# Prove that $\mathbb{R}^n$ cannot be a finite union of its hyperplanes . [duplicate]

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

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.

• Have you tried anything so far? – Matthias Klupsch Jan 21 '15 at 11:52
• @user2867280 The point of asking "have you tried anything" is really more like "what did you try, how far did you get and whera are you stuck?" – 5xum Jan 21 '15 at 11:58
• Start with $n=2$. – lhf Jan 21 '15 at 11:59
• Can you sketch what you have been able to do (for example in the special case $n=2$) and tell where you have problems? – Joonas Ilmavirta Jan 21 '15 at 12:04
• A hyperplane is not necessarily a subspace, unless it passes through the origin. – lhf Jan 21 '15 at 12:07

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$and develop it into a complete proof along the lines you tried, but you need to get the first step correct, and then develop a correct version of the second step. • i am taking a from R* .. – user2867280 Jan 21 '15 at 12:55 • as y doesnt belong to W1 so x+ay doesnt belong to W1 – user2867280 Jan 21 '15 at 13:05 • because x was arbitrary and x+ay to some Wj and y should be in that Wj if not then again x+ay wont be. so -ay also belongs to Wj thus x belongs to Wj. thus W1 is a subset of Wj so W1 is in the union – user2867280 Jan 21 '15 at 13:09 We prove a more general case: Let$\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. 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.