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Math Help - Extended Reals

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    MHF Contributor Mathstud28's Avatar
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    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.
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    Quote Originally Posted by Mathstud28 View Post
    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?
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    Quote Originally Posted by ThePerfectHacker View Post
    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.
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    Moo
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    Hello,

    Another example : how would you define d(\infty,\infty)=|\infty-\infty| ? isn't it supposed to be 0 by definition of a metric ?
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Moo View Post
    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.
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    Moo
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    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 !
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    Quote Originally Posted by Mathstud28 View Post
    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.
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    Moo
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    Quote Originally Posted by ThePerfectHacker View Post
    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
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    Quote Originally Posted by Moo View Post
    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.
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    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    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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