# LARA

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# Sets and Relations

## Sets

Sets are unordered collection of elements.

We denote a finite set containing only elements , and by . The order and number of occurrences does not matter: .

• iff

Empty set: . For every we have .

To denote large or infinite sets we can use set comprehensions: is set of all objects with property . $ y \in \{ x. P(x) \} \ \leftrightarrow\ P(y)$

Notation for set comprehension:

Sometimes the binder can be inferred from context so we write simply . In general there is ambiguity in which variables are bound. (Example: what does the in refer to in the expression: $ \{a \} \cup \{ f(a,b) \mid P(a,b) \}$ does it refer to the outerone as in or is it a newly bound variable? The notation with dot and bar resolves this ambiguity.

Subset: means

$ A \cup B = \{ x. x \in A \lor x \in B \}$ $ A \cap B = \{ x. x \in A \land x \in B \}$ $ A \setminus B = \{ x. x \in A \land x \notin B \}$

Boolean algebra of subsets of some set (we define ):

• are associative, commutative, idempotent
• neutral and zero elements: ,
• absorption: ,
• deMorgan laws: ,
• complement as partition of universal set: ,
• double complement:

Which axioms are sufficient?

### Infinte Unions and Intersections

Note that sets can be nested. Consider, for example, the following set $ S = \{ \{ p, \{q, r\} \}, r \}$ This set has two elements. The first element is another set. We have . Note that it is not the case that

Suppose that we have a set that contains other sets. We define union of the sets contained in as follows: $ \bigcup B = \{ x.\ \exists a. a \in B \land x \in a \}$ As a special case, we have $ \bigcup \{ a_1, a_2, a_3 \} = a_1 \cup a_2 \cup a_3$ Often the elements of the set are computed by a set comprehension of the form . We then write $ \bigcup_{i \in J} f(i)$ and the meaning is $ \bigcup \{ f(i).\ i \in J \}$ Therefore, is equivalent to .

We analogously define intersection of elements in the set: $ \bigcap B = \{ x. \forall a. a \in B \rightarrow x \in a \}$ As a special case, we have $ \bigcap \{ a_1, a_2, a_3 \} = a_1 \cap a_2 \cap a_3$ We similarly define intersection of an infinite family $ \bigcap_{i \in J} f(i)$ and the meaning is $ \bigcap \{ f(i).\ i \in J \}$ Therefore, is equivalent to .

## Relations

Pairs: $ (a,b) = (u,v) \iff (a = u \land b = v)$ Cartesian product: $ A \times B = \{ (x,y) \mid x \in A \land y \in B \}$

Relations is simply a subset of , that is .

Note: $ A \times (B \cap C) = (A \times B) \cap (A \times C)$ $ A \times (B \cup C) = (A \times B) \cup (A \times C)$

#### Diagonal relation

, is given by $ \Delta_A = \{(x,x) \mid x \in A\}$

### Set operations

Relations are sets of pairs, so operations apply.

### Relation Inverse

$ r^{-1} = \{(y,x) \mid (x,y) \in r \}$

### Relation Composition

$ r_1 \circ r_2 = \{ (x,z) \mid \exists y. (x,y) \in r_1 \land (y,z) \in r_2\}$

Note: relations on a set together with relation composition and form a monoid structure: $\begin{array}{l}  r_1 \circ (r_2 \circ r_3) = (r_1 \circ r_2) \circ r_3 \\ r \circ \Delta_A = r = \Delta_A \circ r \end{array}$

Moreover, $ \emptyset \circ r = \emptyset = r \circ \emptyset$ $ r_1 \subseteq r_2 \rightarrow r_1 \circ s \subseteq r_2 \circ s$ $ r_1 \subseteq r_2 \rightarrow s \circ r_1 \subseteq s \circ r_2$

### Relation Image

When and we define image of a set under relation as $ S\bullet r = \{ y.\ \exists x. x \in S \land (x,y) \in r \}$

### Transitive Closure

Iterated composition let . $\begin{array}{l} r^0 = \Delta_A \\ r^{n+1} = r \circ r^n \end{array}$ So, is n-fold composition of relation with itself.

Transitive closure: $ r^* = \bigcup_{n \geq 0} r^n$

Equivalent statement: is equal to the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ (s \circ r)\ \subseteq\ s$ or, equivalently, the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ (r \circ s)\ \subseteq\ s$ or, equivalently, the least relation (with respect to ) that satisfies $ \Delta_A\ \cup\ r \cup (s \circ s)\ \subseteq\ s$

### Some Laws in Algebra of Relations

$ (r_1 \circ r_2)^{-1} = r_2^{-1} \circ r_1^{-1}$ $ r_1 \circ (r_2 \cup r_3) = (r_1 \circ r_2) \cup (r_1 \circ r_3)$ $ (r^{-1})^{*} = (r^{*})^{-1}$

Binary relation can be represented as a directed graph with nodes and edges

• Graphical representation of , , and

Equivalence relation is relation with these properties:

• reflexive:
• symmetric:
• transitive:

Equivalence classes are defined by $ x/r = \{y \mid (x,y) \in r$ The set is a partition:

• each set non-empty
• sets are disjoint
• their union is

Conversely: each collection of sets that is a partition defines equivalence class by $ r = \{ (x,y) \mid \exists c \in P. x \in c \land y \in c \}$

Congruence: equivalence that agrees with some set of operations.

Partial orders:

• reflexive
• antisymmetric:
• transitive

## Functions

Example: an example function for , is $f = \{ (a,3), (b,2), (c,3) \}$

Definition of function, injectivity, surjectivity.

- the set of all functions from to . For it is a strictly bigger set than .

(think of exponentiation on numbers)

Note that is isomorphic to , they are two ways of representing functions with two arguments.

There is also isomorphism between

• n-tuples and
• functions , where

### Function update

Function update operator takes a function and two values , and creates a new function that behaves like in all points except at , where it has value . Formally, $f[a_0 \mapsto b_0](x) = \left\{\begin{array}{l} b_0, \mbox{ if } x=a_0 \\ f(x), \mbox{ if } x \neq a_0$

### Domain and Range of Relations and Functions

For relation we define domain and range of : $ dom(r) = \{ x.\ \exists y. (x,y) \in r \}$ $ ran(r) = \{ y.\ \exists x. (x,y) \in r \}$ Clearly, and .

### Partial Function

Notation: means .

Partial function is relation such that $ \forall x \in A. \exists^{\le 1} y.\ (x,y)\in f$

Generalization of function update is override of partial functions,

### Range, Image, and Composition

The following properties follow from the definitions: $ (S \bullet r_1) \bullet r_2 = S \bullet (r_1 \circ r_2)$ $ S \bullet r = ran(\Delta_S \circ r)$