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
<br /> \bar\rho:X \times X \Rightarrow R_{0}^{+}, (x,y) \Rightarrow \frac{\rho(x,y)}{1+\rho(x,y)}.<br />

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 &gt; 0

that
{\rho}(z,b) &lt; \delta \rightarrow \theta(f(z),f(b)) &lt; \epsilon

I need to show that for some\psi &gt; 0 that

{\bar\rho}(x,a)&lt;\psi \rightarrow \theta(f(x),f(a)) &lt;\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&gt;0, you need to find a number \psi(\epsilon)&gt;0 such that

{\bar\rho}(x,a)=\frac{\rho(x,a)}{1+\rho(x,a)}&lt;\psi(\epsilon) \Rightarrow \theta(f(x),f(a)) &lt;\epsilon

knowing that there exists a number \delta(\epsilon)&gt;0 such that

{\rho}(x,a) &lt; \delta(\epsilon) \Rightarrow \theta(f(x),f(a)) &lt; \epsilon

Notice that, just by algebra, we have that

\frac{\rho(x,a)}{1+\rho(x,a)}&lt;\psi (\epsilon)\Leftrightarrow \rho(x,a)&lt;\frac{\psi(\epsilon)}{1-\psi (\epsilon)}

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

\frac{\psi(\epsilon)}{1-\psi (\epsilon)}&lt;\delta(\epsilon)

then you will have won, yes?