Proving the continuity of functions from one metric to another I'm studying mathematics at university, and am having trouble with some of the continuity questions. The following is a question from a previous assignment that I was unable to complete. The original question asked to first prove that $\bar\rho$ is a metric, but I've completed this part and adjusted the question accordingly. I understand the concept of the $\epsilon$-$\delta$ proofs, but I struggle when it comes to actually proving anything, and I'm never sure where to start. Any assistance on how to complete this would be appreciated. 
[Please note that I have posted this question once previously, but it was put on hold because I hadn't included context. I added context, but couldn't get it taken off hold.]
Let $(X,\rho)$ and $(X,\bar\rho)$ be metric spaces, where
$$
\bar\rho: X\times X\to\Bbb R_0^+\qquad(x,y)\mapsto\frac{\rho(x,y)}{1+\rho(x,y)}.
$$
Let $(Y,\sigma)$ be any metric space. Take functions $f: X\to Y$ and $g: Y\to X$.
Prove that:
(a) $f$ is continuous with respect to $\bar\rho$ if and only if it is continuous with respect to $\rho$; and,
(b) $g$ is continuous with respect to $\bar\rho$ if and only if it is continuous with respect to $\rho$.
 A: To write a proof, you first need to understand what you must show. You have listed four statements to prove. Let's start with the first one:


*

*If $f$ is continuous with respect to $\bar \rho$, then $f$ is continuous with respect to $\rho$.


To show that $f$ is continuous with respect to $\rho$, we must show that $f$ is continuous with respect to $\rho$ at all points in $X$. Start by picking any $x_0 \in X$. We must show that for any $\epsilon > 0$, there exists a $\delta > 0$ such that for all for all $x \in X$
$$\rho(x,x_0) < \delta \text{ implies } \rho(f(x),f(x_0)) < \epsilon$$ 
We start by fixing an $\epsilon > 0$. We now must look for a $\delta$ that makes the above statement true. 
Define $\bar \epsilon = \frac{\epsilon}{1+\epsilon}$ (it will become clear why we make this choice in a second). Because $f$ is continuous with respect to $\bar \rho$, we know that there exists a $\bar \delta > 0$ such that for all $x \in X$ 
$$\bar \rho(x,x_0) = \frac{\rho(x,x_0)}{1+\rho(x,x_0)} < \bar \delta \text{ implies } \bar \rho(f(x),f(x_0)) = \frac{\rho(f(x),f(x_0))}{1+\rho(f(x),f(x_0))} < \bar \epsilon$$
We can rearrange the second part of this expression to obtain
$$\frac{\rho(x,x_0)}{1+\rho(x,x_0)} < \bar \delta \text{ implies } \rho(f(x),f(x_0)) < \frac{\bar \epsilon}{1 - \bar \epsilon} = \epsilon$$
for all $x \in X$ (notice that the convenient choice of $\bar \epsilon$ gives us the expression we want). 
We have to be little bit more careful with the first part of the expression because we cannot simply rearrange when $\bar\delta > 1$. To circumvent this problem, define $\delta^* = \bar \delta$ if $\bar \delta < 1$ and $\delta^* = .5$ if $\bar \delta \geq 1$. Since $\delta^* \leq \bar \delta$, the previous expression means
$$\frac{\rho(x,x_0)}{1+\rho(x,x_0)} < \delta^* \text{ implies } \rho(f(x),f(x_0)) < \epsilon$$
for all $x \in X$. 
But $\delta^* < 1$, so if we define $\delta = \frac{\delta^*}{1-\delta^*}$ we can rearrange to obtain 
$$\rho(x,x_0)< \delta \text{ implies } \rho(f(x),f(x_0)) < \epsilon$$
for all $x\in X$. Similar arguments can be used to prove the other three statements.
