Mixing
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Is the set $mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $mathbb{M}_n(mathbb{R}),n ge 2$?
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Is the set $mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $mathbb{M}_n(mathbb{R}),n ge 2$ ?
My attempt : my answer is False
I take $A = begin{bmatrix} 1& n \0&-1 end{bmatrix}$
I know that set of all invertible matrix is dense. You know that my given matrix $A$ in invertible, so I think statement must be false .
Am I right or wrong?
Thank you.
general-topology matrices trace
edited Jan 1 at 13:57
Brahadeesh
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
$endgroup$
add a comment |
$begingroup$
Is the set $mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $mathbb{M}_n(mathbb{R}),n ge 2$ ?
My attempt : my answer is False
I take $A = begin{bmatrix} 1& n \0&-1 end{bmatrix}$
I know that set of all invertible matrix is dense. You know that my given matrix $A$ in invertible, so I think statement must be false .
Am I right or wrong?
Thank you.
general-topology matrices trace
edited Jan 1 at 13:57
Brahadeesh
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
$endgroup$
1
$begingroup$
But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
$endgroup$
– Arthur
Sep 23 '18 at 7:01
1
$begingroup$
@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
$endgroup$
– Chinnapparaj R
Sep 27 '18 at 3:00
$begingroup$
@ChinnapparajR,,thanks u
$endgroup$
– jasmine
Sep 27 '18 at 9:36
add a comment |
$begingroup$
Is the set $mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $mathbb{M}_n(mathbb{R}),n ge 2$ ?
My attempt : my answer is False
I take $A = begin{bmatrix} 1& n \0&-1 end{bmatrix}$
I know that set of all invertible matrix is dense. You know that my given matrix $A$ in invertible, so I think statement must be false .
Am I right or wrong?
Thank you.
general-topology matrices trace
edited Jan 1 at 13:57
Brahadeesh
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
$endgroup$
Is the set $mathcal{A}$ of all matrices whose trace is $0$ nowhere dense in $mathbb{M}_n(mathbb{R}),n ge 2$ ?
My attempt : my answer is False
I take $A = begin{bmatrix} 1& n \0&-1 end{bmatrix}$
I know that set of all invertible matrix is dense. You know that my given matrix $A$ in invertible, so I think statement must be false .
Am I right or wrong?
Thank you.
general-topology matrices trace
general-topology matrices trace
edited Jan 1 at 13:57
Brahadeesh
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
edited Jan 1 at 13:57
Brahadeesh
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
edited Jan 1 at 13:57
Brahadeesh
6,15742361
edited Jan 1 at 13:57
Brahadeesh
6,15742361
edited Jan 1 at 13:57
Brahadeesh
6,15742361
6,15742361
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
asked Sep 23 '18 at 6:58
jasminejasmine
1,656416
1,656416
1
$begingroup$
But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
$endgroup$
– Arthur
Sep 23 '18 at 7:01
1
$begingroup$
@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
$endgroup$
– Chinnapparaj R
Sep 27 '18 at 3:00
$begingroup$
@ChinnapparajR,,thanks u
$endgroup$
– jasmine
Sep 27 '18 at 9:36
add a comment |
1
$begingroup$
But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
$endgroup$
– Arthur
Sep 23 '18 at 7:01
1
$begingroup$
@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
$endgroup$
– Chinnapparaj R
Sep 27 '18 at 3:00
$begingroup$
@ChinnapparajR,,thanks u
$endgroup$
– jasmine
Sep 27 '18 at 9:36
1
1
$begingroup$
But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
$endgroup$
– Arthur
Sep 23 '18 at 7:01
$begingroup$
But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
$endgroup$
– Arthur
Sep 23 '18 at 7:01
1
1
$begingroup$
@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
$endgroup$
– Chinnapparaj R
Sep 27 '18 at 3:00
$begingroup$
@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
$endgroup$
– Chinnapparaj R
Sep 27 '18 at 3:00
$begingroup$
@ChinnapparajR,,thanks u
$endgroup$
– jasmine
Sep 27 '18 at 9:36
$begingroup$
@ChinnapparajR,,thanks u
$endgroup$
– jasmine
Sep 27 '18 at 9:36
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $mathcal{A}$ is closed
So, In order to prove $mathcal{A}$ is nowhere dense in $M_n(Bbb{R})$, we prove $overline{mathcal{A}}^c=mathcal{A}^c=M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$
To prove $M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$, we prove every point of $M_n(Bbb{R})$ is either a point of $M_n(Bbb{R}) setminus mathcal{A}$ or a limit point of $M_n(Bbb{R}) setminus mathcal{A}$.
