1. ## Extended Reals

Quick question. The metric $d(x,y)=|x-y|$ does not metrize $\mathbb{R}\cup\left\{-\infty,\infty\right\}$ right? Specifically because it fails the triangle inequality.

2. Originally Posted by Mathstud28
Quick question. The metric $d(x,y)=|x-y|$ does not metrize $\mathbb{R}\cup\left\{-\infty,\infty\right\}$ right? Specifically because it fails the triangle inequality.
Well to be a metric space you need $d: \mathbb{R}_{\infty} \to \mathbb{R}$.
How do you define $d(\infty,0)$ for example?

3. Originally Posted by ThePerfectHacker
Well to be a metric space you need $d: \mathbb{R}_{\infty} \to \mathbb{R}$.
How do you define $d(\infty,0)$ for example?
I would define $d(\infty,a)=\infty$ for any $a\in\mathbb{R}$. I believe that is the appropriate extension of the metric defined on $\mathbb{R}$, but then that implies that normal metic for the reals does not metrize the extended reals. Does that make sense?

P.S. Sorry about putting this in the wrong forum, the question arose in a topology setting but I can see how its more of an analysis question.

4. Hello,

Another example : how would you define $d(\infty,\infty)=|\infty-\infty|$ ? isn't it supposed to be 0 by definition of a metric ?

5. Originally Posted by Moo
Hello,

Another example : how would you define $d(\infty,\infty)=|\infty-\infty|$ ? isn't it supposed to be 0 by definition of a metric ?
Another good example! . So that settles it, I firmly believe that the Euclidean metric $d(x,y)=|x-y|$ does not metrize the extended reals.

6. Extended real number line - Wikipedia, the free encyclopedia

There is no metric which is an extension of the ordinary metric on R.
But the sentence before can be interesting too. However, homeomorphisms represent a rude language to me So I'll leave it to you to understand that !

7. Originally Posted by Mathstud28
I would define $d(\infty,a)=\infty$ for any $a\in\mathbb{R}$. I believe that is the appropriate extension of the metric defined on $\mathbb{R}$, but then that implies that normal metic for the reals does not metrize the extended reals. Does that make sense?
No! That is not good. In order to be a metric space it needs to send two points in $\mathbb{R}_{\infty}$ into $\mathbb{R}$ not $\mathbb{R}_{\infty}$.
Thus, $d(\infty,0)$ must be a real number.

8. Originally Posted by ThePerfectHacker
No! That is not good. In order to be a metric space it needs to send two points in $\mathbb{R}_{\infty}$ into $\mathbb{R}^{\color{red}+}$ not $\mathbb{R}_{\infty}$.
Thus, $d(\infty,0)$ must be a real number.
A positive real number

9. Originally Posted by Moo
A positive real number
If $d: X\to \mathbb{R}^+$ then it is still correct to write $d:X\to \mathbb{R}$.
However, I should have been more precise, saying "positive" is better.

10. Originally Posted by ThePerfectHacker
No! That is not good. In order to be a metric space it needs to send two points in $\mathbb{R}_{\infty}$ into $\mathbb{R}$ not $\mathbb{R}_{\infty}$.
Thus, $d(\infty,0)$ must be a real number.
Yes TPH, if you would have actually read my post you would note this is exactly what I was saying, the metric endowed to the reals is not sufficient for the extended reals.