Characteristics of a convex set if its boundary is convex
Clash Royale CLAN TAG#URR8PPP
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If $A$ is a convex set in $mathbb R^n$, when is its boundary convex as well?
I think $partial A$ must be either contained in a hypersurface or must equal $mathbb R^n$.
general-topology metric-spaces convex-analysis
asked Sep 10 at 18:36
Selflearner
286214
 |Â
show 11 more comments
up vote
1
down vote
favorite
If $A$ is a convex set in $mathbb R^n$, when is its boundary convex as well?
I think $partial A$ must be either contained in a hypersurface or must equal $mathbb R^n$.
general-topology metric-spaces convex-analysis
asked Sep 10 at 18:36
Selflearner
286214
Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
1
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
2
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22
 |Â
show 11 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $A$ is a convex set in $mathbb R^n$, when is its boundary convex as well?
I think $partial A$ must be either contained in a hypersurface or must equal $mathbb R^n$.
general-topology metric-spaces convex-analysis
asked Sep 10 at 18:36
Selflearner
286214
If $A$ is a convex set in $mathbb R^n$, when is its boundary convex as well?
I think $partial A$ must be either contained in a hypersurface or must equal $mathbb R^n$.
general-topology metric-spaces convex-analysis
general-topology metric-spaces convex-analysis
asked Sep 10 at 18:36
Selflearner
286214
asked Sep 10 at 18:36
Selflearner
286214
edited Sep 10 at 19:41
asked Sep 10 at 18:36
Selflearner
286214
asked Sep 10 at 18:36
Selflearner
286214
asked Sep 10 at 18:36
Selflearner
286214
286214
Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
1
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
2
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22
 |Â
show 11 more comments
Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
1
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
2
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22
Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
1
1
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
2
2
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22
 |Â
show 11 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Here is a proof for $n=2$. If $partial A = emptyset$ or $partial A = x_1$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2inpartial A$ and assume that there is some $x_3inpartial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $Delta$. Then the whole closed triangle is a subset of $partial A$. That means:
$(*)$ Each neighborhood of each point in $Delta$ contains points in $A$ and $A^c$.
In particular, there are infinitely many point of both $A$ and $A^c$ in $Delta$. Now, choose a point $z$ in the interior of $Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $zin A$, which is a contradiction). Choose points $a,bin AcapDelta$ that form another non-degenerated triangle $Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $Delta'$) contains only points in $A^c$, which contradicts $(*)$.
EDIT: This proof can easily be raised to higher dimensions.
answered Sep 10 at 21:48
amsmath
2,685114
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
add a comment |Â
up vote
0
down vote
Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.
answered Sep 10 at 20:18
Selflearner
286214
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here is a proof for $n=2$. If $partial A = emptyset$ or $partial A = x_1$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2inpartial A$ and assume that there is some $x_3inpartial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $Delta$. Then the whole closed triangle is a subset of $partial A$. That means:
$(*)$ Each neighborhood of each point in $Delta$ contains points in $A$ and $A^c$.
In particular, there are infinitely many point of both $A$ and $A^c$ in $Delta$. Now, choose a point $z$ in the interior of $Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $zin A$, which is a contradiction). Choose points $a,bin AcapDelta$ that form another non-degenerated triangle $Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $Delta'$) contains only points in $A^c$, which contradicts $(*)$.
EDIT: This proof can easily be raised to higher dimensions.
answered Sep 10 at 21:48
amsmath
2,685114
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
add a comment |Â
up vote
1
down vote
accepted
Here is a proof for $n=2$. If $partial A = emptyset$ or $partial A = x_1$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2inpartial A$ and assume that there is some $x_3inpartial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $Delta$. Then the whole closed triangle is a subset of $partial A$. That means:
$(*)$ Each neighborhood of each point in $Delta$ contains points in $A$ and $A^c$.
In particular, there are infinitely many point of both $A$ and $A^c$ in $Delta$. Now, choose a point $z$ in the interior of $Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $zin A$, which is a contradiction). Choose points $a,bin AcapDelta$ that form another non-degenerated triangle $Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $Delta'$) contains only points in $A^c$, which contradicts $(*)$.
EDIT: This proof can easily be raised to higher dimensions.
answered Sep 10 at 21:48
amsmath
2,685114
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here is a proof for $n=2$. If $partial A = emptyset$ or $partial A = x_1$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2inpartial A$ and assume that there is some $x_3inpartial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $Delta$. Then the whole closed triangle is a subset of $partial A$. That means:
$(*)$ Each neighborhood of each point in $Delta$ contains points in $A$ and $A^c$.
