>**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\\).

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\\).

---

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\\).