Order Relations: ranking elements of a set against another (<, \geq, \subseteq)

Partial Order

Definition: relation R on set S is called "partial order" iff R is reflexive, antisymmetric, and transitive. If R is a partial order on a set S then we call (S, R) a partially ordered set (or poset). (和自己玩的比较)

(\mathcal P(S), \subseteq) and (\mathbb{R}, \leq) are poset.

Strict Partial Order

Definition: A relation R on a set S is called a "strict partial order" iff R is irefflexive, antisymmetric, and transitive. If R is a strict partial order on a set S ten we call (S, R) a strict poset. (不和自己玩的比较)
- (S, \subsetneq), (\mathbb{R}, <)

Claim: \subseteq is a partial ordering of \mathcal{P}(\mathbb{N})
Proof: we proceed to show that...

R: let X\in \mathcal{P}(\mathbb{N}), show X\subseteq X.

AS: let X, Y\in \mathcal{P}(\mathbb{N}), let X\subseteq Y, Y\subseteq X, show X=Y

T: let X, Y, Z\in \mathcal{P}(\mathbb{N}), let X\subseteq Y \land Y \subseteq Z, show X \subseteq Z: let x \in X because..., then x\in Y because..., then x\in Z because...

Comparable: when relation holds

Total Order

Total Order: is a partial order and totality
Definition: A binary relation R on a set S is called "total"(linear) order iff (\forall x, y \in S)(x \neq y \implies (x,y)\in R \lor (y, x)\in R) (要么正着能比 要么反着能)

\subseteq is not a total ranking of \mathcal{P}(\mathbb{N}) since \{1\} \in \{1, 2\} but \{1\}\nsubseteq \{2}\}.

Equivalence Relations

Definition: A relation R on a set S is called an "equivalence relation" iff R is reflexive, symmetric, and transitive.

Examples

Equality on any set

Similar triangles (\delta A \sim \delta B iff they ahve the sam einterior angles)

Parallel vectors on \mathbb{R}^n \ 0<a_1,a_2,...a_n> \sim <b_1, b_2,...,b_n> \iff \exists k \in \mathbb{R} such that <a_1,a_2,...a_n> k<b_1, b_2,...,b_n>

Parity on \mathbb{Z}. (a \sim b \iff a and b have the same parity)

Example: let m \in \mathbb{N} be arbitrary and fixed. Define a relation \mathbb{Z} by \forall a, b \in \mathbb{Z}: a \equiv b (mod m) (read as "a is congruent to b modulo m").
Claim: a \equiv b (mod m) iff m|a-b is an equivalence relation.
Proof

R: a-a=0=0m, a \equiv a

S: $m|a-b \implies (\exists k \in \mathbb{Z})(a-b=mk) Then there is (b-a)/m = -k.

T: Let a, b, c, \in \mathbb{Z} such that a\equiv b \land b\equiv c. Then a-c = (a-b)+(b-c)=mk+ml=m(k+l) where k and l are divisor.