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Math Help - Metric Space

  1. #1
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    Metric Space

    The topologies on \mathbb{R}^n induced by the euclidean metric d and the square metric rho are the same as the product topology on \mathbb{R}^n.

    Euclidean metric
    d(x,y)=\sqrt{(x_1-y_1)^2+\cdots +(x_n-y_n)^2}

    Square metric
    \rho(x,y)=\max\{|x_1-y_1|,\cdots ,|x_n-y_n|\}

    So proof says:

    Let x=(x_1,\cdots ,x_n) and y=(y_1,\cdots ,y_n) be two points in \mathbb{R}^n. It is simple algebra to check that

    \rho(x,y)\leq d(x,y)\leq\sqrt{n}\rho(x,y)

    (1) Where and why did they have this inequality?
    (2) I know it is stated that it is simple algebra to check it, but I don't see it. How is checked?
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  2. #2
    Super Member PaulRS's Avatar
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    Re: Metric Space

    Well : \rho (x,y) = \max \{|x_1-y_1| ,..., |x_n - y_n|\} = |x_j - y_j| for some j\in \{1,...,n\} (such a j always exists)

    Now: \rho(x,y) = |x_j - y_j| = \sqrt{(x_j - y_j)^2} \leq \sqrt{ (x_1-y_1)^2 +... + (x_j - y_j)^2 + ... (x_n-y_n)^2} since the square root function is increasing and the squares are non-negative.

    Finally (x_1-y_1)^2 = |x_1 - y_1|^2 \leq |x_j-y_j|^2 = (x_j-y_j)^2 and so on ... (x_n-y_n)^2 \leq (x_j - y_j)^2, thus:

    (x_1-y_1)^2 + ... + (x_n-y_n)^2 \leq n \cdot (x_j  - y_j)^2 and taking square roots on both sides we get the other inequality.

    (1) They probably want to prove that the metrics are equivalent... by showing that the identity between them is a homeomorphism (hence they give rise to the same topology). And going through these inequalities is a simple mean of doing so.
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  3. #3
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    Re: Metric Space

    Quote Originally Posted by PaulRS View Post
    Well : \rho (x,y) = \max \{|x_1-y_1| ,..., |x_n - y_n|\} = |x_j - y_j| for some j\in \{1,...,n\} (such a j always exists)

    Now: \rho(x,y) = |x_j - y_j| = \sqrt{(x_j - y_j)^2} \leq \sqrt{ (x_1-y_1)^2 +... + (x_j - y_j)^2 + ... (x_n-y_n)^2} since the square root function is increasing and the squares are non-negative.

    Finally (x_1-y_1)^2 = |x_1 - y_1|^2 \leq |x_j-y_j|^2 = (x_j-y_j)^2 and so on ... (x_n-y_n)^2 \leq (x_j - y_j)^2, thus:

    (x_1-y_1)^2 + ... + (x_n-y_n)^2 \leq n \cdot (x_j  - y_j)^2 and taking square roots on both sides we get the other inequality.

    (1) They probably want to prove that the metrics are equivalent... by showing that the identity between them is a homeomorphism (hence they give rise to the same topology). And going through these inequalities is a simple mean of doing so.
    Continue the book says:

    The first inequality shows that

    B_d(x,\epsilon)\subset B_{rho}(x,\epsilon)

    How do they get this from the inequality? What is consider the first? Are the reading left to right?
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  4. #4
    Super Member PaulRS's Avatar
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    Re: Metric Space

    They meant the left-most part, yes.

    Pick an element of y\in B_d (x,\epsilon) , then by definition d(x,y)<\epsilon.

    But \rho(x,y)\leq d(x,y) < \epsilon hence y\in B_{\rho}(x,\epsilon) .

    Thus in all : B_d(x,\epsilon) \subset B_{\rho}(x,\epsilon).
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  5. #5
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    Re: Metric Space

    Quote Originally Posted by PaulRS View Post
    They meant the left-most part, yes.

    Pick an element of y\in B_d (x,\epsilon) , then by definition d(x,y) \leq \epsilon.

    But \rho(x,y)\leq d(x,y) \leq \epsilon hence y\in B_{\rho}(x,\epsilon) .

    Thus in all : B_d(x,\epsilon) \subset B_{\rho}(x,\epsilon).
    For the last inequality, why did they divide epsilon by

    B_{\rho}\left(x,\frac{\epsilon}{\sqrt{n}}\right) \subset B_d(x,y)
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  6. #6
    Super Member PaulRS's Avatar
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    Re: Metric Space

    Remember that d(x,y)\leq \sqrt{n} \cdot \rho (x,y), and try to prove the inclusion as I did for the other one.

    (that \sqrt{n} multiplying there obliges us to 'contract' the radius a bit)

    PS: You meant to say \epsilon and not y in B_d(x,y) , I suppose.
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