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Exercise 75 - Chapter 2

Consider \( \mathbb{R} \) as a Lawvere metric space, i.e. as a Cost-category (see Example 2.51). Form the Cost-product \( \mathbb{R} \times \mathbb{R} \). What is the distance from \( (5, 6) \) to \( (−1, 4) \)?

Hint: apply Definition 2.71; the answer is not \( \sqrt{40} \).

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  • 1.
    edited July 2018

    Form the Cost-product \( \mathbb{R} \times \mathbb{R} \):

    1. \( Ob(\mathcal{X}) = \mathbb{R} \times \mathbb{R} \)
    2. \( (\mathcal{X} \times \mathcal{Y})((x_1, y_1), (x_2, y_2)) = \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_1, y_2) = |x_1 - x_2| + |y_1 - y_2| \)

    We end up with L1-distance instead of (perhaps expected) L2-distance: $$ (\mathcal{X} \times \mathcal{Y})((5, 6), (-1, 4)) = 8 $$

    Comment Source:Form the **Cost**-product \\( \mathbb{R} \times \mathbb{R} \\): 1. \\( Ob(\mathcal{X}) = \mathbb{R} \times \mathbb{R} \\) 2. \\( (\mathcal{X} \times \mathcal{Y})((x_1, y_1), (x_2, y_2)) = \mathcal{X}(x_1, x_2) \otimes \mathcal{Y}(y_1, y_2) = |x_1 - x_2| + |y_1 - y_2| \\) We end up with L1-distance instead of (perhaps expected) L2-distance: \[ (\mathcal{X} \times \mathcal{Y})((5, 6), (-1, 4)) = 8 \]
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