Bounded metric on $\mathbb{R}$

I am supposed to give an example of a metric that is bounded on $\mathbb{R}$. In other words, I have to find a function $d:\mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ which satisfies that $d \leq C$ for some $C$ and that $(\mathbb{R},d)$ is a metric space.

I'd really appreciate it if someone could give me an example and explain why it satisfies the criteria. Or simply give me a clue on where to begin.

• Do you know what a metric is? Do you know any examples of metrics? – MJD Aug 27 '14 at 17:52
• Are you interested in any bounded metric, or a bounded metric that induces the usual topology on $\mathbb{R}$? Both are achievable, but the first takes less work than the second. – André Nicolas Aug 27 '14 at 17:56
• Hint: consider a metric which only takes on two values. – Bungo Aug 27 '14 at 18:00
• I'm interested in any bounded metric. @MJD Well, I don't have a really clear idea of what it really is. I know the criteria of a metric (symmetry, triangle inequality and non-degeneracy), and I know some examples. I'm for instance thinking of the discrete metric. Would that one furfill my wishes? – truglr Aug 27 '14 at 18:05
• If $x\ne y$ let distance be $1$, else $0$. – André Nicolas Aug 27 '14 at 18:07

The simplest example of a limited metric is the discret metric, given by $d_0: X\times X\rightarrow X$ s.t. $$d_0(x,y)=\left\{\begin{array}{lr}1&\text{if x\neq y} \\ 0 & \text{if x=y}\end{array}\right.$$
You can see that $d(x,y)\leq 1$ for all $x,y\in X$. Another way is to construct a bounded metric from any given metric is: Given a metric $d: X\times X\rightarrow X$ and $\alpha>0$, consider the new metric $d_\alpha$ given by $$d_\alpha(x,y)=\frac{d(x,y)}{\alpha+d(x,y)}$$ As above, $d_\alpha(x,y)\leq 1$ for all $x,y\in X$. It's very simple to prove that $d_\alpha$ is metric too.
If $d:X\times X\rightarrow \mathbb R$ is a metric on $X$ then so is $d':X\times X\rightarrow \mathbb R$ prescribed by $(x,y)\mapsto d(x,y)$ if $d(x,y)<1$ and $(x,y)\mapsto 1$ otherwise.
Note that $d'$ is a bounded metric.
Metrics $d$ and $d'$ induce the same topology on $X$.