In particular, there are infinitely many point of both $A$ and $A^c$ in $Delta$. Now, choose a point $z$ in the interior of $Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $zin A$, which is a contradiction). Choose points $a,bin AcapDelta$ that form another non-degenerated triangle $Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $Delta'$) contains only points in $A^c$, which contradicts $(*)$.
EDIT: This proof can easily be raised to higher dimensions.
answered Sep 10 at 21:48
amsmath
2,685114
Here is a proof for $n=2$. If $partial A = emptyset$ or $partial A = x_1$, you are done (although I don't think that a singleton can be the boundary of a convex set in two dimensions). Let $x_1,x_2inpartial A$ and assume that there is some $x_3inpartial A$ such that $x_1,x_2,x_3$ form a non-degenerated triangle $Delta$. Then the whole closed triangle is a subset of $partial A$. That means:
$(*)$ Each neighborhood of each point in $Delta$ contains points in $A$ and $A^c$.
In particular, there are infinitely many point of both $A$ and $A^c$ in $Delta$. Now, choose a point $z$ in the interior of $Delta$ which belongs to $A^c$. Draw any line through $z$. Then it is not possible that there are points from $A$ on this line on both sides from $z$ (since then $zin A$, which is a contradiction). Choose points $a,bin AcapDelta$ that form another non-degenerated triangle $Delta'$ with $z$. Then any line from $z$ to a point on $[a,b]$ contains a point from $A$ on one side, hence on the other side there are only points from $A^c$. Therefore, a whole sector between the lines $[a,z]$ and $[b,z]$ (the one opposite to $Delta'$) contains only points in $A^c$, which contradicts $(*)$.
EDIT: This proof can easily be raised to higher dimensions.
answered Sep 10 at 21:48
amsmath
2,685114
edited Sep 10 at 21:55
answered Sep 10 at 21:48
amsmath
2,685114
answered Sep 10 at 21:48
amsmath
2,685114
answered Sep 10 at 21:48
amsmath
2,685114
2,685114
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
add a comment |Â
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
1
1
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
$A = x_1$ is a convex set with boundary $partial A = x_1$.
– gerw
Sep 11 at 6:42
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
@gerw I guess this is supposed to be a counterexample, but it's not.
– amsmath
Sep 11 at 19:12
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
This is only a counterexample for 'although I don't think that a singleton can be the boundary of a convex set in two dimensions'.
– gerw
Sep 12 at 7:03
@gerw Thank you!
– amsmath
Sep 12 at 23:18
@gerw Thank you!
– amsmath
Sep 12 at 23:18
add a comment |Â
up vote
0
down vote
Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.
answered Sep 10 at 20:18
Selflearner
286214
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
add a comment |Â
up vote
0
down vote
Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.
answered Sep 10 at 20:18
Selflearner
286214
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.
answered Sep 10 at 20:18
Selflearner
286214
Yes my guess is correct. In fact according to my book one can prove that a convex set in Euclidean space contains a ball if it is not contained in an (n-1)-dimensional affine subspace.
answered Sep 10 at 20:18
Selflearner
286214
edited Sep 10 at 20:46
answered Sep 10 at 20:18
Selflearner
286214
answered Sep 10 at 20:18
Selflearner
286214
answered Sep 10 at 20:18
Selflearner
286214
286214
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
add a comment |Â
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
1
1
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
I do not have your book, so I cannot verify the claim. Please add the corresponding references ;)
– daw
Sep 10 at 20:39
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
@daw The topology book that I’m reading is “elementary topology, problem book†by Viro et al.
– Selflearner
Sep 10 at 20:43
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
How does that prove that $partial A$ is contained in a hypersurface? You will have to prove as well that the interior of $partial A$ is empty if $A$ is convex.
– amsmath
Sep 10 at 21:42
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
He has not proved it, just a problem. But I will try to prove it and post it here to check if it is correct
– Selflearner
Sep 10 at 21:44
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
I have given a proof in an answer.
– amsmath
Sep 10 at 22:02
add a comment |Â
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Think about a disk in the x-y-plane in $Bbb R^3$. Its boundary is the disk itself. What exactly would you like to prove?
– amsmath
Sep 10 at 18:39
I want to find out when the boundary of a convex set is convex @amsmath
– Selflearner
Sep 10 at 18:42
1
Usually I don't like to repeat myself. You write "how can I prove it?" My question is: what exactly?
– amsmath
Sep 10 at 18:44
I edited my question @amsmath
– Selflearner
Sep 10 at 18:46
2
I don't know what you mean with $sigma A$. Do you mean $partial A$? BTW, take a halfplane in $Bbb R^2$. It is convex and so is its boundary. But it's not contained in a hypersurface. So your statement is wrong.
– amsmath
Sep 10 at 19:22