Lecture 014

Order Relations

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
  1. Equality on any set
  2. Similar triangles ($\delta A \sim \delta B$ iff they ahve the sam einterior angles)
  3. 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 $<a_1,a_2,...a_n> k<b_1, b_2,...,b_n>$
  4. 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.

$\preceq$