# 1.2. What is order?#

## 1.2.1. Review of sets, relations, and functions#

Exercise 1.10

1. Yes

2. No

3. Yes

Exercise 1.11

1. {}, {1}, {2}, {3}, {1,2}, {1,3}, {2,3}, {1,2,3}

2. $${1} \cup {2} = \{1,2\}$$

3. (h,1), (h,2), (h,3), (1,1), (1,2), (1,3)

4. (h,1), (1,1), (1,2), (2,2), (3,2)

5. h, 1, 2, 3

Exercise 1.16.

1. If there were more than one, then the first and second $$p'$$ would share elements.

2. If there were not one, then the $$p$$ would not partition the original set.

Exercise 1.17.

(11,11), (11,12), (12,11), (12,12), (13,13), (21,21), (22,22), (22,23), (23,22), (23,23)

Exercise 1.20.

1. We would not have constructed a subset if there were no elements to put in it. The definition of a $$(\sim)$$-connected subset also included that it was nonempty.

2. Consider the Contraposition. If the union of the two sets was not empty, then there would be some element in both. If that was the case then the $$(\sim)$$-connected property would have unified the two sets.

3. Every element of $$A$$ must end up in at least one partition. Even if it was equivalent to no other element in the set, it would form a $$(\sim)$$-connected and $$(\sim)$$-closed subset on its own.

Exercise 1.24.

1. $$A = \{1\}$$ and $$B = \{a,b\}$$, with $$F(1) = b$$

2. $$B = \{1\}$$ and $$A = \{a,b\}$$, with $$F(a) = 1$$ and $$F(b) = 1$$

3. Yes, No, No, Yes

4. Not injective or surjective, not because one domain element maps to two different codomain elements, not because one domain element does not map to any codomain element, bijective.

Exercise 1.25.

$$A$$ is empty because every domain element must map to one codomain element, and there are no codomain elements.

Exercise 1.27.

In reading order (top to bottom, left to right):

1. $$F(\bullet) = a$$, $$F(\circ) = a$$, $$F(\ast) = a$$

2. $$F(\bullet) = a$$, $$F(\circ) = a$$, $$F(\ast) = b$$

3. $$F(\bullet) = a$$, $$F(\circ) = b$$, $$F(\ast) = a$$

4. $$F(\bullet) = a$$, $$F(\circ) = b$$, $$F(\ast) = b$$

5. $$F(\bullet) = a$$, $$F(\circ) = b$$, $$F(\ast) = c$$

## 1.2.2. Preorders#

Exercise 1.38.

arrow

source

target

a

1

2

b

1

3

c

1

3

d

2

2

e

2

3

Exercise 1.40.

$P = \{a, b, c, d, e\}$

(1,1) (1,2) (1,3) (1,3) (2,2) (2,3) (3,3) (4,4)

Exercise 1.41.

Yes, as long as there is an implicit reflexive relationship.

Exercise 1.42.

There are 5 relations of an element with itself, 3 relations between the first and second rank, 3 relations between the second and third rank, and 1 between the first and third.

For the meaning of the word “rank” here see Attributes | Graphviz.

Exercise 1.44.

Technically no because you can compare elements to themselves.

Exercise 1.46.

Exercise 1.48.

Yes

Exercise 1.51

The Hasse diagram for $$P(\varnothing)$$ is trivial; there are no subsets. The Hasse diagram for $$P\{1\}$$ and $$P\{1,2\}$$ are:

Example 1.52

The author is defining $$Prt(A)$$, which is rather unclear initially. In this definition, it seems like it would be more natural to call a partition $$P$$ finer than a partition $$Q$$ is the cardinality is smaller. In the second paragraph, notice the definition allows for $$P$$ to have the same cardinality as $$Q$$ (both finer and coarser?) if $$h$$ is an injective/bijective function. If $$h$$ is surjective, then you would get the arguably more natural (strict) definition of fine and coarse.

Exercise 1.53.

The coarsest partition corresponds to a function with a codomain of one element. The finest partition corresponds to an identity function, where every unique element gets its own partition.

Example 1.54.

This question has multiple issues. First some visual notes:

The author’s definition of an “upper set” is any of these 4 sets (call each $$U$$) where if $$p \in U$$ and $$p \leq q$$, then $$q \in U$$. He intends us to go through all four of these sets and check if this definition holds for them. The primary issue with this example is that his definition is just plain wrong. Consider the definition applied to the $$\{false\}$$ set (calling it $$U$$); the author is claiming that this set is an upper set if $$false \in \{false\}$$ in and $$false \leq q$$, then $$q \in U$$. When does the author even define $$q$$? If we define $$q$$ to be $$\{false\}$$ then there’s nothing wrong with this set.

To avoid reverse-engineering the definition from the author’s explanation of why $$\{false\}$$ is not an upper set, see Upper set. Using the variables that the author uses rather than Wikipedia’s, the definition of an upper set is then a set where for all $$q \in U$$ and $$p \in P$$, if $$q \leq p$$, then $$p \in U$$. An even better definition for this context is that an upper set is a set where for all $$u \in U$$ and $$p \in P$$, if $$u \leq p$$, then $$p \in U$$. With either of these new definitions, the author’s explanation of why $$\{false\}$$ is not an upper set makes sense.

Exercise 1.55.

A discrete preorder has no relationship between any elements, so every element on its own is an upper set. Adding a second element to any of these sets also creates an upper set, because again there is no relationship between elements. You can continue this with a third and fourth element, in any order, constructing in the end all of the possible subsets of $$X$$.

