TITLE: Given finitely many points in a vector space $V$, is there a basis such that the first coordinate of each point is distinct?
QUESTION [8 upvotes]: Suppose I have some $n$-dimensional vector space $V$ and a finite collection of $m$ distinct points $v_1,\dotsc, v_m\in V$. Is there a basis of $V$ such that the first coordinate of each $v_i$ is distinct?
This obviously fails when the base field is finite, but my intuition over $\mathbb{R}^n$ has convinced me it's true when the base field is infinite. I'm having a hard time proving it though. Any suggestions?

REPLY [3 votes]: Let $V$ be a vector space of dimension $n$ over a finite field.
Lemma: Given a finite set of vectors $F$ there exists $W\leq V$ of dimension $n-1$ with $W\cap F=\varnothing$.
We prove every maximal subspace with $W\cap F=\varnothing$ must have dimension $n-1$.
Suppose not: Let $W$ be a maximal subset with $W\cap F$ and let $U$ be a subspace of dimension $2$ with $W\cap U=\varnothing$.
Notice that for each $u\in U$ we have $f-\alpha u \in W$ for some non-zero scalar $\alpha$ and $f\in F$. Given $f\in F$ let $X_f$ be the set of vectors $u\in U$ such that $f-\alpha u\in W$ for some scalar $\alpha$. We conclude $\bigcup\limits_{f\in F}X_f=U$.
If we define $Y_f$ as the subspace spanned by $X_f$ then it should also be clear that $\bigcup\limits_{f\in F} {Y_f}=U$. This implies that $Y_f=U$ for some $f$ (because a finite union of proper subspaces of a vector space over an infinite field cannot be the entire vector space).
Therefore we can obtain $u_1$ and $u_2$ linearly independent vectors in $X_f$. Notice that we have $f-\alpha u_1\in W$ and $f-\beta u_2\in W$, which implies $\beta u_2-\alpha u_1\in W$, but we also have $\beta u_2-\alpha u_1\in U$. Which implies $\beta u_2-\alpha u_1=0$. A contradiction. Hence the lemma is proved.

Applying the lemma to our particular case is trivial. Let $v_1,v_2\dots v_m$ be the vectors and consider the set $u_1,u_2,\dots u_k$ constisting of all values $v_i-v_j$.
By the lemma there exists a subspace of dimension $n-1$ $W$ such that $u_i\not\in W$ for all $1\leq i \leq k$.
Let $b_2,b_3\dots b_n$ be a basis for $W$.
Then any extension $b_1,b_2\dots b_n$ does the trick.