Math-203-Theory of Sets-Set Functions.pptx

Gerald372164 7 views 12 slides Nov 02, 2025
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About This Presentation

This PowerPoint presentation is about Set Functions.


Slide Content

Further Theory of Functions Chapter 8: Set Functions ARTHUR M. MOSTAZA

OBJECTIVES: At the end of the lesson, students should be able to define and distinguish between different types of functions (injective, surjective, bijective), understand function composition and inverse functions, and illustrate each concept using set theory and arrow diagrams.

OVERVIEW What is a function in set theory? • Domain, codomain, and image • Types of functions • Function composition • Inverse functions • Identity functions

Functions as Sets • A function is a set of ordered pairs (a, b) • Each input a ∈ A maps to exactly one output b ∈ B • Example: A = {1,2,3}, B = { a,b,c }, f = {(1,a), (2,b), (3,c)} Diagram : Arrows from 1→a, 2→b, 3→c

Domain, Codomain, and Image • Domain: Set of all inputs • Codomain: Set of potential outputs • Image: Set of actual outputs • Example: f = {(1,a), (2,b), (3,a), (4,b)} Domain = {1,2,3,4}, Codomain = { a,b,c }, Image = { a,b }

Injective Functions (One-to-One) Each output is mapped by only one input • f(a1) = f(a2) ⇒ a1 = a2 Example: Injective: {(1,a), (2,b), (3,c)} Not Injective: {(1,a), (2,a), (3,b)} • Diagram: Injective: 1→a, 2→b, 3→c; Not Injective: 1→a, 2→a

Surjective Functions (Onto) • Every element in the codomain is used ∀b ∈ B, ∃a ∈ A such that f(a) = b • Example: f = {(1,a), (2,b), (3,a)} Codomain = { a,b } → Surjective Codomain = { a,b,c } → Not Surjective

Bijective Functions Both injective and surjective • One-to-one correspondence • Example: {(1,a), (2,b), (3,c)}

Function Composition If f: A → B and g: B → C, then (g ◦ f)(x) = g(f(x)) • Example: f(x) = 2x, g(x) = x + 3 (g ◦ f)(x) = g(2x) = 2x + 3

Inverse Functions Exist only for bijective functions • f⁻¹(y) = x ⇔ f(x) = y • Example: f(x) = 2x + 5, f⁻¹(x) = (x - 5)/2

Inverse Functions Exist only for bijective functions • f⁻¹(y) = x ⇔ f(x) = y • Example: f(x) = 2x + 5, f⁻¹(x) = (x - 5)/2

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