Proving Metric Space: (X,\bar\rho) is Positive Definite

[boneill3]

Homework Statement

Prove that if (X,\rho)is a metric space then so is (X,\bar\rho), where
<br /> \bar\rho:X \times X \Rightarrow R_{0}^{+}, (x,y) \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}.<br />

Homework Equations

I'm trying to prove the axiom that a metric space is positive definite.

The Attempt at a Solution

because given (X,\rho) is a metric space is it enough to say that (X,\bar\rho), cannot be &lt; 0 because (X,\rho) cannot be &lt; 0 ? ie the limit of (X,\bar\rho) is 0 as (X,\rho) tends to zero ?

 

[VeeEight]

Yes, if x is not y, then <br /> \bar\rho<br /> is greater than 0 since <br /> \rho<br /> is greater than 0 for this case.
If x=y, then it is clear that <br /> \bar\rho<br /> is 0 as you have <br /> \rho(x,x)<br /> <br /> = 0 in the numerator.

 

[boneill3]

Thanks for your help.

I have also got to prove the following
Let (Y, \theta) be a metric space. Prove the following.
f : X \rightarrow Y is continuous with respect to \bar\rho if and only if it is continuous with respect to \rho


I know I have to show that the inverse of Open sets in Y are open in X and visa versa how ever I'm not sure how to represent the open sets in \bar\rho

ie if I was proving (X,\rho) and (Y,\theta)

I would show that for each G \subseteq Y that f^{-1}(G)is an open \subseteq X whenever G isan open subset of Y




As the metric space \bar\rho uses the metric \rho how to I incorporate that into the proof ?

 

[VeeEight]

Let f be continuous wrt <br /> \rho<br /> and let G be open in Y. Then, as you pointed out, <br /> f^{-1}(G) <br /> is open in X. Let a be a point in <br /> f^{-1}(G) <br /> = A. Then, A is open in X implies that neighborhoods of a are in A (under the rho metric). Write out what these neighborhoods look like. They look like {y | ||a-y||_{<br /> \rho<br />} < e }, and these neighborhoods are in X. Now what do the neighborhoods look like under the <br /> \bar\rho<br /> metric and how do they relate to the usual neighborhoods?

 

[boneill3]

Does this look alright ?
I've tried to use Open balls of radius \delta and \psi



So I have three metric spaces

(X,\rho)
(X,\bar\rho)
and
(Y, \theta)
and a function
f : X \rightarrow Y
which is continuous wrt (X,\rho)

So given f : X \rightarrow Ycontinous wrt \rho.

Take a neighbourhood B_{\rho}(a,\delta) = {x \in X \mid \rho(x,a) &lt; \delta }

Then for each \epsilon &gt; 0<br /> there is an \delta &gt; 0


Such that we have B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) &lt; \epsilon As f is continuous



To show that f : X \rightarrow Y is continuous wrt \bar\rho we need to to find a B_{\rho}(b,\psi) \subset B_{\rho}(a,\delta)

So we can take \psi = \delta - \rho(b,a)

if
\rho(x,b) &lt; \psi
then
\rho(x,a) \leq \rho(x,b)+\rho(x,a)
\rho(x,a) &lt; \psi+ \delta - \psi = \delta

So we can know show that:
f : X \rightarrow Y is continuous wrt \bar\rho

0 &lt; B_{\bar\rho}(x,\psi) &lt; \rho(f(x),f(b)) \leq B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) &lt; \epsilon