Suppose $B in M_n(Bbb{R}) setminus mathcal{A}$ ,then we are done! So asume $B notin M_n(Bbb{R}) setminus mathcal{A}$. That is $B in mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(Bbb{R}) setminus mathcal{A}$
take $B=begin{pmatrix}a_{11} & a_{12} &dots&a_{1n} \ a_{21} & a_{22} &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}end{pmatrix}$ with $a_{11}+dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(Bbb{R})setminus mathcal{A}$ $$A_k=begin{pmatrix}a_{11}+1/k & a_{12} &dots&a_{1n} \ a_{21} & a_{22}+1/k &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}+1/kend{pmatrix}$$ Then $A_k rightarrow B$ and so $B$ is a limit point!
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
$endgroup$
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
$begingroup$
@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
$endgroup$
– Brahadeesh
Sep 26 '18 at 16:52
1
$begingroup$
However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
$endgroup$
– Brahadeesh
Sep 27 '18 at 5:43
|
show 12 more comments
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Your set is closed (as it is the inverse image of the closed set ${0}subseteq Bbb R$ under the continuous trace function), and it has empty interior (as any $epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
$endgroup$
add a comment |
$begingroup$
Your set is a finite dimensional (and hence closed) proper subspace of $mathbb{M}_n(mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $mathcal{A}$ is closed
So, In order to prove $mathcal{A}$ is nowhere dense in $M_n(Bbb{R})$, we prove $overline{mathcal{A}}^c=mathcal{A}^c=M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$
To prove $M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$, we prove every point of $M_n(Bbb{R})$ is either a point of $M_n(Bbb{R}) setminus mathcal{A}$ or a limit point of $M_n(Bbb{R}) setminus mathcal{A}$.
Suppose $B in M_n(Bbb{R}) setminus mathcal{A}$ ,then we are done! So asume $B notin M_n(Bbb{R}) setminus mathcal{A}$. That is $B in mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(Bbb{R}) setminus mathcal{A}$
take $B=begin{pmatrix}a_{11} & a_{12} &dots&a_{1n} \ a_{21} & a_{22} &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}end{pmatrix}$ with $a_{11}+dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(Bbb{R})setminus mathcal{A}$ $$A_k=begin{pmatrix}a_{11}+1/k & a_{12} &dots&a_{1n} \ a_{21} & a_{22}+1/k &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}+1/kend{pmatrix}$$ Then $A_k rightarrow B$ and so $B$ is a limit point!
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
$endgroup$
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
$begingroup$
@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
$endgroup$
– Brahadeesh
Sep 26 '18 at 16:52
1
$begingroup$
However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
$endgroup$
– Brahadeesh
Sep 27 '18 at 5:43
|
show 12 more comments
$begingroup$
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $mathcal{A}$ is closed
So, In order to prove $mathcal{A}$ is nowhere dense in $M_n(Bbb{R})$, we prove $overline{mathcal{A}}^c=mathcal{A}^c=M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$
To prove $M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$, we prove every point of $M_n(Bbb{R})$ is either a point of $M_n(Bbb{R}) setminus mathcal{A}$ or a limit point of $M_n(Bbb{R}) setminus mathcal{A}$.
