminus \bigcup\limits_{i=1}^{n-1}V_i $. There exists also an element $v \in \bigcup\limits_{i=1}^{n-1}V_i  \setminus V_n$. 
Now if we look at the infinite set $M=\{ v_n+k \cdot v : k \in K \}$, using Pigeonhole Principle we can see that there exists $i \in \{ 1, \dots, n\}$ such that $|M \cap V_i|\geq 2$ (we can actually say that the cardinal is infinite, but we don't need that for the rest of the proof).
If $V_n \cap M$ contains some element other than $v_n$, then there exists $k \in K \setminus \{ 0 \}$ such that $k \cdot v \in V_n$, which contradicts the choice of $v$.
If $V_i \cap M$ has more than two elements for some $i \in \{1,\dots, n-1\}$ then, looking at their difference, we obtain that $v_n \in V_i$ which contradicts the choice of $v_n$.
Therefore, $V$ is never a finite union of proper subspaces.
Edit: Repetition was not intended, I was posting while BlackAdder posted.