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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?

Arturo Magidin
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Myshkin
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1 Answers1

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Hints:

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.

Myshkin
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Norbert
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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 – Myshkin Apr 20 '12 at 18:27
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    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 – Myshkin 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
  • Honestly; i do not get this one, and could not relate with the definition of dense set. – Myshkin Apr 20 '12 at 18:50
  • So, do you know definition of dense set? – Norbert Apr 20 '12 at 19:02
  • let X be a metric space, A is said to be dense in X if $\bar{A}=X$ or if U is any open set in X then $U\cap A$ is nonempty – Myshkin Apr 20 '12 at 19:05
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    @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
  • $V$ is closed. If it didn't have empty interior, then there would be a ball $B(x,\varepsilon) \subset V$. Translate the ball to get $B(0,\varepsilon) \subset V$. Multiply by a scalar to get $B(0,\alpha) \subset V$ for any scalar $\alpha$. But $X = \bigcup_\alpha B(0,\alpha) \subset V$. – André Caldas Apr 20 '12 at 19:46
  • @AndréCaldas I proposed the same idea, but in a more detailed manner. – Norbert Apr 20 '12 at 19:53
  • @norbert: In your answer? Or in your first comment? I am suggesting a very simple way to get rid of the "dense" thing. As you stated at the end of (2), $V$ is closed. So you could get rid of the "dense" argument at the very beginning. – André Caldas Apr 20 '12 at 20:07
  • @AndréCaldas in the second comment – Norbert Apr 20 '12 at 21:04