Exercise 1.57.

The issues this question has are covered in order theory - Understanding upper set preorders - Math SE and multiple comments in the errata.

## 1.2.3. Monotone maps#

Exercise 1.63.

Taking step 1. from Example 1.50:

Exercise 1.65.

Exercise 1.66.

This question is a notational nightmare; see notation - Making sense of a map, given a new domain and codomain - Math SE for someone else’s helpful comments on it.

In 1. let’s call the ↑ operator the “upper closure” of an element $$p$$ in $$P$$ (following the language in Upper set). The definition constructs an upper set because it includes in its construction all elements that make $$p$$ satisfy the requirements of being an element of an upper set. Because the ordering relationship is transitive, any other element included in the set by this construction will also have the requirements for it to be an element of an upper set satisfied.

Part 2. of the question introduces the notation $$P^{op}$$ out of nowhere to refer to the opposite preorder; see this notation used in Duality (order theory) and Opposite category.

The given construction will map every element $$p \in P^{op}$$ to a set $$s \in U(P)$$ where $$s$$ is an “upper” set that includes $$p$$. With respect to $$P^{op}$$ this set will be the opposite of an “upper” set; what we’ll call a “lower” set with respect to $$p$$.

Remove the first “if” from part 3. of the question to make it understandable/readable. To prove the forward case, let’s use the definition in 1.:

$\begin{split} \uparrow p = A = \{a\in P\mid p\leq a\} \\ \uparrow p' = B = \{b\in P\mid p'\leq b\} \end{split}$

So if $$p \leq p'$$ we have by direct proof that:

$\uparrow p' = B = \{b\in P\mid p\leq p'\leq b\} \subseteq \{a\in P\mid p\leq a\} = A = \uparrow p$

The opposite implication is left as an exercise for the reader.

Part 4. of this question should refer to Exercise 1.57 rather than Example 1.56:

Exercise 1.67.

The monotonicity requirement is that for all $$x,y \in A$$, if $$x \leq_{A} y$$ then $$f(x) \leq_B f(y)$$. The condition of the if statement is always false for a discrete preorder, so any function will satisfy it.

Exercise 1.69.

Define the sets $$X$$ and $$Y$$:

$\begin{split} X = \{0,1,2,3\} \\ Y = \{A,B,C\} \end{split}$

Define the surjective function $$f$$:

$\begin{split} f(0) = A \\ f(1) = A \\ f(2) = B \\ f(3) = C \end{split}$

See Partition of a set for notation. Two different partitions $$P$$ and $$Q$$:

\begin{split} \begin{align} \\ P & = \{\{A,B\},\{C\}\} = A B | C \\ p(A) & = M \\ p(B) & = M \\ p(C) & = N \\ \end{align} \end{split}
\begin{split} \begin{align} \\ Q & = \{\{A\},\{B,C\}\} = A | B C \\ q(A) & = M \\ q(B) & = N \\ q(C) & = N \\ \end{align} \end{split}

The composite $$f;s$$ for $$P$$ is:

\begin{split} \begin{align} \\ f;p(0) & = M \\ f;p(1) & = M \\ f;p(2) & = M \\ f;p(3) & = N \\ f^\ast(P) & = 0 1 2 | 3 \\ \end{align} \end{split}

The composite $$f;s$$ for $$Q$$ is:

\begin{split} \begin{align} \\ f;q(0) & = M \\ f;q(1) & = M \\ f;q(2) & = N \\ f;q(3) & = N \\ f^\ast(Q) & = 0 1 | 2 3 \\ \end{align} \end{split}

Exercise 1.71.

The identity function is monotone because if $$x \leq y$$ then $$id_X(x) \leq id_X(y)$$.

The composition of monotone functions is monotone because if $$x \leq y$$ then $$f;g(x) \leq f;g(y)$$.

Exercise 1.73.

Calling a preorder skeletal removes all equivalence sets (partitions), making every element its own equivalence set (partition). This is a discrete preorder.

Definition 1.75.

Exercise 1.77.

The function:

$\begin{split} \phi(\ast \bullet \circ) = true \\ \phi(\ast | \bullet \circ) = false \\ \phi(\ast \bullet | \circ) = true \\ \phi(\bullet | \ast \circ) = false \\ \phi(\ast | \bullet | \circ) = false \end{split}$

The preorder:

$\begin{split} \phi(\ast | \bullet \circ) \leq \phi(\ast \bullet \circ) \\ \phi(\ast \bullet | \circ) \leq \phi(\ast \bullet \circ) \\ \phi(\bullet | \ast \circ) \leq \phi(\ast \bullet \circ) \\ \phi(\ast | \bullet | \circ) \leq \phi(\ast | \bullet \circ) \\ \phi(\ast | \bullet | \circ) \leq \phi(\ast \bullet | \circ) \\ \phi(\ast | \bullet | \circ) \leq \phi(\bullet | \ast \circ) \\ \end{split}$

By substitution:

$\begin{split} false \leq true \\ true \leq true \\ false \leq true \\ false \leq false \\ false \leq true \\ false \leq false \\ \end{split}$

Exercise 1.79.

The second paragraph should start (remove an “a”):

Viewing upper sets as monotone maps to …

The function $$u$$ here defines an upper set by labeling all elements of $$Q$$ as either belonging to it or not. The composition $$f;u$$ can then be understood as mapping an element of $$P$$ to an element in $$Q$$, and again labeling it as belonging to the upper set or not.