# "Truncated" metric equivalence

I may be having problems with this proof.

Consider the metric on a set $S$ defined by:

$$d^\prime (x,y) = \left\{ \begin{array}{lcr} d(x,y) & \text{if}& d(x,y) \leq 1\\ 1 & \text{if} & d(x,y) > 1 \end{array} \right.$$ where $d$ is a metric on $S$.

I'm trying to determine whether or not $d^\prime$ and $d$ induce the same topology on $S$.

I suspect they do not. Here's my reasoning, via counterexample:

Let $S = \mathbb{R}$, $d(x,y) = |x-y|$. Then the open ball in the $d^\prime$ metric of radius $2$ around any point $x \in \mathbb{R}$ is the entire space $\mathbb{R}$. In order for the two topologies to be equivalent, there would have to be some open ball centered at $x$ in the $d$ metric which contained the open ball in the $d^\prime$ metric. But this cannot be for any ball of finite radius, so the topologies are not the same.

Are there any glaring errors in my logic here?

• I would not call it "partially discrete", it's "truncated by $1$". Informal example: if the metric is the driving time from $A$ to $B$, then we may decide that all driving times longer than $1$ day are the same for practical purposes (i.e., not practical) and record them simply as "$1$". This does not change anything within your neighborhood (city, state...)
– user147263
Sep 28 '14 at 0:28

For $1>r>0$, the open ball (around the point) with radius $r$ under $d$ is the same as under $d'$.
For $r \geq 1$, the open ball with radius $1/2$ (less than $1$) under $d$ is included in open ball with radius $r$ under $d'$. Also, the open ball with radius $1/2$ under $d'$ is included in the open ball with radius $r$ under $d$.
In order to show two topologies generated by metrics are equivalent, you have to show any open ball centered at any point $x$ in one contains an open ball centered at $x$ in the other, and vice versa. Your proof has the containment backwards.