- #1
winter85
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Homework Statement
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?
Homework Equations
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.