How to prove a metric space?

Question

Suppose $(X, \rho)$ is a metric space.
Show that $\forall (x, y) \in X$, $\sigma(x, y) = 2\rho (x, y)$ is also a metric space in X.

Answer 1

We check the conditions of a metric space. (Some of these are redundant, but I wanted to be thorough so you can just look at the conditions below that are in the definition of a metric that you use in your course.)

• Non-negativity: Clearly $\forall x,y\in X$, we see that $\sigma(x,y)=2\rho(x,y) \geq 0$ since $\rho$ is a metric. (This one follows from the following three properties, however, and can probably be left out.)

• Symmetry: Clearly $\forall x,y \in X$, we see that $\sigma(x,y) = 2\rho(x,y) = 2\rho(y,x) = \sigma(y,x)$ since $\rho$ is a metric.

• Identity of indiscernibles: Clearly $\forall x,y\in X$, we see that $\sigma(x,x) = 2\rho(x,x) = 0$ and likewise if $0 = \sigma(x,y) = 2\rho(x,y)$ then $x=y$ since $\rho$ is a metric.

• Triangle inequality: Clearly $\forall x,y,z\in X$, we see that $\sigma(x,z) = 2\rho(x,z) \leq 2\rho(x,y) + 2\rho(y,z) = \sigma(x,y) + \sigma(y,z)$ since $\rho$ is a metric.

Then we have shown that $\sigma$ satisfies all of the necessary conditions to be a metric and hence $(X,\sigma)$ is a metric space.