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Math Help - Metric Spaces

  1. #1
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    Metric Spaces

    Let X=\bf R^3 and define the metric d: \bf R^3 \times \bf R^3 \rightarrow \bf R by
    d((x_1,y_1,z_1),(x_2,y_2,z_2)) = \max\left\{|x_1-x_2|, |y_1-y_2|, |z_1-z_2|\right\}.
    Describe the neighborhood N(0; 1), where  0 is the origin in \bf R^3

    My professor says that it's an open cube with vertices at (1, 1, 1), but I can't see why it's not an open sphere with radius 1.

    I'm thinking N(0; 1) = \left\{y \in X: d(0,y) <1\right\}. Wouldn't there be points whose distance from the origin exceed 1 in an open cube?
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  2. #2
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    Re: Metric Spaces

    "Distance" in this case means the metric function, not "ordinary distance". Let's take a point "as far away as possible" from the origin, say (1,1,1).

    Then d((0,0,0),(1,1,1)) = max{|0 -1|,|0-1|,|0-1|} = max{|-1|,|-1|,|-1|} = max{1,1,1} = 1.

    Convince yourself that if (x,y,z) is any point with two of the 3 coordinates with absolute value less than 1, and the third coordinate with absolute value equal to 1, that d((0,0,0),(x,y,z)) = 1, as well.

    Hopefully this allows you to see that the boundary of N(0,1) is precisely the faces of said cube, so all the interior points comprise N(0,1).
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  3. #3
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    Re: Metric Spaces

    Quote Originally Posted by MadSoulz View Post
    Let X=\bf R^3 and define the metric d: \bf R^3 \times \bf R^3 \rightarrow \bf R by
    d((x_1,y_1,z_1),(x_2,y_2,z_2)) = \max\left\{|x_1-x_2|, |y_1-y_2|, |z_1-z_2|\right\}.
    Describe the neighborhood N(0; 1), where  0 is the origin in \bf R^3
    My professor says that it's an open cube with vertices at (1, 1, 1), but I can't see why it's not an open sphere with radius 1.
    I'm thinking N(0; 1) = \left\{y \in X: d(0,y) <1\right\}. Wouldn't there be points whose distance from the origin exceed 1 in an open cube?
    Consider the vertex $V=(-1,1,-1)$ that is not in the "open cube" ${\frak{B}}(0;1)$$ = \{X| d(0,X)<1\}$

    In that metric $d(0,V)=1$. Correct? So $V$ is on the boundary of that cube. SO?
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    Re: Metric Spaces

    Quote Originally Posted by Plato View Post
    Consider the vertex $V=(-1,1,-1)$ that is not in the "open cube" ${\frak{B}}(0;1)$$ = \{X| d(0,X)<1\}$

    In that metric $d(0,V)=1$. Correct? So $V$ is on the boundary of that cube. SO?
    So any vertex V such that d(0,V)=1 is not in N(0;1). Which is why N(0;1)is an open cube with all such vertices.
    Last edited by MadSoulz; March 2nd 2014 at 08:15 AM.
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    Re: Metric Spaces

    Quote Originally Posted by Deveno View Post
    Convince yourself that if (x,y,z) is any point with two of the 3 coordinates with absolute value less than 1, and the third coordinate with absolute value equal to 1, that d((0,0,0),(x,y,z)) = 1, as well.

    Hopefully this allows you to see that the boundary of N(0,1) is precisely the faces of said cube, so all the interior points comprise N(0,1).
    I can see it now. Thanks for clearing that up.
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