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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.

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  • $\begingroup$ Have you tried anything so far? $\endgroup$ – Matthias Klupsch Jan 21 '15 at 11:52
  • $\begingroup$ @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?" $\endgroup$ – 5xum Jan 21 '15 at 11:58
  • $\begingroup$ Start with $n=2$. $\endgroup$ – lhf Jan 21 '15 at 11:59
  • $\begingroup$ Can you sketch what you have been able to do (for example in the special case $n=2$) and tell where you have problems? $\endgroup$ – Joonas Ilmavirta Jan 21 '15 at 12:04
  • $\begingroup$ A hyperplane is not necessarily a subspace, unless it passes through the origin. $\endgroup$ – lhf Jan 21 '15 at 12:07
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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.

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  • $\begingroup$ i am taking a from R* .. $\endgroup$ – user2867280 Jan 21 '15 at 12:55
  • $\begingroup$ as y doesnt belong to W1 so x+ay doesnt belong to W1 $\endgroup$ – user2867280 Jan 21 '15 at 13:05
  • $\begingroup$ 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 $\endgroup$ – user2867280 Jan 21 '15 at 13:09
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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.

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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.

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