Generalization of the Triangle Inequality with Sets
Definition: The distance between two points $p_1, p_2 in mathbb{R}^n$ is
begin{equation}
d(p_1,p_2)=|| p_1-p_2 ||.
end{equation}
The distance between a point $pin mathbb{R}^n$ and a set $mathcal{C} subset mathbb{R}^n$ is
begin{equation}
d(p,mathcal{C}) = inf{||p-q|| : q in mathcal{C}}.
end{equation}
Given three points $p_i in mathbb{R}^n, i=1,2,3$. The triangle inequality tells us that
begin{equation}
|d(p_1,p_3) - d(p_3,p_2)| le d(p_1, p_2) le d(p_1, p_3) + d(p_3,p_2).
end{equation}
Now the problem is:
Is the triangle inequality still held if one point is replaced by a non-singleton set?
real-analysis inequality
asked Nov 27 at 13:08
winston
507218
add a comment |
Definition: The distance between two points $p_1, p_2 in mathbb{R}^n$ is
begin{equation}
d(p_1,p_2)=|| p_1-p_2 ||.
end{equation}
The distance between a point $pin mathbb{R}^n$ and a set $mathcal{C} subset mathbb{R}^n$ is
begin{equation}
d(p,mathcal{C}) = inf{||p-q|| : q in mathcal{C}}.
end{equation}
Given three points $p_i in mathbb{R}^n, i=1,2,3$. The triangle inequality tells us that
begin{equation}
|d(p_1,p_3) - d(p_3,p_2)| le d(p_1, p_2) le d(p_1, p_3) + d(p_3,p_2).
end{equation}
Now the problem is:
Is the triangle inequality still held if one point is replaced by a non-singleton set?
real-analysis inequality
asked Nov 27 at 13:08
winston
507218
Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10
add a comment |
Definition: The distance between two points $p_1, p_2 in mathbb{R}^n$ is
begin{equation}
d(p_1,p_2)=|| p_1-p_2 ||.
end{equation}
The distance between a point $pin mathbb{R}^n$ and a set $mathcal{C} subset mathbb{R}^n$ is
begin{equation}
d(p,mathcal{C}) = inf{||p-q|| : q in mathcal{C}}.
end{equation}
Given three points $p_i in mathbb{R}^n, i=1,2,3$. The triangle inequality tells us that
begin{equation}
|d(p_1,p_3) - d(p_3,p_2)| le d(p_1, p_2) le d(p_1, p_3) + d(p_3,p_2).
end{equation}
Now the problem is:
Is the triangle inequality still held if one point is replaced by a non-singleton set?
real-analysis inequality
asked Nov 27 at 13:08
winston
507218
Definition: The distance between two points $p_1, p_2 in mathbb{R}^n$ is
begin{equation}
d(p_1,p_2)=|| p_1-p_2 ||.
end{equation}
The distance between a point $pin mathbb{R}^n$ and a set $mathcal{C} subset mathbb{R}^n$ is
begin{equation}
d(p,mathcal{C}) = inf{||p-q|| : q in mathcal{C}}.
end{equation}
Given three points $p_i in mathbb{R}^n, i=1,2,3$. The triangle inequality tells us that
begin{equation}
|d(p_1,p_3) - d(p_3,p_2)| le d(p_1, p_2) le d(p_1, p_3) + d(p_3,p_2).
end{equation}
Now the problem is:
Is the triangle inequality still held if one point is replaced by a non-singleton set?
real-analysis inequality
real-analysis inequality
asked Nov 27 at 13:08
winston
507218
asked Nov 27 at 13:08
winston
507218
edited Nov 27 at 13:22
asked Nov 27 at 13:08
winston
507218
asked Nov 27 at 13:08
winston
507218
asked Nov 27 at 13:08
winston
507218
507218
Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10
add a comment |
Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10
Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10
add a comment |
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Suppose part 3 held true. Do you see that you get parts 1 and 2 from here by taking some sets as singletons? Try to work therefore with three sets at first(I am not telling you if it's true or not, only that if it were true life is easier).
– астон вілла олоф мэллбэрг
Nov 27 at 13:15
@астонвіллаолофмэллбэрг how about the edits? (non-singleton sets?)
– winston
Nov 27 at 13:22
You have edited the question. Is this how it will finally stand?
– астон вілла олоф мэллбэрг
Nov 27 at 13:28
@астонвіллаолофмэллбэрг I try to simplify the problem first. I know that the right part of the inequality is not correct because if this set contains the two points, then $d(p_1,p_2) ge 0$ (assume that $p_3$ is replaced by a non-singleton set).
– winston
Nov 27 at 13:49
Correct. So, for example, if we take a set $P$ with at least two points, and $p_1 neq p_2$ in $P$, then $d(p_1,p_2) > 0$ although $d(p_1,P)$ and $d(p_2,P)$ are both zero. However, if we replace $p_1$ by a set $P$, then does the right side of the inequality work? We will figure the left side later.
– астон вілла олоф мэллбэрг
Nov 27 at 14:10