MHB Properties of Metric $\sigma(x,y)$

[evinda]

Hello! (Wave)
I want to show that if $\rho(x,y)$ is a metric on $X$, then $\sigma (x,y)= \min \{ 1, \rho(x,y) \}$ is a metric.

I have thought the following:

$\rho(x,y)$ is a metric on $X$, so:

• $\rho(x,y) \geq 0, \forall x,y \in X$
• $\rho(x,y)=0$ iff $x=y$
• $\rho(x,y)=\rho(y,x) \forall x,y \in X$
• $\rho(x,z) \leq \rho(x,y)+\rho(y,z), \forall x,y,z \in X$

• $\sigma(x,y)= \min \{ 1, \rho(x,y) \} \geq 0 \forall x,y \in X$ since $1 \geq 0$ and $\rho(x,y) \geq 0, \forall x,y \in X$
• $\sigma(x,y)=0 \Leftrightarrow \min \{ 1, \rho(x,y) \} =0 \Leftrightarrow \rho(x,y)=0 \Leftrightarrow x=y$
• $\sigma(x,y)= \min \{ 1, \rho(x,y) \}= \min \{ 1, \rho(y,x) \}=\sigma(y,x) \forall x,y \in X$
• Suppose that $1< \rho(x,y), \forall x,y \in X$. Then $\sigma(x,z)=1 \leq 1+1= \sigma(x,y)+ \sigma(y,z)$.
  Suppose that $1> \rho(x,y), \forall x,y \in X$. Then $\sigma(x,z)= \rho(x,z) \leq \rho(x,y)+ \rho(x,z)= \sigma(x,y)+ \sigma(x,z)$

Could I improve something? Could we also show the last inequalities without distinguishing cases? (Thinking)

 

[Ackbach]

I would say you're solid on everything except the triangle inequality. I don't see any particular reason to make such a strong assumption as $1>\rho(x,y)$ for all $x,y$. How do you know there aren't two pairs of numbers, one of which has a $\rho$ distance greater than $1$, and the other pair less than $1$? Your breakdown of cases doesn't handle this case. Let me see:
$$\sigma(x,z)=\min\{1,\rho(x,z)\}\le\min\{1,\rho(x,y)+\rho(y,z)\}.$$
If you could say that
$$\min\{1,\rho(x,y)+\rho(y,z)\}\le\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\},$$
then you'd be done. Is that the case? Here's where you would need a breakdown by cases - I don't see any way around it. But the cases are the following:

1. Both $\rho(x,y)$ and $\rho(y,z)$ are less than $1$.
2. One of the two $\rho$ distances is less than $1$, and one is greater.
3. Both the $\rho$ distances are greater than $1$.

By symmetry, we can ignore, in the second case, which $\rho$ distance is greater than $1$. I would do a "without loss of generality", and pick one to be bigger, and one smaller.

Does that help?

 

[evinda]

I would say you're solid on everything except the triangle inequality. I don't see any particular reason to make such a strong assumption as $1>\rho(x,y)$ for all $x,y$. How do you know there aren't two pairs of numbers, one of which has a $\rho$ distance greater than $1$, and the other pair less than $1$? Your breakdown of cases doesn't handle this case. Let me see:
$$\sigma(x,z)=\min\{1,\rho(x,z)\}\le\min\{1,\rho(x,y)+\rho(y,z)\}.$$
If you could say that
$$\min\{1,\rho(x,y)+\rho(y,z)\}\le\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\},$$
then you'd be done. Is that the case? Here's where you would need a breakdown by cases - I don't see any way around it. But the cases are the following:

1. Both $\rho(x,y)$ and $\rho(y,z)$ are less than $1$.
2. One of the two $\rho$ distances is less than $1$, and one is greater.
3. Both the $\rho$ distances are greater than $1$.

By symmetry, we can ignore, in the second case, which $\rho$ distance is greater than $1$. I would do a "without loss of generality", and pick one to be bigger, and one smaller.

Does that help?

At the first case, doesn't it always hold that $\min\{1,\rho(x,y)+\rho(y,z)\}=\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\}=\rho(x,y)+\rho(y,z)$ ? (Thinking)

Also do we show the last case for example as follows? (Thinking)
$\min\{1,\rho(x,y)+\rho(y,z)\}=1 \leq 1+1=\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\}$

 

[Ackbach]

At the first case, doesn't it always hold that $\min\{1,\rho(x,y)+\rho(y,z)\}=\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\}=\rho(x,y)+\rho(y,z)$ ? (Thinking)

What about $1=\min(1,0.75+0.75)\overset{?}{=}\min(1,0.75)+\min(1,0.75)=1.5$?

Also do we show the last case for example as follows? (Thinking)

$\min\{1,\rho(x,y)+\rho(y,z)\}=1 \leq 1+1=\min\{1,\rho(x,y)\}+\min\{1,\rho(y,z)\}$

Sure, that works!

 

[evinda]

What about $1=\min(1,0.75+0.75)\overset{?}{=}\min(1,0.75)+\min(1,0.75)=1.5$?

Ah, I see... But how can we explain the first case formally? (Thinking)

Sure, that works!

Great... (Smirk)

 

[Ackbach]

Ah, I see... But how can we explain the first case formally? (Thinking)

Hmm. Well, maybe we broke the cases down incorrectly. What if the cases should be, instead,

1. $\rho(x,y)+\rho(y,z)<1$
2. $\rho(x,y)+\rho(y,z)>1$

I would try a few actual numbers in there to see if you need to further subdivide these cases.