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Prove two metric spaces have the same open balls

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data

    Let X be a metric space with metric d, and let [tex] d_{1} [/tex] be the metric defined by [tex] d_{1}(x,y) = \frac{d(x,y)}{1+d(x,y)} [/tex] . Show that the two metric spaces [tex] (X,d) [/tex] and [tex] (X,d_{1}) [/tex] have precisely the same open sets. Hint: show that they have the same open balls with one exception. What is this exception?

    2. Relevant equations

    3. The attempt at a solution

    First I note that the function [tex] f(x) = \frac{x}{1+x} [/tex] is continuous and monotically increasing, and that [tex] d_{1}(x,y) < d(x,y) [/tex] for all distinct x and y.

    I reasoned like this: Let [tex] S(x) [/tex] be an open ball in [tex] (X,d) [/tex] of center x and radius r. Then for all y in S(x), we have [tex] d(x,y) < r [/tex] , therefore [tex] d_{1}(x,y) < \frac{r}{1+r}[/tex] and so y is in an open ball centered at x with radius at most [tex] \frac{r}{1+r} [/tex]. Let S'(x) be the open ball in [tex] (X,d_{1}) [/tex] centered at x with radius [tex] \frac{r}{1+r} [/tex]. So we have [tex] S(x) \subseteq S'(x) [/tex]
    Similarly, for all y in S'(x) we have [tex] d_{1}(x,y) < \frac{r}{1+r} [/tex], so [tex] d(x,y) < r [/tex] hence y in S(x), therefore [tex] S'(x) \subseteq S(x) [/tex]. So we have that S(x) = S'(x) which means the two metric spaces have the same open balls.

    It's easy to conclude that the two metric spaces have thus the same open sets, because every open set is the union of open balls.

    Now i'm not very sure about this argument, so I hope anyone would point out the weaknesses or points that are wrong or need further justification. Also any hint about what this exception might be?

    Thank you.
  2. jcsd
  3. Nov 29, 2008 #2


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    (minor aside: did you verify that r/(r+1) exists, for all r?)

    An oddity about your proof is that you used a different naming scheme for the open balls in (X,d) and (X,d1). (In what follows, I've deliberately chosen different names for my indeterminate variables) In (X,d), you chose a point a and a real number q, and used that to denote the ball
    {v | d(q,v) < a}.​
    However, in (X,d1), you chose a point c and a real number e, and used that to denote the ball
    {w | d(c,w) < e/(e+1)}.​

    Any particular reason for doing so? (Was it a conscious decision?) Do you have an argument that all balls in (X,d1) can be named by such a scheme?
  4. Nov 29, 2008 #3
    I'm sorry but i don't see what you mean by this:

    in (X,d), i chose a point a and a real number q and denoted the ball

    S = {v | d(a,v) < q}. I picked a point v from this set. Then i chose an open ball S' of center a but under the metric d1, that contains the point v. Then I proved that if v' is a point in S', then it must be a point of S also. At least that was my working plan.

    for an argument that all balls in (X,d1) can be named by

    {w | d(c,w) < e/(e+1)},

    e/(e+1) can be any arbitrarily chosen real number in the interval [0,1). the diameter of any set under the d1 metric is in the interval [0,1]. So the radius of any ball in (X,d1) must be in the set [0,1]. If it's less than 1, then it can be represented as e/(e+1) for some real number e. However if it is 1 then it cannot be represented in this way. This would happen when then ball we're considering in (X,d) is the entire set X, and the set X has infinite radius, right? But we know that the set X is an open ball in any metric, it is an open ball in both (X,d) and (X,d1).
    I hope now my reasoning is clearer? I'm having trouble organizing my thoughts about this problem :(
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