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Continuity characterization (metric spaces)

  1. May 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Let (X,d) and (Y,d') be metric spaces and f: X-> Y a continuous map.
    Suppose that for each a>0 there exists b>0 such that for all x in X
    we have:

    B(f(x), b) is contained in closure( f(B(x,a))).

    Here B(f(x),b) represents the open ball with centre f(x) and radius b.
    Similarly B(x,a) represents the open ball with centre x and radius a.

    Prove that for all x in X and for every c > a :

    B(f(x), b) is contained in f(B(x,c)).

    3. The attempt at a solution

    No clue here, I took an y element in B(f(x),b) so d(f(x),y) < b.
    Then by assumption B(f(x),b) is contained in closure(f(B(x,a)) so y
    is in closure(f(B(x,a)), and then?
     
  2. jcsd
  3. May 30, 2009 #2

    Hao

    User Avatar

    If you don't get an answer, this forum

    http://www.mathhelpforum.com/math-help/ [Broken]

    should help you immensely.

    Questions in their section on analysis are typically very well answered.
     
    Last edited by a moderator: May 4, 2017
  4. May 31, 2009 #3

    matt grime

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    Science Advisor
    Homework Helper

    This is ripe for a proof by contradiction approach (which may even yield a direct proof).

    If there is a c such that B(f(x),b) is not contained in f(B(x,c)) what does that mean?
     
  5. May 31, 2009 #4
    Hi Matt, thanks for your reply. Assume there exists a point q in B(f(x),b) such
    that q is not in f(B(x,c)).

    Now since c>a it follows that B(x,a) is contained in B(x,c).
    Hence f(B(x,a)) is contained in f(B(x,c)).

    Since q is not in f(B(x,c)) then q is not in f(B(x,a)).

    But f is continuous so f(closure(B(x,a))) is contained in closure(f(B(x,a)).

    I'm stuck here. What else should I do?
     
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