Recent content by nirajkadiyan6

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

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

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.

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)

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+...