Does A a subset of B imply d(B,C)<=d(A,C)

[gottfried]

Homework Statement

Does A \subsetB imply d(B,C)<=d(A,C)?

The Attempt at a Solution

So my feeling is yes. But trying to prove this is difficult.
se we know A \subsetB
Suppose C intersection B isn't empty then d(B,C)=0<=d(A,C) is true since distance is always non-negative.

Suppose C intersection B is empty.
How to prove it in this case I can't do.
Any pointers?

 

[voko]

What's the definition of d(x, y)?

 

[clamtrox]

How do you define the distance between two sets?

 

[gottfried]

Sorry I should have defined that d(A,B)=inf{d(a,b):a element of A and b element of B}
and d is a distance function in a metric space

 

[voko]

So you have to compare two infima.

 

[clamtrox]

Perhaps you are able to prove that if S \subset T, then \inf(x \in S) \ge \inf(x \in T).

 

[gottfried]

Yes so I got to show inf{d(b,c):b element of B and c element of C}<=inf{d(c,a):c element of C and a element of A}.But how?

 

[gottfried]

What if I just suppose

inf{d(a,c)}<inf{d(b,c)} implies a isn't in B. this can't be and therefore
inf{d(a,c)}>=inf{d(b,c)} since a is in B.

 

[voko]

That will work if you can prove that "implies" thing.

 

[gottfried]

Let inf{d(b,c)}=x then x<=b' for all b' element of {d(b,c)} but x>inf{d(a,c)} implying {d(a,c)} isn't in {d(b,c)}

fine?

 

[gottfried]

I said imply again. Always dangerous.

 

[voko]

Yes, "implying" is dangerous. I think at this stage you can forget about d(x, y) and just focus on one set of real numbers B with subset A. You have to prove that inf A >= inf B.

Prove by contradiction: assume inf A < inf B. That means there is some a in A such that...