# Metric space proof

1. Feb 26, 2010

### boneill3

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. The attempt at a solution

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

Last edited: Feb 26, 2010
2. Feb 26, 2010

### VeeEight

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

3. Feb 27, 2010

### boneill3

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 ?

4. Feb 27, 2010

### VeeEight

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

5. Mar 1, 2010

### 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 continous wrt $(X,\rho)$

So given $f : X \rightarrow Y$continous wrt $\rho$.

Take a neighbourhood $B_{\rho}(a,\delta) = {x \in X \mid \rho(x,a) < \delta }$

Then for each $\epsilon > 0$ there is an $\delta > 0$

Such that we have $B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) < \epsilon$ 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
$\rho(x,b) < \psi$
then
$\rho(x,a) \leq \rho(x,b)+\rho(x,a)$
$\rho(x,a) < \psi+ \delta - \psi = \delta$

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

$0 < B_{\bar\rho}(x,\psi) < \rho(f(x),f(b)) \leq B_{\rho}(a,\delta) \leq \theta(f(x),f(a)) < \epsilon$