countable union of proper subspaces












1












$begingroup$


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?






linear-algebra general-topology analysis metric-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't see a question...
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:22






  • 1




    $begingroup$
    @AsafKaragila I think the question is how to proceed from results that OP came to
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:25












  • $begingroup$
    See this thread and the links inside: math.stackexchange.com/questions/10760
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:32
















1












$begingroup$


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?






linear-algebra general-topology analysis metric-spaces







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don't see a question...
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:22






  • 1




    $begingroup$
    @AsafKaragila I think the question is how to proceed from results that OP came to
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:25












  • $begingroup$
    See this thread and the links inside: math.stackexchange.com/questions/10760
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:32














1












1








1





$begingroup$


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?






linear-algebra general-topology analysis metric-spaces







share|cite|improve this question











$endgroup$




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?






linear-algebra general-topology analysis metric-spaces




linear-algebra general-topology analysis metric-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 20 '12 at 19:58









Arturo Magidin

266k34590920




266k34590920










asked Apr 20 '12 at 18:16









Ding DongDing Dong

17.4k1060183




17.4k1060183








  • 1




    $begingroup$
    I don't see a question...
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:22






  • 1




    $begingroup$
    @AsafKaragila I think the question is how to proceed from results that OP came to
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:25












  • $begingroup$
    See this thread and the links inside: math.stackexchange.com/questions/10760
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:32














  • 1




    $begingroup$
    I don't see a question...
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:22






  • 1




    $begingroup$
    @AsafKaragila I think the question is how to proceed from results that OP came to
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:25












  • $begingroup$
    See this thread and the links inside: math.stackexchange.com/questions/10760
    $endgroup$
    – Asaf Karagila♦
    Apr 20 '12 at 18:32








1




1




$begingroup$
I don't see a question...
$endgroup$
– Asaf Karagila♦
Apr 20 '12 at 18:22




$begingroup$
I don't see a question...
$endgroup$
– Asaf Karagila♦
Apr 20 '12 at 18:22




1




1




$begingroup$
@AsafKaragila I think the question is how to proceed from results that OP came to
$endgroup$
– Norbert
Apr 20 '12 at 18:25






$begingroup$
@AsafKaragila I think the question is how to proceed from results that OP came to
$endgroup$
– Norbert
Apr 20 '12 at 18:25














$begingroup$
See this thread and the links inside: math.stackexchange.com/questions/10760
$endgroup$
– Asaf Karagila♦
Apr 20 '12 at 18:32




$begingroup$
See this thread and the links inside: math.stackexchange.com/questions/10760
$endgroup$
– Asaf Karagila♦
Apr 20 '12 at 18:32










1 Answer
1






active

oldest

votes


















1












$begingroup$

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 $Vsubsetmathbb{R}^n$ such that is not nowhere dense in $mathbb{R}^n$. Then there exist some ball $B(x,r)subsetmathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $xin V$, (otherwise we can allways make a small shift). This means that for all $yin B(x,r)$ and $varepsilon>0$ we can find $v_yin V$ such that $|y-v_y|leqvarepsilon$. Take arbitrary $zinmathbb{R}^n$ and $varepsilon>0$. Consider vector $y=x+r|z|^{-1}z$. Since $|y-x|<r$, then $yin B(x,r)$. Hence we can find $v_yin V$ 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 $zin mathbb{R}^n$ and $varepsilon>0$ we found $v_zin V$ such that $|z-v_z|leqvarepsilon$. 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:27








  • 1




    $begingroup$
    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 $mathbb{R}^n$
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:44












  • $begingroup$
    what do you mean by 'dense at some point'? i dont know the definition of dense at some point
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:47










  • $begingroup$
    Saying this I meant that closure of $V$ contains some ball at this point.
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:49






  • 1




    $begingroup$
    @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 $Bbb R^n$.
    $endgroup$
    – David Mitra
    Apr 20 '12 at 19:29












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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 $Vsubsetmathbb{R}^n$ such that is not nowhere dense in $mathbb{R}^n$. Then there exist some ball $B(x,r)subsetmathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $xin V$, (otherwise we can allways make a small shift). This means that for all $yin B(x,r)$ and $varepsilon>0$ we can find $v_yin V$ such that $|y-v_y|leqvarepsilon$. Take arbitrary $zinmathbb{R}^n$ and $varepsilon>0$. Consider vector $y=x+r|z|^{-1}z$. Since $|y-x|<r$, then $yin B(x,r)$. Hence we can find $v_yin V$ 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 $zin mathbb{R}^n$ and $varepsilon>0$ we found $v_zin V$ such that $|z-v_z|leqvarepsilon$. 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:27








  • 1




    $begingroup$
    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 $mathbb{R}^n$
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:44












  • $begingroup$
    what do you mean by 'dense at some point'? i dont know the definition of dense at some point
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:47










  • $begingroup$
    Saying this I meant that closure of $V$ contains some ball at this point.
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:49






  • 1




    $begingroup$
    @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 $Bbb R^n$.
    $endgroup$
    – David Mitra
    Apr 20 '12 at 19:29
















