countable union of proper subspaces
$begingroup$
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
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
$endgroup$
add a comment |
$begingroup$
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
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
$endgroup$
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
add a comment |
$begingroup$
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
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
$endgroup$
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
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
edited Apr 20 '12 at 19:58
Arturo Magidin
266k34590920
266k34590920
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
asked Apr 20 '12 at 18:16
Ding DongDing Dong
17.4k1060183
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
$endgroup$
$begingroup$
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
|
show 7 more comments
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
$endgroup$
$begingroup$
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
|
show 7 more comments
$begingroup$
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.
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
$endgroup$
$begingroup$
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
|
show 7 more comments
$begingroup$
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.
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
$endgroup$
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.
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
edited Aug 11 '12 at 22:38
Ding Dong
17.4k1060183
17.4k1060183
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
answered Apr 20 '12 at 18:25
NorbertNorbert
46k774162
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
|
show 7 more comments
$begingroup$
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
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– Ding Dong
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 $mathbb{R}^n$
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– Norbert
Apr 20 '12 at 18:44
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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 $mathbb{R}^n$
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– Norbert
Apr 20 '12 at 18:44
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what do you mean by 'dense at some point'? i dont know the definition of dense at some point
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– Ding Dong
Apr 20 '12 at 18:47
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what do you mean by 'dense at some point'? i dont know the definition of dense at some point
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– Ding Dong
Apr 20 '12 at 18:47
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Saying this I meant that closure of $V$ contains some ball at this point.
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– Norbert
Apr 20 '12 at 18:49
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Saying this I meant that closure of $V$ contains some ball at this point.
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– Norbert
Apr 20 '12 at 18:49
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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 $Bbb R^n$.
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– David Mitra
Apr 20 '12 at 19:29
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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 $Bbb R^n$.
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– David Mitra
Apr 20 '12 at 19:29
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I don't see a question...
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– Asaf Karagila♦
Apr 20 '12 at 18:22
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@AsafKaragila I think the question is how to proceed from results that OP came to
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– Norbert
Apr 20 '12 at 18:25
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See this thread and the links inside: math.stackexchange.com/questions/10760
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– Asaf Karagila♦
Apr 20 '12 at 18:32