Mixing
If $0<rneq 1$ and $U$ is convex then $partial Ucappartial (rU)=emptyset$
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Let $UsubseteqmathbbR^n$ be convex and closed. I'm trying to find conditions for which the following happens:
If $0<rneq 1$ then $partial Ucappartial (rU)=emptyset$.
It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=(0,y):yinmathbbR$ or $U=(x,y):yge $. In fact, in both examples $U=rU$.
So, let us suppose that $0intextInt (U)$. Does this suffice to prove that $partial Ucappartial (rU)=emptyset$? I couldn't find counterexamples so far.
What do you think?
Thank you.
real-analysis convex-analysis
asked Jul 23 at 20:15
Tanius
46427
add a comment |Â
up vote
2
down vote
favorite
Let $UsubseteqmathbbR^n$ be convex and closed. I'm trying to find conditions for which the following happens:
If $0<rneq 1$ then $partial Ucappartial (rU)=emptyset$.
It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=(0,y):yinmathbbR$ or $U=(x,y):yge $. In fact, in both examples $U=rU$.
So, let us suppose that $0intextInt (U)$. Does this suffice to prove that $partial Ucappartial (rU)=emptyset$? I couldn't find counterexamples so far.
What do you think?
Thank you.
real-analysis convex-analysis
asked Jul 23 at 20:15
Tanius
46427
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
@Math1000 yes I do
– Tanius
Jul 23 at 22:44
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $UsubseteqmathbbR^n$ be convex and closed. I'm trying to find conditions for which the following happens:
If $0<rneq 1$ then $partial Ucappartial (rU)=emptyset$.
It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=(0,y):yinmathbbR$ or $U=(x,y):yge $. In fact, in both examples $U=rU$.
So, let us suppose that $0intextInt (U)$. Does this suffice to prove that $partial Ucappartial (rU)=emptyset$? I couldn't find counterexamples so far.
What do you think?
Thank you.
real-analysis convex-analysis
asked Jul 23 at 20:15
Tanius
46427
Let $UsubseteqmathbbR^n$ be convex and closed. I'm trying to find conditions for which the following happens:
If $0<rneq 1$ then $partial Ucappartial (rU)=emptyset$.
It seems to me that this is not true if $0$ is in the frontier of $U$, for instance $U=(0,y):yinmathbbR$ or $U=(x,y):yge $. In fact, in both examples $U=rU$.
So, let us suppose that $0intextInt (U)$. Does this suffice to prove that $partial Ucappartial (rU)=emptyset$? I couldn't find counterexamples so far.
What do you think?
Thank you.
real-analysis convex-analysis
asked Jul 23 at 20:15
Tanius
46427
edited Jul 23 at 20:22
asked Jul 23 at 20:15
Tanius
46427
asked Jul 23 at 20:15
Tanius
46427
asked Jul 23 at 20:15
Tanius
46427
46427
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
@Math1000 yes I do
– Tanius
Jul 23 at 22:44
add a comment |Â
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
@Math1000 yes I do
– Tanius
Jul 23 at 22:44
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
@Math1000 yes I do
– Tanius
Jul 23 at 22:44
@Math1000 yes I do
– Tanius
Jul 23 at 22:44
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = frac1r$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $xin partial (rU)$ such that $xinpartial U$. Since $xin partial (rU)$, $y = r^-1xin partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0intextInt(U)$ there is an open neighborhood of $0$ also in $textInt(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $Isubset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $yinpartial U$. Thus, if $0intextInt(U)$ then the intersection must be empty.
If $0notintextInt(U)$ then $0 in partial U$ or $0 in textExt(U)$. If $0inpartial U$ then obviously $0 in partial (rU)$ so the intersection will be nonempty. If $0 in textExt(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<fracRM$, then $rUcap U = emptyset$. However, if $U$ is a line segment not through the origin, then $rUcap U = emptyset$ for all $rneq 1$.
Thus, $partial U cap partial (rU) = emptyset$ for all $r$ if $0intextInt(U)$ and $partial U cap partial (rU) neq emptyset$ for all $r$ if $0in partial U$.
answered Jul 24 at 0:15
AlexanderJ93
4,067421
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
add a comment |Â
up vote
0
down vote
I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$partial U$
can be described as
$(t, f(t))$
for
$0 le t lt 2pi$.
Suppose $(0, 0) in U$
$partial U$.
Suppose
$partial U cap partial(rU) ne emptyset$
where $0 < r < 1$.