1












$begingroup$

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 $Vsubsetmathbb{R}^n$ such that is not nowhere dense in $mathbb{R}^n$. Then there exist some ball $B(x,r)subsetmathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $xin V$, (otherwise we can allways make a small shift). This means that for all $yin B(x,r)$ and $varepsilon>0$ we can find $v_yin V$ such that $|y-v_y|leqvarepsilon$. Take arbitrary $zinmathbb{R}^n$ and $varepsilon>0$. Consider vector $y=x+r|z|^{-1}z$. Since $|y-x|<r$, then $yin B(x,r)$. Hence we can find $v_yin V$ 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 $zin mathbb{R}^n$ and $varepsilon>0$ we found $v_zin V$ such that $|z-v_z|leqvarepsilon$. 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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:27








  • 1




    $begingroup$
    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 $mathbb{R}^n$
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:44












  • $begingroup$
    what do you mean by 'dense at some point'? i dont know the definition of dense at some point
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:47










  • $begingroup$
    Saying this I meant that closure of $V$ contains some ball at this point.
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:49






  • 1




    $begingroup$
    @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 $Bbb R^n$.
    $endgroup$
    – David Mitra
    Apr 20 '12 at 19:29














1












1








1





$begingroup$

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 $Vsubsetmathbb{R}^n$ such that is not nowhere dense in $mathbb{R}^n$. Then there exist some ball $B(x,r)subsetmathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $xin V$, (otherwise we can allways make a small shift). This means that for all $yin B(x,r)$ and $varepsilon>0$ we can find $v_yin V$ such that $|y-v_y|leqvarepsilon$. Take arbitrary $zinmathbb{R}^n$ and $varepsilon>0$. Consider vector $y=x+r|z|^{-1}z$. Since $|y-x|<r$, then $yin B(x,r)$. Hence we can find $v_yin V$ 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 $zin mathbb{R}^n$ and $varepsilon>0$ we found $v_zin V$ such that $|z-v_z|leqvarepsilon$. 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.






share|cite|improve this answer











$endgroup$



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 $Vsubsetmathbb{R}^n$ such that is not nowhere dense in $mathbb{R}^n$. Then there exist some ball $B(x,r)subsetmathbb{R}^n$ such that $V$ is dense in $B(x,r)$. Without loss of generality we can assume that $xin V$, (otherwise we can allways make a small shift). This means that for all $yin B(x,r)$ and $varepsilon>0$ we can find $v_yin V$ such that $|y-v_y|leqvarepsilon$. Take arbitrary $zinmathbb{R}^n$ and $varepsilon>0$. Consider vector $y=x+r|z|^{-1}z$. Since $|y-x|<r$, then $yin B(x,r)$. Hence we can find $v_yin V$ 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 $zin mathbb{R}^n$ and $varepsilon>0$ we found $v_zin V$ such that $|z-v_z|leqvarepsilon$. 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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 11 '12 at 22:38









Ding Dong

17.4k1060183




17.4k1060183










answered Apr 20 '12 at 18:25









NorbertNorbert

46k774162




46k774162












  • $begingroup$
    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
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:27








  • 1




    $begingroup$
    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 $mathbb{R}^n$
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:44












  • $begingroup$
    what do you mean by 'dense at some point'? i dont know the definition of dense at some point
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:47










  • $begingroup$
    Saying this I meant that closure of $V$ contains some ball at this point.
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:49






  • 1




    $begingroup$
    @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 $Bbb R^n$.
    $endgroup$
    – David Mitra
    Apr 20 '12 at 19:29


















  • $begingroup$
    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
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:27








  • 1




    $begingroup$
    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 $mathbb{R}^n$
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:44












  • $begingroup$
    what do you mean by 'dense at some point'? i dont know the definition of dense at some point
    $endgroup$
    – Ding Dong
    Apr 20 '12 at 18:47










  • $begingroup$
    Saying this I meant that closure of $V$ contains some ball at this point.
    $endgroup$
    – Norbert
    Apr 20 '12 at 18:49






  • 1




    $begingroup$
    @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 $Bbb R^n$.
    $endgroup$
    – David Mitra
    Apr 20 '12 at 19:29
















$begingroup$
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
$endgroup$
– Ding Dong
Apr 20 '12 at 18:27






$begingroup$
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
$endgroup$
– Ding Dong
Apr 20 '12 at 18:27






1




1




$begingroup$
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 $mathbb{R}^n$
$endgroup$
– Norbert
Apr 20 '12 at 18:44






$begingroup$
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 $mathbb{R}^n$
$endgroup$
– Norbert
Apr 20 '12 at 18:44














$begingroup$
what do you mean by 'dense at some point'? i dont know the definition of dense at some point
$endgroup$
– Ding Dong
Apr 20 '12 at 18:47




$begingroup$
what do you mean by 'dense at some point'? i dont know the definition of dense at some point
$endgroup$
– Ding Dong
Apr 20 '12 at 18:47












$begingroup$
Saying this I meant that closure of $V$ contains some ball at this point.
$endgroup$
– Norbert
Apr 20 '12 at 18:49




$begingroup$
Saying this I meant that closure of $V$ contains some ball at this point.
$endgroup$
– Norbert
Apr 20 '12 at 18:49




1




1




$begingroup$
@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 $Bbb R^n$.
$endgroup$
– David Mitra
Apr 20 '12 at 19:29




$begingroup$
@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 $Bbb R^n$.
$endgroup$
– David Mitra
Apr 20 '12 at 19:29


















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