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Suppose $d$ is a metric on the set $X$. For $x$, $y \in X$ define the function $c$ by $c(x,y) = \lceil d(x,y) \rceil$. Show that $c$ is a metric on $X$.

(The reason I put the question here is because I was wondering if it was possible to "discretise" a metric, and the simplest way that I thought of was to use the ceiling.)

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    $\begingroup$ I am confused as to why you asked and then answered your own question at the same time. If you know the solution, why did you ask? $\endgroup$ – N. Owad Nov 17 '14 at 16:59
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    $\begingroup$ Asking and answering your own question is explicitly encouraged on Stack Exchange: blog.stackoverflow.com/2011/07/… My own reason is because I thought it was interesting enough that someone else may want to know the answer in the future. $\endgroup$ – DavidButlerUofA Nov 17 '14 at 22:10
  • $\begingroup$ Comment after 5 years ;-) Your metric $c$ generates the discrete topology on $X$. If $x \ne y$, then $c(x,y) \ge 1$. Hence the open ball with radius $1$ and center $x$ only contains $x$. $\endgroup$ – Paul Frost Sep 30 at 23:24
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To be a metric, $c$ must satisfy the following three conditions:

  1. $c(x,y) \geq 0$ for all $x$, $y \in X$.
  2. $c(x,y) = 0$ if and only if $x=y$.
  3. $c(x,y) = c(y,x)$ for all $x$, $y\in X$.
  4. $c(x,z) \leq c(x,y) + c(y,z)$ for all $x$, $y$, $z \in X$ (the triangle inequality)

Before we start considering these four conditions, let us consider the properties of the ceiling function $t \mapsto \lceil t \rceil$ for $t \geq 0$.

The ceiling function is non-decreasing, so if $s \leq t$ then $\lceil s \rceil \leq \lceil t\rceil$.

Also, given two numbers $s$ and $t$ with $s, t \geq 0$, $\lceil t \rceil$ is the smallest integer greater than or equal to $t$, so we have that $t \leq \lceil t \rceil$ and similarly $s \leq \lceil s \rceil$. Thus $s + t \leq \lceil s \rceil + \lceil t \rceil$. However, $\lceil s + t \rceil$ is the smallest integer greater than or equal to $s+t$ and $\lceil s \rceil + \lceil t \rceil$ is an integer greater than or equal to $s+t$. Therefore it must be that $\lceil s+t \rceil \leq \lceil s\rceil + \lceil t\rceil$.

Now we consider the four conditions in turn:

  1. Since $d(x,y) \geq 0$, $\lceil d(x,y) \rceil \geq \lceil 0 \rceil = 0$. So $c(x,y) \geq 0$.

  2. $c(x,y) = 0$ if and only if $\lceil d(x,y) \rceil = 0$, if and only if $d(x,y) = 0$ (since $d(x,y)$ is nonnegative), if and only if $x=y$.

  3. $c(x,y) = \lceil d(x,y) \rceil = \lceil d(y,x) \rceil = c(y,x)$ for all $x, y \in X$.

  4. Given $x, y, z \in X$, $$ \begin{align} d(x,z) &\leq d(x,y) + d(y,z)\\ \lceil d(x,z) \rceil &\leq \lceil d(x,y) + d(y,z) \rceil\\ &\leq \lceil d(x,y) \rceil + \lceil d(y,z) \rceil \end{align} $$ So $c(x,z) \leq c(x,y) + c(y,z)$.

Therefore $c$ is a metric on $X$.

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