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Math Help - Euclidean Spaces

  1. #1
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    Euclidean Spaces

    Let's say if we have  \bold{x} \in \mathbb{R}^{n} and we want to show that  B_{r}(\bold{x}) (open ball of radius  r >0 ) is convex. We know that  d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}| (e.g. the norm).

    We can still apply the triangle inequality right? So we have:  |t(\bold{y}-\bold{x})+(1-t)(\bold{z}-\bold{x})| \leq t|\bold{y}-\bold{x}| + (1-t)|\bold{z}-\bold{x}| < r .


    In other words, does  d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}| only apply to  \mathbb{R} ? Or does it apply to general  \mathbb{R}^{n} so that we can invoke the triangle inequality? We can use the triangle inequality to prove the general case (which is what I did) right?
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    In other words, does  d(\bold{x}, \bold{y}) = |\bold{x}- \bold{y}| only apply to  \mathbb{R} ? Or does it apply to general  \mathbb{R}^{n} so that we can invoke the triangle inequality? We can use the triangle inequality to prove the general case (which is what I did) right?
    Perhaps I should not attempt an answer because I may have missed your point.
    But one of the requirements for a metric is the triangle inequality holds.
    That is, if d is a metric on a space then d(x,y) \leqslant d(x,z) + d(z,y) for all x, y, & z in the space.
    Does that get at what you are asking?
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  3. #3
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    yes, thanks.
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  4. #4
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    how about the euclidean norm? does it hold for that?
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  5. #5
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    Quote Originally Posted by particlejohn View Post
    how about the euclidean norm? does it hold for that?
    Yes it does. Here is how that goes.
    \left\{ {y,z} \right\} \subseteq B_r (x)\; \Rightarrow \;\left\| {y - x} \right\| < r\;\& \;\left\| {z - x} \right\| < r
    \begin{array}{*{20}c}<br />
   {\left\| {\lambda y\_(1 - \lambda )z - x} \right\|} &  =  & {\left\| {\lambda (y - x) + (1 - \lambda )(z - x)} \right\|}  \\<br />
   {} &  \leqslant  & {\lambda \left\| {y - x} \right\| + (1 - \lambda )\left\|{z - x} \right\|}  \\   {} &  <  & {\lambda r + (1 - \lambda )r}  \\   {} &  =  & r  \\ \end{array}
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  6. #6
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    so the triangle inequality applies to the following:  |\bold{x}| = \sqrt{x_{1}^{2} + \cdots x_{n}^{2}} ?

    The triangle inequality works for all metrics on Euclidean spaces? But not on say L^{p} spaces?
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  7. #7
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    I think that you are confusing yourself: the idea of a norm on a linear space with the distance function or metric.
    Have a look at these pages:
    Norm -- from Wolfram MathWorld
    Vector Norm -- from Wolfram MathWorld
    In the case of a ‘norm’ as with a metric, the triangle inequality is a requirement.
    I think that you need to go to the instructor and clear away this confusion.
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  8. #8
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    yeah even Rudin agrees.
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