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

  1. Mar 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Does A [itex]\subset[/itex]B imply d(B,C)<=d(A,C)?

    3. The attempt at a solution

    So my feeling is yes. But trying to prove this is difficult.
    se we know A [itex]\subset[/itex]B
    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?
     
  2. jcsd
  3. Mar 4, 2013 #2
    What's the definition of d(x, y)?
     
  4. Mar 4, 2013 #3
    How do you define the distance between two sets?
     
  5. Mar 4, 2013 #4
    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
     
  6. Mar 4, 2013 #5
    So you have to compare two infima.
     
  7. Mar 4, 2013 #6
    Perhaps you are able to prove that if [itex] S \subset T [/itex], then [itex] \inf(x \in S) \ge \inf(x \in T) [/itex].
     
  8. Mar 4, 2013 #7
    Yes so I gotta 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?
     
  9. Mar 4, 2013 #8
    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.
     
  10. Mar 4, 2013 #9
    That will work if you can prove that "implies" thing.
     
  11. Mar 4, 2013 #10
    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?
     
  12. Mar 4, 2013 #11
    I said imply again. Always dangerous.
     
  13. Mar 4, 2013 #12
    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...
     
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