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

Is the following just a matter of showing the 3 properties that make up a metric??

Define d on $\Bbb R\times\Bbb R$ by $d(x,y)=\min \{1,|x-y|\}$. Show that $d$ is a metric on $\Bbb R$

1. $d(x,y)=0$ if $x=y$
2. $d(x,y)=d(y,x)$ for every $x,y \in X$
3. $d(x,y)\le d(x,z)+d(z,y)$ for every $x,y,z \in X$
• Yes. Note that for every metric $d'$ the construction $d(x,y)=\min\{1,d'(x,y)\}$ yields a metric. – Stefan Hamcke Oct 20 '13 at 22:32
• Which of the three conditions are you having trouble showing? – Brian M. Scott Oct 20 '13 at 22:36
• well the triangle inequality always seems to be the more difficult one to show. – cele Oct 20 '13 at 22:37
• You're missing the property that $d(x,y)>0$ if $x\neq y$. – Stefan Smith Oct 20 '13 at 22:41

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.
• @StefanSmith: It is a structure between metric and topology where you have a kind of neighborhoods which are uniform over the whole space, they are called entourages. So you can take an entourage $U$ and put it around a point $x$ to get the neighborhood $U(x)$, the same way as you can choose an $ϵ$ and then put the ball $B_ϵ$ around $x$ to get the neighborhood $B_ϵ(x)$ in a metric space. So a metric induces a uniform structure, and a uniform structure induces a topology. – Stefan Hamcke Oct 20 '13 at 22:47