Suppose $B in M_n(Bbb{R}) setminus mathcal{A}$ ,then we are done! So asume $B notin M_n(Bbb{R}) setminus mathcal{A}$. That is $B in mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(Bbb{R}) setminus mathcal{A}$
take $B=begin{pmatrix}a_{11} & a_{12} &dots&a_{1n} \ a_{21} & a_{22} &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}end{pmatrix}$ with $a_{11}+dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(Bbb{R})setminus mathcal{A}$ $$A_k=begin{pmatrix}a_{11}+1/k & a_{12} &dots&a_{1n} \ a_{21} & a_{22}+1/k &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}+1/kend{pmatrix}$$ Then $A_k rightarrow B$ and so $B$ is a limit point!
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
$endgroup$
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
$begingroup$
@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
$endgroup$
– Brahadeesh
Sep 26 '18 at 16:52
1
$begingroup$
However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
$endgroup$
– Brahadeesh
Sep 27 '18 at 5:43
|
show 12 more comments
$begingroup$
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $mathcal{A}$ is closed
So, In order to prove $mathcal{A}$ is nowhere dense in $M_n(Bbb{R})$, we prove $overline{mathcal{A}}^c=mathcal{A}^c=M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$
To prove $M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$, we prove every point of $M_n(Bbb{R})$ is either a point of $M_n(Bbb{R}) setminus mathcal{A}$ or a limit point of $M_n(Bbb{R}) setminus mathcal{A}$.
Suppose $B in M_n(Bbb{R}) setminus mathcal{A}$ ,then we are done! So asume $B notin M_n(Bbb{R}) setminus mathcal{A}$. That is $B in mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(Bbb{R}) setminus mathcal{A}$
take $B=begin{pmatrix}a_{11} & a_{12} &dots&a_{1n} \ a_{21} & a_{22} &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}end{pmatrix}$ with $a_{11}+dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(Bbb{R})setminus mathcal{A}$ $$A_k=begin{pmatrix}a_{11}+1/k & a_{12} &dots&a_{1n} \ a_{21} & a_{22}+1/k &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}+1/kend{pmatrix}$$ Then $A_k rightarrow B$ and so $B$ is a limit point!
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
$endgroup$
Result: If $A$ is a subset of a metric space $(X,d)$, then $A$ is nowhere dense in $X$ if and only if $overline{A}^c$ is dense in $X$
[For a proof, see this]
Note that $mathcal{A}$ is closed
So, In order to prove $mathcal{A}$ is nowhere dense in $M_n(Bbb{R})$, we prove $overline{mathcal{A}}^c=mathcal{A}^c=M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$
To prove $M_n(Bbb{R}) setminus mathcal{A}$ is dense in $M_n(Bbb{R})$, we prove every point of $M_n(Bbb{R})$ is either a point of $M_n(Bbb{R}) setminus mathcal{A}$ or a limit point of $M_n(Bbb{R}) setminus mathcal{A}$.
Suppose $B in M_n(Bbb{R}) setminus mathcal{A}$ ,then we are done! So asume $B notin M_n(Bbb{R}) setminus mathcal{A}$. That is $B in mathcal{A}$. In this case we prove $B$ is a limit point of $M_n(Bbb{R}) setminus mathcal{A}$
take $B=begin{pmatrix}a_{11} & a_{12} &dots&a_{1n} \ a_{21} & a_{22} &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}end{pmatrix}$ with $a_{11}+dots+a_{nn}=0$
Then consider the sequence of elements of $M_n(Bbb{R})setminus mathcal{A}$ $$A_k=begin{pmatrix}a_{11}+1/k & a_{12} &dots&a_{1n} \ a_{21} & a_{22}+1/k &dots&a_{2n} \ vdots \a_{n1} & a_{n2} &dots&a_{nn}+1/kend{pmatrix}$$ Then $A_k rightarrow B$ and so $B$ is a limit point!
