Extended Reals

**Mathstud28** · January 22nd 2009, 04:43 PM

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.

**ThePerfectHacker** · January 22nd 2009, 08:50 PM

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?

**Mathstud28** · January 22nd 2009, 09:40 PM

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.

**Moo** · January 22nd 2009, 11:08 PM

Hello,

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

**Mathstud28** · January 22nd 2009, 11:20 PM

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.

**Moo** · January 22nd 2009, 11:49 PM

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 !

**ThePerfectHacker** · January 23rd 2009, 07:14 AM

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.

**Moo** · January 23rd 2009, 10:06 AM

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

**ThePerfectHacker** · January 23rd 2009, 10:14 AM

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.

**Mathstud28** · January 23rd 2009, 01:00 PM

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.

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