Then there are
$a, b in [0, 2pi)$
such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c in mathbbR$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\
=c(1-r)(b, f(b))+r(b, f(b))\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$partial U$
also passes through
a third point,
which would be a contradiction.
answered Jul 23 at 22:43
marty cohen
69.1k446122
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
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
WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = frac1r$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $xin partial (rU)$ such that $xinpartial U$. Since $xin partial (rU)$, $y = r^-1xin partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0intextInt(U)$ there is an open neighborhood of $0$ also in $textInt(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $Isubset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $yinpartial U$. Thus, if $0intextInt(U)$ then the intersection must be empty.
If $0notintextInt(U)$ then $0 in partial U$ or $0 in textExt(U)$. If $0inpartial U$ then obviously $0 in partial (rU)$ so the intersection will be nonempty. If $0 in textExt(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<fracRM$, then $rUcap U = emptyset$. However, if $U$ is a line segment not through the origin, then $rUcap U = emptyset$ for all $rneq 1$.
Thus, $partial U cap partial (rU) = emptyset$ for all $r$ if $0intextInt(U)$ and $partial U cap partial (rU) neq emptyset$ for all $r$ if $0in partial U$.
answered Jul 24 at 0:15
AlexanderJ93
4,067421
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
add a comment |Â
up vote
1
down vote
accepted
WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = frac1r$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $xin partial (rU)$ such that $xinpartial U$. Since $xin partial (rU)$, $y = r^-1xin partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0intextInt(U)$ there is an open neighborhood of $0$ also in $textInt(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $Isubset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $yinpartial U$. Thus, if $0intextInt(U)$ then the intersection must be empty.
If $0notintextInt(U)$ then $0 in partial U$ or $0 in textExt(U)$. If $0inpartial U$ then obviously $0 in partial (rU)$ so the intersection will be nonempty. If $0 in textExt(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<fracRM$, then $rUcap U = emptyset$. However, if $U$ is a line segment not through the origin, then $rUcap U = emptyset$ for all $rneq 1$.
Thus, $partial U cap partial (rU) = emptyset$ for all $r$ if $0intextInt(U)$ and $partial U cap partial (rU) neq emptyset$ for all $r$ if $0in partial U$.
answered Jul 24 at 0:15
AlexanderJ93
4,067421
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = frac1r$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $xin partial (rU)$ such that $xinpartial U$. Since $xin partial (rU)$, $y = r^-1xin partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0intextInt(U)$ there is an open neighborhood of $0$ also in $textInt(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $Isubset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $yinpartial U$. Thus, if $0intextInt(U)$ then the intersection must be empty.
If $0notintextInt(U)$ then $0 in partial U$ or $0 in textExt(U)$. If $0inpartial U$ then obviously $0 in partial (rU)$ so the intersection will be nonempty. If $0 in textExt(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<fracRM$, then $rUcap U = emptyset$. However, if $U$ is a line segment not through the origin, then $rUcap U = emptyset$ for all $rneq 1$.
Thus, $partial U cap partial (rU) = emptyset$ for all $r$ if $0intextInt(U)$ and $partial U cap partial (rU) neq emptyset$ for all $r$ if $0in partial U$.
answered Jul 24 at 0:15
AlexanderJ93
4,067421
WLOG assume that $r>1$ (if $0<r<1$ then let $U' = rU$ and $r' = frac1r$, so $r'U' = U$ and $r'>1$). Suppose that the intersection is nonempty. Then, there is some $xin partial (rU)$ such that $xinpartial U$. Since $xin partial (rU)$, $y = r^-1xin partial U$. Thus, both $x$ and $y$ are on the boundary of $U$. We can draw a line segment from $0$ through $x$, which will also pass through $y$. If $0intextInt(U)$ there is an open neighborhood of $0$ also in $textInt(U)$ since the interior is always open. Then, take the collection of line segments (not including endpoints) from points in that neighborhood to $x$. The union $I$ of all these line segments is an open set containing $y$. Since $U$ is convex, $Isubset U$. Thus, there is an open subset of $U$ containing $y$, contradicting that $yinpartial U$. Thus, if $0intextInt(U)$ then the intersection must be empty.
If $0notintextInt(U)$ then $0 in partial U$ or $0 in textExt(U)$. If $0inpartial U$ then obviously $0 in partial (rU)$ so the intersection will be nonempty. If $0 in textExt(U)$, then let the shortest distance between $0$ and $U$ be $R$ and the largest distance be $M$. If $0<r<fracRM$, then $rUcap U = emptyset$. However, if $U$ is a line segment not through the origin, then $rUcap U = emptyset$ for all $rneq 1$.