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
edited Sep 23 '18 at 13:17
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
answered Sep 23 '18 at 7:05
Chinnapparaj RChinnapparaj R
5,2931828
5,2931828
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
$begingroup$
@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
$endgroup$
– Brahadeesh
Sep 26 '18 at 16:52
1
$begingroup$
However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
$endgroup$
– Brahadeesh
Sep 27 '18 at 5:43
|
show 12 more comments
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
$begingroup$
@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
$endgroup$
– Brahadeesh
Sep 26 '18 at 16:52
1
$begingroup$
However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
$endgroup$
– Brahadeesh
Sep 27 '18 at 5:43
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
$begingroup$
beautiful answer......simple and easy to understand , thanks u @Chinnapparaj
$endgroup$
– jasmine
Sep 23 '18 at 7:12
1
1
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
$begingroup$
You are welcome!
$endgroup$
– Chinnapparaj R
Sep 23 '18 at 7:13
2
2
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
$begingroup$
@CinnapparajR Your approach would also prove that $Bbb Q$ is nowhere dense in $Bbb R$ (take any rational number and add $sqrt2/k$ to it). It therefore cannot be right.
$endgroup$
– Arthur
Sep 23 '18 at 7:26
1
1
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@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
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– Brahadeesh
Sep 26 '18 at 16:52
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@ChinnapparajR Your proof is perfectly alright. There is nothing wrong with it, Arthur is incorrect.
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– Brahadeesh
Sep 26 '18 at 16:52
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However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
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– Brahadeesh
Sep 27 '18 at 5:43
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However, it is true that the answer is correct now, as it stands. There are no errors in the logic; in particular, the proof does not show that $mathbb{Q}$ is nowhere dense in $mathbb{R}$.
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– Brahadeesh
Sep 27 '18 at 5:43
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show 12 more comments
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Your set is closed (as it is the inverse image of the closed set ${0}subseteq Bbb R$ under the continuous trace function), and it has empty interior (as any $epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
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add a comment |
$begingroup$
Your set is closed (as it is the inverse image of the closed set ${0}subseteq Bbb R$ under the continuous trace function), and it has empty interior (as any $epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
$endgroup$
add a comment |
$begingroup$
Your set is closed (as it is the inverse image of the closed set ${0}subseteq Bbb R$ under the continuous trace function), and it has empty interior (as any $epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
$endgroup$
Your set is closed (as it is the inverse image of the closed set ${0}subseteq Bbb R$ under the continuous trace function), and it has empty interior (as any $epsilon$-ball around a matrix with trace $0$ will contain a matrix with non-zero trace). So it is nowhere dense.
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
answered Sep 23 '18 at 7:12
ArthurArthur
112k7107190
112k7107190
add a comment |
add a comment |
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Your set is a finite dimensional (and hence closed) proper subspace of $mathbb{M}_n(mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
Your set is a finite dimensional (and hence closed) proper subspace of $mathbb{M}_n(mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
$endgroup$
add a comment |
$begingroup$
Your set is a finite dimensional (and hence closed) proper subspace of $mathbb{M}_n(mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
$endgroup$
Your set is a finite dimensional (and hence closed) proper subspace of $mathbb{M}_n(mathbb{R})$ so it is nowhere dense because every proper subspace of a normed space has empty interior.
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
answered Jan 1 at 14:04
mechanodroidmechanodroid
27.1k62446
27.1k62446
add a comment |
add a comment |
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But your matrix doesn't represent the set of all invertible matrices. So that doesn't help. Integers are rationals and rationals are dense in the real numbers. That doesn't mean the integers are dense in the reals.
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– Arthur
Sep 23 '18 at 7:01
1
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@jasmine: There are lot of comments under my answer and the user Arthur saying my answer is incorrect. Actually Arthur was misunderstanding my answer. Also many other users are misunderstanding my answer. Thats why they downvote my answer. I also ask my answer is whether right or wrong in this post and peoples are responding 'my answer is correct'. So don't confuse yourself. Kindly read again my answer. It was correct!
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– Chinnapparaj R
Sep 27 '18 at 3:00
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@ChinnapparajR,,thanks u
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– jasmine
Sep 27 '18 at 9:36