Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Understanding Functions and Relations in Mathematics - Prof. Macgillivray, Lecture notes of Mathematical logic

A comprehensive overview of functions and relations in mathematics. It covers topics such as binary relations, domains, codomains, ranges, injective, surjective, and bijection functions. The document also explains how to determine if a function is total, univalent, or equal to another function. It includes examples and proofs to illustrate these concepts.

Typology: Lecture notes

2022/2023

Uploaded on 04/11/2024

lisa-yu-1
lisa-yu-1 🇨🇦

6 documents

1 / 5

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Relations are subsets of Catesian products (i.e. R A x B)
Functions are binary relations and binary relations are sets.
Let f: A --> B be a function. We write f(a) = b when a A, b B and (a,b) f.
Note: A is the domain, B is the codomian, f(A) is the range
To be a function must satisfying the following two:
- for each element in domain, there's a unique element in codomain (pass vertical-line test)
- each element in domain must exist and must appear only once in the binary relation f
Note:
- the preimage may not be unique (when this is the case, the preimage of an element b is the subset S of
the domain A such that f(S)={b})
- There is not necessarily a preimage of an element of the codomain (i.e. f: D --> C, |D| < |C| is ok)
e.g.:
binary relations from A = {1,2,3} to B = {a,b,c}
R1 = {(1,a),(2,a),(3,c)}: this is a function
R2 = {(1,a),(1,b),(2,b),(2,c)}: this is not a function, since 3 does not appear in any pair and 1 and 2 both
appear twice each
R3 = {(1,c),(2,a)}: this is not a function, since 3 does not appear in any pair
y = f(x) (x A) y f(A)
y f(A) y = f(x) (x A); likewise, y f(A B) y = f(x) (x A and x B)
pf3
pf4
pf5

Partial preview of the text

Download Understanding Functions and Relations in Mathematics - Prof. Macgillivray and more Lecture notes Mathematical logic in PDF only on Docsity!

Relations are subsets of Catesian products (i.e. R ⊆ A x B) Functions are binary relations and binary relations are sets. Let f: A --> B be a function. We write f(a) = b when a ∈ A, b ∈ B and (a,b) ∈ f. Note: A is the domain, B is the codomian, f(A) is the range To be a function must satisfying the following two:

  • for each element in domain, there's a unique element in codomain (pass vertical-line test)
  • each element in domain must exist and must appear only once in the binary relation f Note:
  • the preimage may not be unique (when this is the case, the preimage of an element b is the subset S of the domain A such that f(S)={b})
  • There is not necessarily a preimage of an element of the codomain (i.e. f: D --> C, |D| < |C| is ok) e.g.: binary relations from A = {1,2,3} to B = {a,b,c} R1 = {(1,a),(2,a),(3,c)}: this is a function R2 = {(1,a),(1,b),(2,b),(2,c)}: this is not a function, since 3 does not appear in any pair and 1 and 2 both appear twice each R3 = {(1,c),(2,a)}: this is not a function, since 3 does not appear in any pair y = f(x) (x ∈ A) ⇒ y ∈ f(A) y ∈ f(A) ⇒ y = f(x) (x ∈ A); likewise, y ∈ f(A ∩ B) ⇒ y = f(x) (x ∈ A and x ∈ B)

Equal functions: f: A --> B and g: A --> B are equal if and only if ∀a∈A, f(a) = g(a), i.e., their domains and codomains are the same, and images of every element of the domain is the same under both functions. To show f is a function, we must show it's total and univalent. To show f is not a function, it's sufficient to show there exists x ∈ domain such that f(x) is undefined (i.e., f(x) ∉ codomain) Injective: every element in the range of the function has a unique preimage, i.e., for function f: A-->B, ∀a1,a2 ∈ A, a1≠a2 --> f(a1)≠f(a2), i.e., ∀b ∈ B, if f(a1) = b = f(a2), then a1 = a For an injective function f: A-->B, |A| ≤ |B| To prove injective: ∀a1,a2 ∈ A, a1 = a2 --> f(a1) = f(a2); or let f(a1) = f(a2) --> a1 = a2; or to prove the function f is strictly monotonic Surjective: its codomain and range are equal, i.e., for function f: A-->B, ∀b ∈ B, ∃a ∈ A such that f(a) = b For an surjective function f: A-->B, |A| ≥ |B| To prove surjective: for f(x)=..., solve for x=... and show x ∈ domain Bijection / invertible functions/ one-to-one correspondence: both injective (one-to-one function) and surjective. (note one-to-one function is not the same as one-to-one correspondence) For an bijective function f: A-->B, |A| = |B| e.g: Whether the following functions are injective and surjective?

Constant function: the value of the function is always the same value Inverse function: a bijection f from set A to se B, then the inverse of f, denoted f' is a function from B to A defined as f'(b) = a ↔ f(a) = b Note: the inverse of an inverse is the original function: (f')' = f Composition: the order of the composition matters. If f: A --> B, g: B --> C are functions, then f ∘ g is not even defined, unless C = A Any invertible function over a set A. f: A --> A, it holds that: f ∘ f' = f' ∘ f = idA f = f ∘ IdA f(Id(a)) = f(a) Id(f(a)) = f(a) An invertible function f: A --> B and its inverse f': B --> A: (f ∘ f')(x) = f(f'(x)) = idB (f' ∘ f)(x) = f'(f(x)) = idA Prove invertible:

  • invertible <--> bijective, i.e., injective & surjective
  • or use: f ∘ f' = f' ∘ f = idA Floor: ⌊x⌋ = z ∈ Z | z ≤ x < z+1 (⌊-4.3⌋ = - 5) Ceiling: ⌈x⌉ = z ∈ Z | z-1 < x ≤ z (⌈-4.3⌉ = - 4) Note: the floor / ceiling of an integer is itself

non-decreasing / increasing: x ≤ y --> f(x) ≤ f(y); strictly increasing: x < y --> f(x) < f(y) non-increasing / decreasing: x ≤ y --> f(x) ≥ f(y); strictly decreasing: x < y --> f(x) > f(y) monotonic: either increasing or decreasing; strictly monotonic: either strictly increasing or strictly decreasing Cardinality:

  • If there exists a bijection between sets A and B, then |A| = |B|
  • A set A is finite if its cardinality is a natural number n, i.e., |A| < |N|
  • A set A is countably infinite if |A| = |N|
  • A set A is uncountably infinite if |A| > |N| To prove |A| = |B| is to find/prove a function s.t. f: A --> B is bijective (i.e., to prove there's a bijection), and to prove bijection, use one of the following:
  • find the function's inverse (f'), and then check if f ∘ f' = id (i.e., f(f'(x)) = x)
  • directly prove injective and surjective To prove a set A is countable, is to prove the set to set N (i.e., the set of all natural numbers) has a bijective function (i.e., |A| = |N|), or |A| < |N| To prove a set / an interval is uncountably infinite, can proceed by contradiction, i.e., assume it's countable, then list all the elements in the set, then find some new element which is not in the list. A binary relation R on a set S can be represented as a square zero-one matrix A = (a_i,j) of order |S|:
  • The relation is reflexive if and only if its main diagonal is all 1s.
  • The relation is symmetric if and only if A = A ^(T)
  • The relation is antisymmetric if and only if a_i,j = 0 or a_j,i = 0 for i != j (i.e. a_i,j = 1 --> a_j,i = 0 for all i != j)
    • thus, if a matrix is a triangular matrix (either upper or lower), then it must be antisymmetric