Definition: Left S be a nonempty set
Proof: 1. Homework 2. Let F = \{S_i\}_{i\in I} be a partition of S. Define relation R on S By (x, y) \in R \iff (\exists i \in I)(x \in S_i \land y \in S_i) - WTS R is equivalence relation and S/R = F - Reflexive: x \in S_i for some i \in I. Therefore (x,x)\in R - Symmetric: Let x, y \in S \land (x, y) \in R. Then x \in S_i \land y \in S_i (meaning they are in the same partition). Then (y, x) \in R because conjunction is commutative. - Transitive: Let x, y, z \in S \land (\exists i,j \in I)((x \in S_i \land y \in S_i)\land (y\in S_j \land z \in S_j)) Because partition, i = j must hold. Thus x \in S_i \land z \in S_i and (x, z) \in R. - Show S/R = F by double containment. - [x]_R = \{y\in R | y \in S_i \} for precisely one i \in I, since [x]_R = \{y\ \in S | y \in S_i\} = S_i, (x\in S_i is not here because we already defined) therefore [x]_R \in F. - Let S_i \in F. Since partition, S_i \neq \emptyset. Fix x \in S_i, then for i \neq j, S \notin S_j. Then [x]_R = \{y \in S | y \in S_i \} since [x]_R = \{y\ \in S | y \in S_i\} = S_i (x\in S_i is not here because we already defined), therefore S_i \in S/R.
Definition
(\forall [x]_R \in S/R)([x]_R \neq \emptyset)
(\forall [x]_R, [y]_R \in S/R)([x]_R \neq [y]_R \implies [x]_R \cup [y]_R = \emptyset)
\bigcup_{[x]_R \in S/R} [x]_R = S
Definition: Let A and B be set and let f be a binary relation between A and B. Then f is a function from A to B iff (\forall a \in A)(\exists b \in B)(a, b \in f). And b exists for exactly 1 time. b = f(a)
if f:A \to B, then A is the domain of f and B is codomain of f
if f \subseteq A \times B, then A is the domain of f and B is codomain of f
Identity function: Let S be any set. We define identity function on S, denoted Ids (or Id) is the domain is clear from context. by Id_S: S \to S \text{ s.t. } (\forall x \in S)(Id_S(x) = x) (Meaning that the identity function maps each element to itself)
Ill-defined: f: \mathbb{Q} \to \mathbb{Z} by f(\frac{p}{q} = p+q). Then one value is mapped to two different values. That's not a function.
Well-defined: A mapping f: A \to B is a well-defined function iff f satisfy:
(\forall a \in A)(\exists b \in B)(f(a) = b)
(\forall a, a' \in A)(a=a' \implies f(a)=f(a'))
Function Equality: Let A and B be sets and f : A\to B, g:A\to B be functions. f = g iff (\forall a \in A)(f(a) = g(a)). (since it's based on set equality, different rule may produce the same function)
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