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In an interview I was asked to solve a question by using Baire Category Theorem (a complete metric space can not be written as union of nowhere dense subsets), the question was:

"Is the vector space $\mathbb{R}^n$ can be written as countable union of its proper subspaces?"

My approach was: first I show that $\mathbb{R}^2$ cannot be written as countable union of straight lines passing through the origin and as lines in $\mathbb{R}^2$ are nowhere dense sets and $\mathbb{R}^2$ is a complete metric space, so I concluded from the Baire Category Theorem, and for higher dimension I claimed and showed that hyperplanes are also nowhere dense set but later I couldn't conclude the final result.

Could you please help me in this regard? Or any other way to solve this one?

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I don't see a question... – Asaf Karagila Apr 20 '12 at 18:22
@AsafKaragila I think the question is how to proceed from results that OP came to – Norbert Apr 20 '12 at 18:25
See this thread and the links inside: – Asaf Karagila Apr 20 '12 at 18:32
up vote 1 down vote accepted


a) It is well known that $\mathbb{R}^n$ is complete.

b) Every proper subspace of $\mathbb{R}^n$ is nowhere dense in $\mathbb{R}^n$. To prove this assume that there exist some proper subspace $V\subset\mathbb{R}^n$ such that is not nowhere dense in $\mathbb{R}^n$. Then there exist some ball $B(x,r)\subset\mathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $x\in V$, (otherwise we can allways make a small shift). This means that for all $y\in B(x,r)$ and $\varepsilon>0$ we can find $v_y\in V$ such that $\|y-v_y\|\leq\varepsilon$. Take arbitrary $z\in\mathbb{R}^n$ and $\varepsilon>0$. Consider vector $y=x+r\|z\|^{-1}z$. Since $\|y-x\|<r$, then $y\in B(x,r)$. Hence we can find $v_y\in V$ such that $\|y-v_y\|\leq \varepsilon \|z\|^{-1}r$. Now consider $v_z=\|z\| r^{-1}(v_y-x)\in V$, then we get $$ \|z-v_z\|=\|z\|r^{-1}\|\|z\|^{-1}rz-(v_y-x)\|=\|z\|r^{-1}\|\|z\|^{-1}rz+x-v_y\|\leq $$ $$ \|z\|r^{-1}\|y-v_y\|\leq\|z\|r^{-1}\varepsilon\|z\|^{-1}r=\varepsilon $$ Thus for each $z\in \mathbb{R}^n$ and $\varepsilon>0$ we found $v_z\in V$ such that $\|z-v_z\|\leq\varepsilon$. This means that $V$ is dense in $\mathbb{R}^n$. Since $V$ is finite dimensional then it is closed. Since $V$ is closed and dense in $\mathbb{R}^n$, then $V=\mathbb{R}^n$. Contradiction, because $V$ is a proper subspace. Hence every proper subspace is nowhere dense in $\mathbb{R}^n$.

c) Now we apply Baire category theorem and get the desired result.

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will you explain 2)? a little? I guess If I chhose any vector then dividing by its norm and multiply by $\epsilon/2$ it can be shrunk into the subspace? is that the way to show that? I mean I can grab any vecotor inside the open $\epsilon$ ball – Un Chien Andalou Apr 20 '12 at 18:27
Even if grab, this doesn't prove anything. To prove 2) you can say the following. If subspace $V$ is dense in some points, then it is dense in the origin. Since $V$ is finite dimensional its closure coincide with $V$. Hence $V$ contians some ball at the origin. Hence $V$ coincides with $\mathbb{R}^n$ – Norbert Apr 20 '12 at 18:44
what do you mean by 'dense at some point'? i dont know the definition of dense at some point – Un Chien Andalou Apr 20 '12 at 18:47
Saying this I meant that closure of $V$ contains some ball at this point. – Norbert Apr 20 '12 at 18:49
@Makuasi A set is nowhere dense if its closure has empty interior. So the closure of a set that is not nowhere dense contains an open set. But a subspace $V$ of $\Bbb R^n$ is closed; so if $V$ is not nowhere dense, then $V$ contains an open ball, and thus $V$ must be all of $\Bbb R^n$. – David Mitra Apr 20 '12 at 19:29

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