Prove that $\mathbb{R}^n$ cannot be a finite union of its hyperplanes . Prove that $\mathbb{R}^n$ cannot be a finite union of its hyperplanes . I want a prove using linear algebra only and not functional analysis
i tried by contradiction
we know R^n is a vector space over R.
let R^n = U Wi (i from 1-k) Wi's are hyperplanes so are proper subspaces,
let x belongs to W1. and take y belongs to R^n-W1.
so there are infinitely many x+ay for a belongs to R.
x+ay doesnt belong to W1
as R^n=U Wi so x+ay belongs to some Wj, j not equal to 1.
so Wj contains x and y.
so W1 is a subset of U Wi (i from 2 to k)
now applying induction we get
R^n=Wk which is a contradiction as Wk is a proper subspace.
but my prof says there are gaps in the proof which i am not able to find. 
 A: We prove a more general case:
Let $\mathbb{F}$ be an infinite field,then $\mathbb{F}^{n}$ cannot be the union of finitely many hyperplanes.
Indeed,suppose the finitely many hyperplanes are given by equations:
$$\sum_{i=1}^{n} a_{i}^{(k)} x_{i} = b^{(k)} , k=1,2,...,N$$
where,for fixed $k$,$\left(a_{i}^{(k)}\right)_{i}$ are not all zeros.
Choose $(x_{i})_{i=1}^{n}$ be $(t^{i})_{i=1}^{n},t \in \mathbb{F}$，
Since a polynomial has only finitely many roots, and $\mathbb{F}$ is infinite,there must be a t$\in \mathbb{F}$ s.t 
$$\sum_{i=1}^{n} a_{i}^{(k)} t^{i} \neq b^{(k)} , \forall k=1,2,...,N$$
and we are done.
A: I expect that by hyperplane you mean codimension 1. Then each hyperplane $P_j$ has an associated normal vector $v_j$.
Pick two points in $P_j$ to determine a vector $w_j\neq v_j$. Among the infinitely many linear combinations of the $w_j$ you can find a vector $w$ different from $\{v_1,\ldots,v_m\}$.
Call $L$ the hyperplane normal to $w$. Its intersection with any $P_j$ is codimension 1 in $L$, so you reduced the problem to the corresponding statement in dimension $n-1$. When the dimension of the ambient space is 1, the statement is true because $\mathbb{R}$ is infinite.
A: The gap in your proof comes where you write $x+ a y \not\in W_1$, hence $x + ay \in W_j$ for some $j  = 2, \ldots, n$, and hence conclude that $W_1$ is contained
in $W_2 \cup \cdot \cup W_n$.
First: you don't explain why $x + a y \not\in W_1$.  In fact, it's not always true; if you set $a = 0$ then $x \in W_1$ (by your very choice of $x$).  If $a \neq 0$,then
you are right that $x + ay \not\in W_1 (can you say why?).  
Second: you don't explain how you conclude that $W_1 \subseteq W_2 \cup \cdot \cup W_n$.  My guess is that you are applying your first conclusion (that $x + ay \not\in W_1$) in the case when $a = 0$; but as I already noted, in this case it's not true.
Finally: I think that you should be able to take your idea of looking at the line $x + ay$ and develop it into a complete proof along the lines you tried, but you need to get the first step correct, and then develop a correct version of the second step.
