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Thread: metrics - exam tomorrow!

  1. #1
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    metrics - exam tomorrow!

    Let d denote the usual metric on R and consider the metric e on R
    given by e(x, y) = |x − y|/(1 + |x − y|). (You may assume e is a
    metric.)

    In (R, e), show that B(0, 1) = R, and find B(0,1/3).

    Can someone please explain simply what I would have to do to get 4 marks?

    Thanks in advance!
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    Have you drawn a graph of $\displaystyle \frac{{\left| x \right|}}{{1 + \left| x \right|}}$?
    It is essentially the same as that of the metric $\displaystyle e$, WHY?
    What are the upper and lower bounds?
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by hunkydory19 View Post
    Let d denote the usual metric on R and consider the metric e on R
    given by e(x, y) = |x − y|/(1 + |x − y|). (You may assume e is a
    metric.)

    In (R, e), show that B(0, 1) = R, and find B(0,1/3).

    Can someone please explain simply what I would have to do to get 4 marks?

    Thanks in advance!
    Notice that $\displaystyle \forall x, \forall y$, $\displaystyle e(x,y)<1$. Thus, $\displaystyle \forall~ x\in\mathbb{R}, e(0,x)<1$ and it follows that $\displaystyle B(0,1)$ will contain every point in $\displaystyle \mathbb{R}$.

    For the second part, $\displaystyle y=0$ and you need to find all the $\displaystyle x$ such that $\displaystyle \frac{|x|}{1+|x|}<\frac{1}{3}$.

    $\displaystyle 3|x|<1+|x| \implies 2|x|<1 \implies |x|<\frac{1}{2}$

    So $\displaystyle B\left(0,\frac{1}{3}\right)=\left(-\frac{1}{2},\frac{1}{2}\right)$
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