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Proving that two metrics are topologically equivalent

  1. Oct 6, 2013 #1
    The problem statement, all variables and given/known data.

    Let ##(X,d)## a metric space and ##f: X→X## an injective, continuous and open function. We define ##d' : XxX→ \mathbb R_{≥0}## as ##d'(x,y)=d(f(x),f(y))##. Prove that ##d'## defines a distance and that ##d'## is topologically equivalent to ##d##.

    The attempt at a solution.

    Here's the first part: using the fact that d is a distance, we have that
    1)##d'(x,y)=d(f(x),f(y))≥0##
    2)##d'(x,y)=0## iff ##d(f(x),f(y))=0## iff ##f(x)=f(y)##, ##f## is injective, so ##x=y##.
    3)##d'(x,y)=d(f(x),f(y))=d(f(y),f(x))=d'(y,x)##
    4)##d'(x,y)=d(f(x),f(y))≤d(f(x),f(z)+d(f(z),f(y))=d'(x,z)+d(z,y)##

    As you can see, this is the easy part. I got stuck trying to prove that ##d'## and ##d## are topologically equivalent. To prove this, I should take an open set in ##(X,d)## and prove it is also open in ##(X,d')##. Then, I should take an open set in ##(X,d')## and prove it's open with the metric ##d##. So let ##U## be an open set with ##d'##. Let ##x \in U##, then, there exists ##r_x>0## such that ##B_d(x,r_x) \subset U##. How can I get the appropiate radius ##r'_x## to find a ball ##B_d'(x,r'_x)## is included in ##U##?
     
  2. jcsd
  3. Oct 7, 2013 #2

    pasmith

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    Homework Helper

    Two metrics d and d' are topologically equivalent if and only if the identity function is both (d,d')-continuous and (d',d)-continuous.

    The identity is (d,d')-continuous at [itex]x \in X[/itex] if and only if for every [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that for all [itex]y \in X[/itex], if [itex]d(x,y) < \delta[/itex] then [itex]d'(x,y) < \epsilon[/itex].

    The identity is (d',d)-continuous at [itex]x \in X[/itex] if and only if for every [itex]\epsilon > 0[/itex] there exists [itex]\delta > 0[/itex] such that for all [itex]y \in X[/itex], if [itex]d'(x,y) < \delta[/itex] then [itex]d(x,y) < \epsilon[/itex].

    At some point you should expect to use the facts that f is continuous and open (both with respect to d), neither of which you have yet used.
     
    Last edited: Oct 7, 2013
  4. Oct 7, 2013 #3
    I think I could do it with the property you've suggested:

    First, I'll show that ##Id:(X,d)→(X,d')## is continuous.
    Let ##x \in X## and let ##ε>0##. We know that ##f## is continuous, so, for that given ##ε##, there exists ##δ_{x,ε}## such that ##d(x,y)< δ_{x,ε}##→##d(f(x),f(y))<ε##. But then, ##d'(x,y)=d(f(x),f(y))<ε##. This shows the identity from ##(X,d)## to ##(X,d')## is continuous.

    Now it remains to prove that ##Id:(X,d')→(X,d)## is continuous. So let ##x \in X## and let ##ε>0##. We know f is open, so ##f(B(x,ε))## is an open set in ##(X,d)##. This means that there exists ##δ'## such that ##B_{d}(f(x),δ') \subset f(B_{d}(x,ε))##. If ##d'(x,y)=d(f(x),f(y))<δ'####→ f(y) \in f(B_{d}(x,ε)) → y \in B_{d}(x,ε) → d(x,y)<ε##
     
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