Proving the triangle inequality property of the distance between sets

In summary, to prove the "triangle inequality" property of the distance between sets, one must show that for measurable subsets A, B, and C of a measurable set X, the outer measure of the symmetric difference between A and C is less than or equal to the sum of the outer measures of the symmetric differences between A and B and B and C. This can be achieved by using the fact that the symmetric difference between A and C is a subset of the union of the symmetric differences between A and B and B and C, and then applying the definition of Lebesgue Outer Measure.
  • #1
jdinatale
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Proving the "triangle inequality" property of the distance between sets

Here's the problem and how far I've gotten on it:

Untitled-3.png


If you are unfamiliar with that notation, S(A, B) = (A \ B) U (B \ A), which is the symmetric difference.

And D(A, B) = m^*(S(A, B)), which is the outer measure of the symmetric difference.
 
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  • #2


Hm, can you show that [itex]m^*(A\backslash B)+m^*(B\backslash C)\ge m^*(A \backslash C)[/itex], same for [itex]m^*(B\backslash A)+[/itex]...

One question, is [itex]m^*[/itex] outer Jordan measure?
 
  • #3


Note that A and B cannot be just "arbitrary subsets of an arbitrary set X". They must be measurable subsets of the measurable set X.
 
  • #4


Karamata said:
Hm, can you show that [itex]m^*(A\backslash B)+m^*(B\backslash C)\ge m^*(A \backslash C)[/itex], same for [itex]m^*(B\backslash A)+[/itex]...

One question, is [itex]m^*[/itex] outer Jordan measure?

m^* is Lebesgue Outer Measure.

HallsofIvy said:
Note that A and B cannot be just "arbitrary subsets of an arbitrary set X". They must be measurable subsets of the measurable set X.

I can see your point, but I'm copying the problem verbatim out of my textbook, and it uses "arbitrary subsets of an arbitrary set X." Would you say that is a mistake on the author's part?
 
  • #5


jdinatale said:
m^* is Lebesgue Outer Measure.

Ok.

Well, if [itex]A[/itex] and [itex]B[/itex] are disjoint sets, then [itex]m^*(A\cup B)\le m^*(A)+m^*(B)[/itex].

You can prove this by definition Lebesgue Outer Measure.
 
  • #6


Well, I got really close. Using the fact that S(A, C) subset S(A, B) U S(B, C), I obtained the following:

close.png
 

FAQ: Proving the triangle inequality property of the distance between sets

1. What is the triangle inequality property of the distance between sets?

The triangle inequality property states that the distance between any two points is always less than or equal to the sum of the distances between those points and a third point. In other words, it ensures that the shortest distance between two points is always a straight line.

2. Why is the triangle inequality property important in mathematics and science?

The triangle inequality property is important because it allows us to define the concept of distance and measure it accurately. This property is used in various fields of science, including physics, chemistry, and engineering, to analyze and solve problems involving distance and space.

3. How is the triangle inequality property proven?

The triangle inequality property can be proven using the concept of the Pythagorean theorem. By using the Pythagorean theorem, we can show that the sum of the squares of the two shorter sides of a right triangle is always equal to the square of the hypotenuse, which is the longest side. This is the foundation of the triangle inequality property.

4. Can the triangle inequality property be extended to more than three points?

Yes, the triangle inequality property can be extended to any number of points. For example, if we have four points A, B, C, and D, the distance between points A and B will always be less than or equal to the sum of the distances between A and C, C and D, and D and B. This can be extended to any number of points, making it a useful property in multi-dimensional spaces.

5. How is the triangle inequality property used in real-life applications?

The triangle inequality property has many real-life applications, such as in navigation systems, where it is used to calculate the shortest distance between two points. It is also used in computer graphics to create realistic images by accurately measuring the distance between objects. Additionally, this property is crucial in fields like economics and finance, where it is used to calculate the shortest routes for transportation or to optimize investments.

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