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Thread: Use the triangle inequality to show...

  1. #1
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    Use the triangle inequality to show...

    Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

    I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

    ----- (-e)-------a-------(e)------

    -----(-e)--------b-------(e)-----
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    Use the triangle inequality to show that if a and be are in R, with a not equal to b, then there exists an open interval U centered at a and V centered at b, both of radius epsilon= 1/2 abs[a-b] , with U/\V= 0 (intersection)

    I really don't know what I wanna do here. I'm imagining a line centered at A and another at b with radius epsilon something like this =>

    ----- (-e)-------a-------(e)------

    -----(-e)--------b-------(e)-----
    We may assume wlog that $\displaystyle a<b$. Thus the open interval around $\displaystyle a$ is $\displaystyle U=\left(a-\frac{1}{2}(b-a),a+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(3a-b),\frac{1}{2}(a+b)\right)$

    The open interval around $\displaystyle b$ is $\displaystyle V=\left(b-\frac{1}{2}(b-a),b+\frac{1}{2}(b-a)\right)=\left(\frac{1}{2}(a+b),\frac{1}{2}(-a+3b)\right)$

    From this it's clear that $\displaystyle U$ and $\displaystyle V$ are disjoint, so $\displaystyle U\cap V=\emptyset$.
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  3. #3
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    what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
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  4. #4
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    what about showing the interval for both U and V to be of radius epsilon= 1/2*absolute value[a-b]?
    It doesn't have to be; take any $\displaystyle c\leq\frac{1}{2}$, and taking the radius to be $\displaystyle c|a-b|$, will guarantee $\displaystyle U\cap V=\emptyset$.
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  5. #5
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    I don't really follow...how come we can assume that a < b ?
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  6. #6
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    I don't really follow...how come we can assume that a < b ?
    One of them has to be greater than the other. If it's not $\displaystyle a$, then it's $\displaystyle b$. The reason we can assume wlog is that if we assumed $\displaystyle b<a$, we would still get the same result.
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  7. #7
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    and we make this assumption so we can relate U and V?
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  8. #8
    Super Member redsoxfan325's Avatar
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    I made the assumption so I didn't have to deal with the absolute value sign.
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  9. #9
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    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R

    then absloutevalue(x-a) < e = 1/2absloute(a-b)
    absloute value(x-b) < e = 1/2absloute(a-b)

    is this correct? and if so....how does he go from abs(c-a) < e to
    abs(x-a) < e
    and then to = 1/2 abs(a-b)
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  10. #10
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R
    $\displaystyle U$ is a set; $\displaystyle |c-a|$ is a number. How can those be equal?
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  11. #11
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    ok i should have written
    { c in R : abs(c-x) < e }
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  12. #12
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by mtlchris View Post
    ok I follow this, but if I need to use the triangle property how do i go about using it?
    A friend told me:
    let u= (a-e, a+e) = absolute value(c-a) < e for c in R
    let v= (b-e, b+e) = absolute value(d-a) < e for d in R

    then absloutevalue(x-a) < e = 1/2absloute(a-b)
    absloute value(x-b) < e = 1/2absloute(a-b)

    is this correct? and if so....how does he go from abs(c-a) < e to
    abs(x-a) < e
    and then to = 1/2 abs(a-b)
    Your friend introduced more variables than are necessary (namely $\displaystyle c$ and $\displaystyle d$). If $\displaystyle U=\{x\in\mathbb{R}:|a-x|<\epsilon\}$ and $\displaystyle V=\{x\in\mathbb{R}:|x-b|<\epsilon\}$, then assume for a contradiction that $\displaystyle U\cap V\neq\emptyset$; that is, $\displaystyle \exists x$ such that $\displaystyle x\in U$ and $\displaystyle x\in V$. Then $\displaystyle |a-x|<\epsilon$ and $\displaystyle |x-b|<\epsilon$.

    Then by the triangle inequality $\displaystyle |a-b|\leq|a-x|+|x-b|<\epsilon+\epsilon=2\epsilon=|a-b|$.

    So we have $\displaystyle |a-b|<|a-b|$, which is a contradiction. So $\displaystyle U\cap V=\emptyset$.
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