Functions Chapter 4 (Modified) PDF

Fundamentals Proofs Sets Informal Logic Functions Relations Strategies for Proofs Infinite and Finite Sets Countable sets and Cardinal numbers  Functions Functions...

Fundamentals Proofs Sets Informal Logic Functions Relations Strategies for Proofs Infinite and Finite Sets Countable sets and Cardinal numbers  Functions Functions 2  Fundamentals Functions Contents 4.1 Functions 4.2 Image and Inverse Image 4.3 Composition and Inverse Functions 4.4 Injectivity, Surjectivity and Bijectivity 3 4.1 Functions Function is like a machine f (x ) = x 2 𝑖𝑓 𝑤𝑒 𝑝𝑢𝑡 𝑎 = 5 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑚𝑎𝑐ℎ𝑖𝑛𝑒 , 𝑡ℎ𝑒𝑛 𝑖𝑡 𝑤𝑖𝑙𝑙 𝑠𝑝𝑖𝑡 𝑜𝑢𝑡 𝑓 𝑎 = 25 y = f (x ) Input Output Name of Function Definition 4.1.1 Let A and B be sets. A function (also called a map) f from A to B, denoted f : A→B, is a subset F ⊆ A×B such that for each a ϵ A, there is one and only one pair in F of the form (a , b). The set A is called the domain of f and the set B is called the codomain of f. f : A → B is read: “f is a function from A to B”  Variable x is called independent variable Variable y is called dependent variable For convenience, we use f(x) instead of y. The ordered pair in new notation becomes: (x, y) = (x, f(x))  for example; 𝑙𝑒𝑡 𝐴 = 𝑎 , 𝑏 , 𝑐 , 𝑑 𝑎𝑛𝑑 𝐵 = {1 , 2 , 3 , 4} (i) is Function (ii) Is Not Function { 𝑎 , 2 , 𝑏 , 1 , 𝑐, 4 , 𝑑 , 4 } ⊆ 𝐴 × 𝐵 { 𝑎 , 1 , 𝑎 , 2 , 𝑏, 3 , 𝑐 , 4 } Ex. 4.1.1 𝐿𝑒𝑡 𝐴 = 𝑎 , 𝑏 , 𝑐 𝑎𝑛𝑑 𝐵 = 1 , 2 , 3. 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝐴 × 𝐵 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐴 → 𝐵 ? 1 𝑏, 1 , 𝑐, 2 , 𝑎, 3. 2 𝑎 ,3 , 𝑐 ,2 , 𝑎 ,1. 3 𝑐 ,1 , 𝑏 ,1 , 𝑎 ,2.  (Group work) Ex. 4.1.3 𝑊ℎ𝑖𝑐ℎ 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑖𝑎𝑔𝑟𝑎𝑚𝑠 𝑖𝑛 𝐹𝑖𝑔𝑢𝑟𝑒 4.1.4 𝑟𝑒𝑝𝑟𝑒𝑠𝑒𝑛𝑡 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 ? for example; Consider the formula 𝑓 𝑥 = 𝑥 2 − 5𝑥 + 6 Does it define a function? We Can’t determine if it define a function or not. Why??? We must first given the domain and codomain If the domain and codomain is ℝ , then f ( x ) is not a function If the domain is −∞ , 2 ∪ [3 , ∞) and codomain is ℝ , then f ( x ) is a function  Note Not all function , even