Bounding the Diameter of Union of Two Sets in a Metric Space

[nirajkadiyan6]

Hi, I am stuck with the following proofs. In metric space
here, A,B,C are subset of metric space (X,d) and C is bounded



Problem 1.) d(A,B) <=d(A,C)+d(B,C)+diam(C)
Problem 2.)|d(b,A)-d(c,A)| <= d(b,c) where 'b' belongs to 'B' and 'c' belongs to 'C'.

Problem 3)- diam(A U B)<= diam A+ diam B+ d(A,B) ,Here d(A,B)= inf {d(a,b)| a belong to A and b to B}

I think if i get some clue even about one i can handle other one.

Thanks in advance..:)

 

[micromass]

Hi nirajkadiyan6!

For the first one. Take $a\in A,b\in B, c\in C,c^\prime \in C$ arbitrary. It suffices to show

$$d(A,B)\leq (a,c)+d(b,c^\prime)+d(c,c^\prime)$$

Can you show that?

 

[nirajkadiyan6]

This was the thing i proved in very first attempt but it is not helping me
This can be proved easily but, for arbitrary a and c, d(a,c) can't be simply written is d(A,C) as d(A,C) is smallest distance between some a',c'
so d(a,c)< d(A,C) for arbitrary a,c
Same holds for d(A,B)

 

[nirajkadiyan6]

Do you mean me to take c and c' to be those elements of C which makes d(A,C) and d(B,C) shortest.

If it is the case then i am done with my problem.
Can you suggest something about others.

 

[micromass]

Do you mean me to take c and c' to be those elements of C which makes d(A,C) and d(B,C) shortest.

If it is the case then i am done with my problem.
Can you suggest something about others.

No, such a c and c' won't exist in general. But if you show the inequality in my post for all c and c', then you've shown it for the infimum too.

 

[nirajkadiyan6]

That's what the problem is. I am not able to show it for all c and c'.

 

[micromass]

just apply the triangle inequality for a certain choice of c and c'. This works for every choice of c and c'...

 

[nirajkadiyan6]

It works for all c and c' but then how can we prove the original problem with all c and c'