Show that $d(x,y)=\min \{1,|x-y|\}$ is a metric on $\mathbb R$

Solution 1:

That's correct (though your first condition should read "if and only if $x=y$"). This is what is known sometimes as the standard bounded metric induced by the metric $\rho(x,y)=|x-y|$. More generally, if $\langle X,\rho\rangle$ is any metric space, then the function $$d(x,y)=\min\{1,\rho(x,y)\}$$ is again a metric on $X$, called the standard bounded metric induced by $\rho$, and induces the same metric topology.

To prove that it is in fact a metric, the first two properties are relatively straightforward. For the triangle inequality, note that if $\rho(x,y)\ge1$ or $\rho(y,z)\ge 1,$ then $$d(x,y)+d(y,z)\ge1\ge d(x,z).$$ Otherwise, we have $$d(x,y)+d(y,z)=\rho(x,y)+\rho(y,z)\ge\rho(x,z)\ge d(x,z)$$ by definition of $d,$ since $\rho$ is a metric.