with domain and codomain equal to ℝ , can be defined by a numerical formula. Even when a function is defined by a formula , it is not always Consider, for example , the function 𝑓: ℝ → ℝ defined by 𝑥, 𝑖𝑓 𝑥 ≥ 0 𝑓 𝑥 =ቊ −1 , 𝑖𝑓 𝑥 < 0 𝐹𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑢𝑐ℎ 𝑥, 𝑡ℎ𝑒 𝑑𝑒𝑓𝑖𝑛𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑜𝑛𝑒 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑣𝑎𝑙𝑢𝑒 𝑓 𝑥 𝜖 𝑅. 𝐻𝑒𝑛𝑐𝑒, 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑥 𝜖 𝑅 𝑡ℎ𝑒𝑟𝑒 𝑖𝑠 𝑜𝑛𝑒 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑜𝑛𝑒 𝑝𝑜𝑖𝑛𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ 𝑜𝑓 𝑓 𝑡ℎ𝑎𝑡 𝑖𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑙𝑖𝑛𝑒 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑥. A function consists of three things: 1. 𝑑𝑜𝑚𝑎𝑖𝑛 2. 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛 3. 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑎𝑛𝑑 𝑡ℎ𝑒 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛 𝑠𝑎𝑡𝑖𝑠𝑓𝑦𝑖𝑛𝑔 𝑎 𝑐𝑒𝑟𝑡𝑎𝑖𝑛 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛. Definition 4.1.2 𝐿𝑒𝑡 𝑓 ∶ 𝐴 → 𝐵 𝑎𝑛𝑑 𝑔: 𝐶 → 𝐷 𝑏𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. 𝑇𝑜 𝑠𝑎𝑦 𝑡ℎ𝑎𝑡 “ 𝑓 = 𝑔” 𝑚𝑒𝑎𝑛𝑠 𝑡ℎ𝑎𝑡 𝐴 = 𝐶 , 𝑡ℎ𝑎𝑡 𝐵 = 𝐷 𝑎𝑛𝑑 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝑐𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑢𝑏𝑠𝑒𝑡 𝑜𝑓 𝐴 × 𝐵. 𝑇ℎ𝑖𝑠 𝑙𝑎𝑠𝑡 𝑠𝑡𝑎𝑡𝑒𝑚𝑒𝑛𝑡 𝑐𝑎𝑛 𝑏𝑒 𝑟𝑒𝑝ℎ𝑟𝑎𝑠𝑒𝑑 𝑏𝑦 𝑠𝑎𝑦𝑖𝑛𝑔 𝑡ℎ𝑎𝑡 𝑓 (𝑥) = 𝑔(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 𝜖𝐴. A proof that f = g has the following form. 𝑃𝑟𝑜𝑜𝑓. 𝐴𝑟𝑔𝑢𝑚𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛 ⋮ 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑔. ⋮ (𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛) ⋮ 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑡ℎ𝑒 𝑐𝑜𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑔. 𝐿𝑒𝑡 𝑎 𝑏𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑜𝑚𝑎𝑖𝑛 𝑜𝑓 𝑓 𝑎𝑛𝑑 𝑔. ⋮ (𝑎𝑟𝑔𝑢𝑚𝑒𝑛𝑡𝑎𝑡𝑖𝑜𝑛) ⋮ 𝑇ℎ𝑒𝑛 𝑓 𝑎 = 𝑔 𝑎. 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑓 = 𝑔  Definition 4.1.3. 𝐿𝑒𝑡 𝐴 𝑎𝑛𝑑 𝐵 𝑏𝑒 𝑠𝑒𝑡𝑠, 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑆 ⊆ 𝐴 𝑏𝑒 𝑎 𝑠𝑢𝑏𝑠𝑒𝑡. 1. 