Brute force way to show that $\rho(x,y) = \min\{1, d(x,y)\}$ is a metric

Question

Following a hint in Short proof that $\rho^\prime(x,y) = \min\{1,\rho(x,y)\}$ is a metric

I would like to use the brute force method to show that the standard bounded metric is a metric

$$\rho(x,y) = \min\{1, d(x,y)\}$$

By brute force I mean to substitute the definition of $\min$ operator which is:

$$\min(a,b) = \dfrac{a+b - |a-b|}{2}$$

So let $x,y,z \in (X, \rho)$ \begin{align} \rho(x,z) &= \min\{1, d(x,z)\}\\ &= \dfrac{1+d(x,z) - |1-d(x,z)|}{2}\\ & = \dfrac{1}{2} + \dfrac{d(x,z)}{2} - \dfrac{|1-d(x,z)|}{2}\\ & \leq \dfrac{1}{2} + \dfrac{d(x,y)}{2} + \dfrac{1}{2} + \dfrac{d(y,z)}{2} - \dfrac{|1-d(x,z)|}{2} \tag{obvious} \end{align}

If I can show that

$$- \dfrac{|1-d(x,z)|}{2} \leq - \dfrac{|1-d(x,y)|}{2} + - \dfrac{|1-d(y,z)|}{2}$$

Then I am done.

However, this seems to be really messy and difficult to show. How can I show that the above inequality holds?

Answer 1

you have to check three things.

We have $\rho(x,y)=\rho(y,x)$ trivially (since $\delta(x,y)=\delta(y,x)$)

We also have $\rho(x.y)=\min(1,\delta(x,y))=0\iff\delta(x,y)=0\iff x=y$

The only tricky part is the triangle inequality.

We have to show $\rho(x,y)+\rho(y,z)\leq \rho(x,z)$.

if $\rho(x,y)\neq \delta(y,z)$ or $\rho(y,z)\neq \delta(y,z)$ we have $\rho(x,y)+\rho(x,z)\geq 1 \geq \rho(x,z)$ Otherwise:

$\rho(x,y)+\rho(y,z)=\delta(x,y)+\delta(y,z)\geq\delta(x,z)\geq\min(1,\delta(x,z))=\rho(x,z)$

Answer 2

In this case worth to calculate step by step. We have eight (8) possibilites.

Look: we want to show that $\rho^\prime(x,z) \leq \rho^\prime(x,y) + \rho^\prime(y,z)$, ok?!

This is that $\min\{1, d(x,z)\} \leq \min\{1, d(x,y)\} + \min\{1, d(y,z)\}$.

• Notice that, if $\rho^\prime(x,z)=d(x,z)=\min\{1, d(x,z)\}$ (this hypothesis say that $d(x,z)\leq 1$ it is very important for you understand the solution following) than we have four (4) possibilites:

  1. $d(x,z) \leq 1+ 1$;
  2. $d(x,z) \leq 1+ d(y,z);$
  3. $d(x,z) \leq d(x,y) +1;$
  4. $d(x,z) \leq d(x,y) + d(y,z);$ (it's true by hypothesis that $d$ is a metric)

• Now, if $\rho^\prime(x,z)=1=\min\{1, d(x,z)\}$ we have:

  1. $1 \leq 1+ 1;$
  2. $1 \leq d(x,y) + 1;$
  3. $1 \leq 1+ d(y,z);$
  4. $1 \leq d(x,y)+ d(y,z);$ this is true by hipothesis $\rho^\prime(x,z)=1\leq d(x,z)\leq d(x,y)+d(y,z);$

However, by 1-8 we proof triangle inequality.