Thus, $partial U cap partial (rU) = emptyset$ for all $r$ if $0intextInt(U)$ and $partial U cap partial (rU) neq emptyset$ for all $r$ if $0in partial U$.
answered Jul 24 at 0:15
AlexanderJ93
4,067421
edited Jul 24 at 1:10
answered Jul 24 at 0:15
AlexanderJ93
4,067421
answered Jul 24 at 0:15
AlexanderJ93
4,067421
answered Jul 24 at 0:15
AlexanderJ93
4,067421
4,067421
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
add a comment |Â
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
1
1
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
First of all, thank you very much. Are you sure you meant $I$ is the intersection of the line segments joining points of the neighborhood to $x$? Seems to me their intersection is just the point $x$. Didn't you want to mean the union? Also, $x$ is not an interior point so it's not open. Maybe you wanted to say $y$ is an interior point of $I$?
– Tanius
Jul 24 at 1:05
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
Yes, I meant the union of line segments not including their endpoints, so that $xnotin I$. I've edited the answer to reflect this.
– AlexanderJ93
Jul 24 at 1:12
1
1
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
Thank you, you helped me so much.
– Tanius
Jul 24 at 1:17
add a comment |Â
up vote
0
down vote
I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$partial U$
can be described as
$(t, f(t))$
for
$0 le t lt 2pi$.
Suppose $(0, 0) in U$
$partial U$.
Suppose
$partial U cap partial(rU) ne emptyset$
where $0 < r < 1$.
Then there are
$a, b in [0, 2pi)$
such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c in mathbbR$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\
=c(1-r)(b, f(b))+r(b, f(b))\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$partial U$
also passes through
a third point,
which would be a contradiction.
answered Jul 23 at 22:43
marty cohen
69.1k446122
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
add a comment |Â
up vote
0
down vote
I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$partial U$
can be described as
$(t, f(t))$
for
$0 le t lt 2pi$.
Suppose $(0, 0) in U$
$partial U$.
Suppose
$partial U cap partial(rU) ne emptyset$
where $0 < r < 1$.
Then there are
$a, b in [0, 2pi)$
such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c in mathbbR$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\
=c(1-r)(b, f(b))+r(b, f(b))\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$partial U$
also passes through
a third point,
which would be a contradiction.
answered Jul 23 at 22:43
marty cohen
69.1k446122
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$partial U$
can be described as
$(t, f(t))$
for
$0 le t lt 2pi$.
Suppose $(0, 0) in U$
$partial U$.
Suppose
$partial U cap partial(rU) ne emptyset$
where $0 < r < 1$.
Then there are
$a, b in [0, 2pi)$
such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c in mathbbR$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\
=c(1-r)(b, f(b))+r(b, f(b))\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$partial U$
also passes through
a third point,
which would be a contradiction.
answered Jul 23 at 22:43
marty cohen
69.1k446122
I have an approach
that seems promising to me
but I can't push it through.
I'll enter it here
in the hope that
someone else can complete it.
$partial U$
can be described as
$(t, f(t))$
for
$0 le t lt 2pi$.
Suppose $(0, 0) in U$
$partial U$.
Suppose
$partial U cap partial(rU) ne emptyset$
where $0 < r < 1$.
Then there are
$a, b in [0, 2pi)$
such that
$(a, f(a)) = r(b, f(b))$
so that
$a = rb$ and
$f(a) = rf(b)$.
Consider the line from
$(b, f(b))$
to
$(rb, rf(b))$.
This has equation
$c(b, f(b))+(1-c)(rb, rf(b))$
for $c in mathbbR$
or
$c(b-rb, f(b)-rf(b))+(rb, rf(b))\
=c(1-r)(b, f(b))+r(b, f(b))\
=c(1-r)(b, f(b))+(a, f(a))
$.
Frustrating.
I can't see where
this leads to a contradiction.
What I would like to do
is show that this line
which passes through
two points on
$partial U$
also passes through
a third point,
which would be a contradiction.
answered Jul 23 at 22:43
marty cohen
69.1k446122
answered Jul 23 at 22:43
marty cohen
69.1k446122
answered Jul 23 at 22:43
marty cohen
69.1k446122
answered Jul 23 at 22:43
marty cohen
69.1k446122
69.1k446122
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
add a comment |Â
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Thank you for your help. I'm still trying to grasp why there "must" be a third point, though...
– Tanius
Jul 24 at 0:02
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
Because the line goes inside the contract set and so must leave it. But I'm not able to find a proof to match that intuition.
– marty cohen
Jul 24 at 2:39
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860745%2fif-0r-neq-1-and-u-is-convex-then-partial-u-cap-partial-ru-emptyset%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
By "frontier of $U$" do you mean the boundary $$partial U equiv overline Ucap overline mathbb R^nsetminus U mathrm ?$$
– Math1000
Jul 23 at 22:07
@Math1000 yes I do
– Tanius
Jul 23 at 22:44