𝐴 𝒄𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒎𝒂𝒑 𝑓: 𝐴 → 𝐵 𝑖𝑠 𝑎𝑛𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚 𝑓 𝑥 = 𝑏 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐴, 𝑤ℎ𝑒𝑟𝑒 𝑏 ∈ 𝐵 𝑖𝑠 𝑠𝑜𝑚𝑒 𝑓𝑖𝑥𝑒𝑑 𝑒𝑙𝑒𝑚𝑒𝑛𝑡. 2. 𝑇ℎ𝑒 𝒊𝒅𝒆𝒏𝒕𝒊𝒕𝒚 𝒎𝒂𝒑 𝑜𝑛 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 1𝐴 : 𝐴 → 𝐴 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 1𝐴 𝑥 = 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐴. 3. 𝑇ℎ𝑒 𝒊𝒏𝒄𝒍𝒖𝒔𝒊𝒐𝒏 𝒎𝒂𝒑 𝑓𝑟𝑜𝑚 𝑆 𝑡𝑜 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑗: 𝑆 → 𝐴 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑗 𝑥 = 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑆. 4. 𝐼𝑓 𝑓 ∶ 𝐴 → 𝐵 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑡ℎ𝑒 𝒓𝒆𝒔𝒕𝒓𝒊𝒄𝒕𝒊𝒐𝒏 𝑜𝑓 𝑓 𝑡𝑜 𝑆, 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑓⃓𝑆 , 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓⃓𝑆 : 𝑆 → 𝐵 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓⃓𝑆 = 𝑓 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑆. 5. 𝐼𝑓 𝑔 ∶ 𝑆 → 𝐵 𝑖𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 , 𝑎𝑛 𝒆𝒙𝒕𝒆𝒏𝒔𝒊𝒐𝒏 𝑜𝑓 𝑔 𝑡𝑜 𝐴 𝑖𝑠 𝑎𝑛𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝐺 ∶ 𝐴 → 𝐵 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝐺⃓𝑆 = 𝑔. 6. 𝑇ℎ𝑒 𝒑𝒓𝒐𝒋𝒆𝒄𝒕𝒊𝒐𝒏 𝒎𝒂𝒑𝒔 𝑓𝑟𝑜𝑚 𝐴 × 𝐵 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 𝜋1 : 𝐴 × 𝐵 → 𝐴 𝑎𝑛𝑑 𝜋2 ∶ 𝐴 × 𝐵 → 𝐵 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝜋1 𝑎, 𝑏 = 𝑎 𝑎𝑛𝑑 𝜋2 𝑎, 𝑏 = 𝑏 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎, 𝑏 ∈ 𝐴 × 𝐵. 𝐹𝑜𝑟 𝑎𝑛𝑦 𝑓𝑖𝑛𝑖𝑡𝑒 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑒𝑡𝑠 𝐴1 ,... , 𝐴𝑝 , 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑝𝑠 𝜋𝑖 : 𝐴1 × ⋯ × 𝐴𝑝 → 𝐴𝑖 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑖 ∈ 1, ⋯ , 𝑝 𝑐𝑎𝑛 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦.  Example 4.1.4. 3 𝑤𝑒 𝑐𝑎𝑛 𝑡ℎ𝑖𝑛𝑘 𝑜𝑓 𝑎𝑠 ℝ2 𝑎𝑠 ℝ × ℝ. 𝑊𝑒 𝑡ℎ𝑒𝑛 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑡𝑤𝑜 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑝 𝜋1 : ℝ2 → ℝ 𝑎𝑛𝑑 𝜋2 : ℝ2 → ℝ 𝑡ℎ𝑎𝑡 𝑎𝑟𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝜋1 𝑥, 𝑦 = 𝑥 𝑎𝑛𝑑 𝜋2 𝑥, 𝑦 = 𝑦 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦 ∈ ℝ2. 