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# Exercise 5 - Chapter 2

edited June 2018

Consider again the preorder $$( \mathbb{R} , \le)$$.

Someone proposes 1 as a monoidal unit and $$*$$ (usual multiplication) as a monoidal product. But an expert walks by and says “that won’t work.”

Figure out why, or prove the expert wrong!

Example 2.4. There is a well-known preorder structure, denoted $$\le$$, on the set $$\mathbb{R}$$ of real numbers; e.g. $$−5 \le 2$$ .

Definition 2.2. A symmetric monoidal structure on a preorder $$(X, \le)$$ consists of two constituents:

1. an element $$I \in X$$, called the monoidal unit, and
2. a function $$\otimes : X \times X \rightarrow X$$, called the monoidal product.

These constituents must satisfy the following properties:

1. (a) for all $$x_1 , x_2 , y_1 , y_2 \in X, \text{ if } x_1 \le y_1 \text{ and } x_2 \le y_2 , \text{ then } x_1 \otimes x_2 \le y_1 \otimes y_2$$
2. (b) for all $$x \in X$$, the equations $$I \otimes x = x$$ and $$x \otimes I = x$$ hold
3. (c) for all $$x, y, z \in X$$, the equation $$(x \otimes y) \otimes z = x \otimes (y \otimes z)$$ holds, and
4. (d) for all $$x, y \in X$$, the equivalence $$x \otimes y \cong y \otimes x$$ holds.

A preorder equipped with a symmetric monoidal structure, $$(X, \le, I, \otimes)$$, is called a symmetric monoidal preorder.

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1.

Puzzle FE-1 Would $$I = 1$$ and $$\otimes = *$$ with the preorder $$(\mathbb{R}^+, \le)$$ form a symmetric monoidal preorder?

Comment Source:**Puzzle FE-1** Would \$$I = 1 \$$ and \$$\otimes = \* \$$ with the preorder \$$(\mathbb{R}^+, \le) \$$ form a symmetric monoidal preorder?
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2.
edited May 2018

Let $$x_1, x_2, y_1, y_2$$ be $$-2, -3, 2, 1$$, respectively. We have $$−2 \le 2$$ and $$-3 \le 1$$. However, $$(-2) ∗ (-3)$$ is greater than $$2 ∗ 1$$: thus, condition 1 of Definition 2.1 is not satisfied. This is similar to Puzzle 61. I'm tempted to say that, if we consider positive reals, the answer is yes -- I did not find any counterexample yet.

Comment Source:Let \$$x_1, x_2, y_1, y_2\$$ be \$$-2, -3, 2, 1\$$, respectively. We have \$$−2 \le 2 \$$ and \$$-3 \le 1\$$. However, \$$(-2) ∗ (-3) \$$ is greater than \$$2 ∗ 1 \$$: thus, condition 1 of Definition 2.1 is not satisfied. This is similar to Puzzle 61. I'm tempted to say that, if we consider positive reals, the answer is yes -- I did not find any counterexample yet.