How Does Hausdorff Distance Determine Equality of Compact Sets?

[diracy]

Homework Statement

Given a compact set A\subset\Re^{n} and a point x\in\Re^{n} define the distance from x to A as the quantity:

d(x, A)=inf({\left\|x-y\right\|: y\inA})

Given two compact sets A, B \subset\Re^{n}, define the Hausdorff distance between them to be:

d(A, B)=max(sup{d(x, B) : x\inA}, sup{d(x, A) : x\inB})

a. For any compact set A, prove that the function f : \Re^{n}\rightarrow\Re given by: f(x)=d(x, A) is continuous.

b. For any two compact sets A, B it's true that: d(A, B)<\infty

c. For any two compact sets A, B it's true that d(A, B)=0 if and only if A=B.

Homework Equations

The Attempt at a Solution

I think I handled part a. I'm just completely lost on b and c. Any help?

 

[diracy]

The \left\ and \right\ should be norm.

 

[diracy]

Edited for a typo.

 

[micromass]

For (b), you need to show that

\sup_{x\in A}{d(x,B)}

is bounded and equivalently that

\sup_{x\in B}{d(x,A)}

is bounded. Now, can you use (a) to show this?

 

[diracy]

For (b), you need to show that

\sup_{x\in A}{d(x,B)}

is bounded and equivalently that

\sup_{x\in B}{d(x,A)}

is bounded. Now, can you use (a) to show this?

I got it! I really appreciate the help. I have one more question.

How do I show the following:

For any three compact sets: A, B, C we have a triangle inequality: d(A, C)\leq d(A,B) + d(B, C)

 

[micromass]

Let

a(A,B)=\sup_{a\in A}{d(a,B)}

(then d(A,B)=\max{a(A,B),a(B,A)}).

Try to show first that

a(A,B)\leq a(A,C)+a(C,B)

Thus, show the triangle inequality first for a.