Let $V$ be a vector space (can be finite or infinite) over finite field $K$, such that $\dim V > 1$ and $|K| = q < \infty$. Prove there exist proper subspaces $V_0, \dots, V_q$ such that $V = V_0 \cup \dots \cup V_q$.
I have no idea where to start from.
Hint: If $(x_1,x_2,\ldots)\in V$ then either $x_1=0$ or there exists $c\in K$ with $x_2=cx_1$.
We claim that if $V$ is a vector space over a finite field $K$ of order $q$ such that $\dim V>1$, and $m$ is a non-negative integer, then $V$ can be written as a union of $m$ proper subspaces of $V$ if and only if $m\geq q+1$. (From the proof below, it also follows that if $K$ is not finite, then there is no way to cover a vector space $V$ over $K$ with $\dim V>1$ by finitely many proper subspaces.)
First suppose that $m\geq q+1$. It suffices to assume that $m=q+1$. Pick a basis $\mathcal{B}$ of $V$. Let $a,b\in\mathcal{B}$ be two distinct elements (noting that $|\mathcal{B}|>1$ since $\dim V>1$). For each $k\in K$, we define $V_k$ to be the span of $\{a+kb\}\cup\big(\mathcal{B}\setminus\{a,b\}\big)$, and $U$ is the span of $\mathcal{B}\setminus\{a\}$. Show that $V=U\cup \bigcup_{k\in K}V_k$.
Conversely, suppose that $V$ can be written as a union of $m$ proper subspaces $W_1,W_2,\ldots,W_m$ with $m$ being smallest possible (from the previous paragraph we know $m$ exists, so taking the smallest one is possible). It is easy to see that $m>1$. By minimality of $m$, for any $i$, we have $W_i\not\subseteq \bigcup_{j\neq i}W_j$.
Take $u\in W_1\setminus\bigcup_{j\neq1}W_j$ and $v\in W_2\setminus\bigcup_{j\neq 2}W_j$. Since $u+sv\in V$ for all $s\in K$ such that $s\neq 0$, we must have $u+sv\in W_j$ for some $j$. We claim that the assignment $s\in K\setminus\{0\}$ to the smallest $j$ such that $u+sv\in W_j$ is an injective function from $K\setminus\{0\}$ to $\{3,4,\ldots,m\}$. From here, it follows that $$q-1=\big|K\setminus\{0\}\big|\leq \big|\{3,4,\ldots,m\}\big|=m-2,$$ establishing our claim.
Now, to prove the assertion in the previous paragraph, we first note that $u+sv\notin W_1$ and $u+sv\notin W_2$ for $s\ne 0$. If $u+sv\in W_1$, then $v=s^{-1}\big((u+sv)-u\big)\in W_1$ since $u\in W_1$, which is a contradiction. If $u+sv\in W_2$, then $u=(u+sv)-sv\in W_2$ since $v\in W_2$, which is also a contradiction. So, $u+sv\in W_j$ for some $j\in\{3,4,\ldots,m\}$.
Now, suppose that there are two non-zero $s,t\in K$ such that $u+sv$ and $u+tv$ are in the same $W_i$, where $i\in\{3,4,\ldots,m\}$. Then, $$v=(s-t)^{-1}\big((u+sv)-(u+tv)\big)\in W_i.$$ But $v\in W_2\setminus \bigcup_{j\neq 2}W_j$, so we have another contradiction. The assertion is now proven.
Let $V$ be a vector space (can be finite or infinite) over finite field $K$, such that $\dim V > 1$ and $|K| = q < \infty$. Prove there exist proper subspaces $V_0, \dots, V_q$ such that $V = V_0 \cup \dots \cup V_q$.
Inspired by discussion here on my Quora space, I'll present a method to construct these $V_i$'s.
Let us label the elements of the field as $k_1$, $k_2$, $\ldots$, $k_q$ where $k_1$ is the zero element and other $k_i$'s are non-zero.
Let $W$ be a subspace of $V$ of codimension $2$ i.e., $\dim(V/W)=2$. We note that $|V/W| = q^2$.
$\exists \,v_0,v_1,\ldots,v_q\in V$ s.t.
$\begin{array}{ll}V/W =& \big\{ W,\\ & \quad k_2v_0+W& k_3v_0+W,&\ldots,& k_qv_0+W,\\ & \quad k_2v_1+W,& k_3v_1+W,&\ldots,& k_qv_1+W,\\ &\quad\quad\vdots\\& \quad k_2v_q+W, &k_3v_q+W,&\ldots, & k_qv_q+W\big\}\end{array} $
Choosing $V_i=\text{span}(v_i)\, \oplus \, W$, we are done.
It is interesting to note that if we put some restrictions on the type of subspaces we consider, then we can recover the "Baire-like properties" ("whole space can't be written as (finite in this case) union of 'small' subspaces") of proper subspaces of a VS over infinite fields, or infinite-codim subspaces of a VS over any field.
Thm: for any finite dimensional subspace $V \subseteq \mathbb F^I := \{(x_i)_{i\in I} : x_i\in \mathbb F\}$ (where $\mathbb F$ is any field), $V$ contains a vector with $\geq \dim V$ non-zero entries.
I.e. $V$ is not covered by any union of axes-parallel subspaces of strictly smaller dimension, where axes-parallel subspaces $A_I$ (picture it in $\mathbb R^3$ for why I call it "axes parallel") are $A_I:= \{x\in \mathbb F^I: i\in I \implies x_i=0\}$.
This is used to prove that the slice rank of 3D diagonal tensors is $\geq$ the number of non-zero diagonal entries (see e.g. Lemma 2.2.9 in this PDF), which was in turn used in a breakthrough result on 3-term arithmetic progressions in $\mathbb F_3^n$.
The proof of Thm is not difficult. Suppose that all $v\in V$ have support $S_v:= \{i\in I: v_i\neq 0\}$ of size $<\dim V$. Choose $v^*$ so that $|S_{v^*}|$ is maximal. We aim to get a contradiction to this maximality, by perturbing $v$ (staying in $V$) so that entries $v_i$ for $i\in S_v$ stay non-zero, but some entry outside of $S_v$ goes from zero to non-zero.
(General problem solving tip: we all know about the power of induction in finite settings; this is basically equivalent to the power of being able to get the maximum/minimum of a set of integers bounded above/below resp.)
Consider the axes-parallel subspace $A_{S_{v^*}}$. It's codimension in $\mathbb F^I$ is $|S_{v^*}|<\dim V$. So, we must have that $V$ and $A_{S_{v^*}}$ intersect at some non-zero vector $w \in A_{S_{v^*}} \cap V$ ("witness" of the non-trivial intersection).
But then $v^*+w \in V$ has strictly larger support than $v^*$ ($w$ being zero on $S_{v^*}$ means adding it to $v^*$ it can't change those non-zero values; and $w$ being non-zero means it changes at least one entry outside of $S_{v^*}$ to be non-zero); contradiction.
Summary: vector spaces over finite fields can generally be covered by finitely many subspaces of strictly smaller dimension (as the other answers to this MSE post show), but can NOT be covered by any union of axes-parallel subspaces of strictly smaller dimension (which I call a "Baire-like property"). And this Baire-like property can be used to prove an amazing result in additive combinatorics.
