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# Exercise 75 - Chapter 2

edited June 2018

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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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: $(\mathcal{X} \times \mathcal{Y})((5, 6), (-1, 4)) = 8$