Part (a): Showing \((A, \rho_1)\) is a metric space
Since \((A, \rho)\) is a metric space, the distance function \(\rho\) satisfies the four properties of a metric for all \(u, v, w \in A\). Now, we will show \(\rho_1\) satisfies these properties as well:
1. Non-negativity:
Since \(\rho(u, v) \ge 0\), it follows that:
$$
\rho_1(u, v)=\frac{\rho(u, v)}{1+\rho(u, v)} \ge 0
$$
2. Identity of indiscernibles:
If \(u = v\), then \(\rho(u, v) = 0\). Therefore, \(\rho_1(u, v) = \frac{\rho(u, v)}{1+\rho(u, v)} = 0\). Conversely, if \(\rho_1(u, v) = 0\), we have:
$$
0 = \frac{\rho(u, v)}{1+\rho(u, v)} \Rightarrow \rho(u, v) = 0
$$
This implies that \(u = v\).
3. Symmetry:
Since \(\rho(u, v) = \rho(v, u)\), it follows that:
$$
\rho_1(u, v)=\frac{\rho(u, v)}{1+\rho(u, v)} = \frac{\rho(v, u)}{1+\rho(v, u)} = \rho_1(v, u)
$$
4. Triangle inequality:
We need to show that:
$$
\rho_1(u, v) + \rho_1(v, w) \ge \rho_1(u, w)
$$
Let \(a = \rho(u, v)\), \(b = \rho(v, w)\), and \(c = \rho(u, w)\). Since \((A, \rho)\) is a metric space, we have, by the triangle inequality:
$$
a + b \ge c \Rightarrow 1 - \frac{1}{a+1} + 1 - \frac{1}{b+1} \ge 1 - \frac{1}{c+1}
$$
Now, we want to show that:
$$
\frac{a}{a+1} + \frac{b}{b+1} \ge \frac{c}{c+1}
$$
This inequality is equivalent to the previous one. Thus, the triangle inequality is satisfied for \(\rho_1\). Therefore, \((A, \rho_1)\) is a metric space.