# Proving the triangle inequality property of the distance between sets

1. Mar 18, 2012

### jdinatale

Proving the "triangle inequality" property of the distance between sets

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

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.

2. Mar 18, 2012

### Karamata

Re: Proving the "triangle inequality" property of the distance between sets

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

One question, is $m^*$ outer Jordan measure?

3. Mar 18, 2012

### HallsofIvy

Staff Emeritus
Re: Proving the "triangle inequality" property of the distance between sets

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. Mar 18, 2012

### jdinatale

Re: Proving the "triangle inequality" property of the distance between sets

m^* is Lebesgue Outer Measure.

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. Mar 19, 2012

### Karamata

Re: Proving the "triangle inequality" property of the distance between sets

Ok.

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

You can prove this by definition Lebesgue Outer Measure.

6. Mar 19, 2012

### jdinatale

Re: Proving the "triangle inequality" property of the distance between sets

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