Continuous functions in metric spaces

[boneill3]

Hi guy's I know this is more of a homework question, I posted a similar thread earlier on but I think I ended up confusing myself.

I need to show that a function is continuous between metric spaces. I'll post the question and what I've done any tips on moving forward would be great.

I have any metric spaces
$(X,\rho)$
and
$(Y, \theta)$

And a metric space
$(X,\bar\rho)$
where
$\bar\rho:X \times X \Rightarrow R_{0}^{+}, (x,y) \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}.$

I have got to show the following

Let $(Y, \theta)$
be a metric space.
Prove that.
$f : X \rightarrow Y$
is continuous with respect to $\bar\rho$ if and only if it is continuous with respect to $\rho$


I have been given that $f : X \rightarrow Y$
is continuous with respect to $(X,\rho)$

So I know that for some $\delta$ and $\epsilon > 0$

that
${\rho}(z,b) < \delta \rightarrow \theta(f(z),f(b)) < \epsilon$

I need to show that for some$\psi > 0$ that

${\bar\rho}(x,a)<\psi \rightarrow \theta(f(x),f(a)) <\epsilon$

Can some one please show me how to go about finding $\psi$ ?

 

[quasar987]

Fix a in X. To show continuity at a, given $\epsilon>0$, you need to find a number $\psi(\epsilon)>0$ such that

${\bar\rho}(x,a)=\frac{\rho(x,a)}{1+\rho(x,a)}<\psi(\epsilon) \Rightarrow \theta(f(x),f(a)) <\epsilon$

knowing that there exists a number $\delta(\epsilon)>0$ such that

${\rho}(x,a) < \delta(\epsilon) \Rightarrow \theta(f(x),f(a)) < \epsilon$

Notice that, just by algebra, we have that

$$\frac{\rho(x,a)}{1+\rho(x,a)}<\psi (\epsilon)\Leftrightarrow \rho(x,a)<\frac{\psi(\epsilon)}{1-\psi (\epsilon)}$$

So, surely, if you can chose $$\psi(\epsilon)$$ such that

$$\frac{\psi(\epsilon)}{1-\psi (\epsilon)}<\delta(\epsilon)$$

then you will have won, yes?