A vector space over an infinite field is not a finite union of proper subspaces? Show that if $V$ is a vector space over an infinite field $\mathbb{F}$, then $V$ cannot be written as
set-theoretic union of a finite number of proper subspaces.
 A: If your original vector space is finite-dimensional, say dimension $d$, then consider the vectors that lie on the power curve $(1,\alpha,\alpha^2,\ldots,\alpha^{d-1})$ where $\alpha$ is an arbitrary element in your field. Show that all of these vectors are distinct for distinct $\alpha$ in your field and that any $d$ of them generate your entire vector space if your vector space is $d$-dimensional. (Hint: Vandermonde determinant). Then if your space is a union of finitely many subspaces, one of them must contain infinitely many vectors on the power curve so...
A: By contraposition: if $V$ is the set-theoretic union of $n$ proper subspaces $W_i\,$ ($1\le i\le n$), then $\lvert\mkern2mu F\,\rvert\le n-1$.
Proof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u\in W_i,\enspace u\notin \bigcup\limits_{j\neq i}W_j$ and $v\notin W_i$.
Then $(v+Fu)\cap W_i=\varnothing$ and $(v+Fu)\cap W_j\enspace(j\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence
$$\lvert\mkern2mu v+ Fu\, \rvert=\lvert\mkern2mu F\,\rvert\le n-1.$$
Corollary: Avoidance lemma for vector spaces.
Let $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.
Note: There exists a similar (and better known) Avoidance lemma for prime ideals in commutative rings.
A: Assume $V \subset \bigcup_{i=1}^n V_i$.  In the infinite-dimensional situation, reduce to the finite dimensional situation by choosing one vector $v_i$ from each $V\setminus V_i$.  Then intersect all spaces with $\text{span}(v_1,\dots,v_n)$.  The hypothesis are still satisfied (and even $V = \bigcup V_i$ now), but we can assume all involved spaces to be finite-dimensional.
For each $V_i$ choose a linear form (the coefficients are "normal vector") that vanishes on $V_i$.  The product of those linear forms is a degree $n$ polynomial that vanishes on $V$ which is a contradiction $-$ over an infinite field the zero polynomial is the only polynomial that vanishes on all of $V$.
