>**Puzzles 1 and 3.** What is a "poset" according to the book and what mathematicians usually call it?

A set \$$S\$$ with a relation \$$R \subseteq S \times S\$$, denoted \$$\le\$$, such that:

- For all \$$x \in S\$$, \$$x \le x\$$, that is, the relation is _reflexive_
- For all \$$x, y, z \in S\$$, if \$$x \le y\$$ and \$$y \le z\$$, then \$$x \le z\$$, that is, the relation is _transitive_

>**Puzzle 2.** How does their definition differ from the usual definition?

A set \$$S\$$ with a relation \$$R \subseteq S \times S\$$, denoted \$$\le\$$, such that:

- For all \$$x \in S\$$, \$$x \le x\$$, that is, the relation is _reflexive_
- For all \$$x, y, z \in S\$$, if \$$x \le y\$$ and \$$y \le z\$$, then \$$x \le z\$$, that is, the relation is _transitive_
- For all \$$x, y \in S\$$, if \$$x \le y\$$ and \$$y \le x\$$, then \$$x = y\$$, that is, the relation is _antisymmetric_

>**Puzzle 4.** List an interesting example of a _preordered_ set.

Consider the set \$$\Lambda\$$ of untyped lambda calculus terms under the reflexive, transitive closure of the following relation:

Let \$$a, b \in \Lambda\$$, then \$$a \sim b\$$ if \$$a\$$ beta reduces to \$$b\$$.

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A total order is a partial order with the additional property:

- For all \$$x, y \in S\$$, \$$x \le y\$$ or \$$y \\le x\$$ (or both!).

>**Puzzle 5.** Why is this property called "trichotomy"?

Define \$$x < y\$$ as \$$x \le y \land x \ne y\$$. Then, by the property above, \$$x < y\$$, \$$x = y\$$, or \$$x > y\$$ for all \$$x, y \in S\$$ (note that the "or" is exclusive here!).

>**Puzzle 11.** Show that if the monotone function \$$f : A \to B\$$ has an inverse \$$g : B \to A\$$ that is also a monotone function, then \$$g\$$ is both a right adjoint and a left adjoint of \$$f\$$.

Let \$$a \in A\$$ and \$$b \in B\$$ such that \$$f(a) \le b\$$. Since \$$g\$$ is monotone, we have \$$g(f(a)) \le g(b)\$$. Since \$$g\$$ is the inverse of \$$f\$$, we have \$$a \le g(b)\$$.

Let \$$a \in A\$$ and \$$b \in B\$$ such that \$$a \le g(b)\$$. Since \$$f\$$ is monotone, we have \$$f(a) \le f(g(b))\$$. Since \$$f\$$ is the inverse of \$$g\$$, we have \$$f(a) \le b\$$.