𝑇ℎ𝑒𝑛 𝑖𝑠 , 𝑡ℎ𝑒 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑚𝑎𝑝 𝜋1 𝑝𝑖𝑐𝑘𝑠 𝑜𝑢𝑡 𝑡ℎ𝑒 𝑓𝑖𝑟𝑠𝑡 𝑐𝑜𝑜𝑟𝑑𝑖𝑛𝑎𝑡𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 𝑥, 𝑦 , 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥, 𝑦 ∈ ℝ2 , 𝑎𝑛𝑑 𝑠𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦 𝑓𝑜𝑟 𝜋2. 4.2 Image and Inverse Image Definition 4.2.1. 𝑳𝒆𝒕 𝑨 𝒂𝒅𝒏 𝑩 𝒃𝒆 𝒔𝒆𝒕𝒔, 𝒂𝒏𝒅 𝒍𝒆𝒕 𝒇: 𝑨 → 𝑩 𝒃𝒆 𝒂 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏. 1. 𝐿𝑒𝑡 𝑃 ⊆ 𝐴. 𝑇ℎ𝑒 𝒊𝒎𝒂𝒈𝑒 𝑜𝑓 𝑃 𝑢𝑛𝑑𝑒𝑟 𝑓, 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑓 𝑃 , 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓 𝑃 = {𝑏 ∈ 𝐵 ⃓ 𝑏 = 𝑓 𝑝 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑝 ∈ 𝑃 𝑇ℎ𝑒 𝒓𝒂𝒏𝒈 𝑜𝑓 𝑓 𝑎𝑙𝑠𝑜 𝑐𝑎𝑙𝑙𝑒𝑑 𝑡ℎ𝑒 𝒊𝒎𝒂𝒈𝒆 𝑜𝑓 𝑓 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑓 𝐴. 2. 𝐿𝑒𝑡 𝑄 ⊆ 𝐵. 𝑇ℎ𝑒 𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝒊𝒎𝒂𝒈𝑒 𝑜𝑓 𝑄 𝑢𝑛𝑑𝑒𝑟 𝑓, 𝑑𝑒𝑛𝑜𝑡𝑒𝑑 𝑓 −1 𝑄 , 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑒𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓 −1 𝑄 = 𝑎 ∈ 𝐴⃓𝑓 𝑎 ∈ 𝑄.  𝑑𝑟𝑎𝑤𝑖𝑛𝑔 𝑜𝑓 𝑓 𝑃 𝑎𝑛𝑑 𝑓 −1 𝑄 𝑓 𝑃 = 𝑓 𝑝 𝑝 ∈ 𝑃}, 𝑓 −1 (𝑄) = {𝑎 ∈ 𝐴 |𝑓 𝑎 = 𝑞 𝑓𝑜𝑟 𝑠𝑜𝑚𝑒 𝑞 ∈ 𝑄  Example 4.2.2. 𝐹𝑜𝑟 𝑓: ℝ → ℝ 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑓 𝑥 = 𝑥 2 𝐹𝑖𝑛𝑑 𝑓 0 , 𝑓 2 , 𝑓 2, −2 , 𝑓 −1,3 , 𝑓 −1 1 , 𝑓 −1 −2,1 , 𝑓 −1 ( 1,2 ) f (0 ) = 0 As element f ( 2 ) =  4  As set f ( 2 , − 2 ) =  4  f ( − 1, 3)) =  0 , 9 ) f −1 ( 1  ) =  − 1,1  f ( − 2 , 1 ) =  − 1 , 1  −1 f −1 ( 1, 2 ) = − 2 , − 1 , 1 , 2   Example 4.2.3. 𝐿𝑒𝑡 𝑓: ℝ → ℝ 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓 𝑥 = 𝑥 2 − 6𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ ℝ 𝐹𝑖𝑛𝑑 𝑓 6,7 , 𝑓 −1 0,16 , 𝑓 −1 𝑓 6,7 , 𝑓 𝑓 −1 −12,0 f (  6 , 7  ) = 0 , 7  f −1 ( f (  6 , 7 ) ) = f −1 (  0 , 7  ) =  − 1, 0    6 , 7  f −1 ( f (  6 , 7 ) )   6 , 7 f −1 ( − 12 , 0 ) =  0 , 6  ( ) ( 0 , 6 ) =  − 9 , 0  f f −1 ( − 12 , 0 ) = f f ( f ( − 12 , 0 ))   − 12 , 0  −1  Theorem 4.2.4. 𝐿𝑒𝑡 𝐴 𝑎𝑛𝑑 𝐵 𝑏𝑒 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛. 𝑙𝑒𝑡 𝐶, 𝐷 ⊆ 𝐴 𝑎𝑛𝑑 𝑆, 𝑇 ⊆ 𝐵 𝑏𝑒 𝑠𝑢𝑏𝑠𝑒𝑡𝑠, 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑓: 𝐴 → 𝐵 𝑏𝑒 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛. 𝐿𝑒𝑡 𝐼 𝑎𝑛𝑑 𝐾 𝑏𝑒 𝑛𝑜𝑛‐ 𝑒𝑚𝑝𝑡𝑦 𝑠𝑒𝑡𝑠, 𝑙𝑒𝑡 {𝑈𝑖 }𝑖∈𝐼 𝑏𝑒 𝑎 𝑓𝑎𝑚𝑖𝑙𝑦 𝑜𝑓 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝐴 𝑖𝑛𝑑𝑒𝑥𝑒𝑑 𝑏𝑦 𝐼, 𝑎𝑛𝑑 𝑙𝑒𝑡 {𝑉𝑘 }𝑘∈𝐾 𝑏𝑒 𝑎 𝑓𝑎𝑚𝑖𝑙𝑦 𝑜𝑓 𝑠𝑢𝑏𝑠𝑒𝑡𝑠 𝑜𝑓 𝐵 𝑖𝑛𝑑𝑒𝑥𝑒𝑑 𝑏𝑦 𝐾. 1. 𝑓 ∅ = ∅ 𝑎𝑛𝑑 𝑓 −1 ∅ = ∅. 2. 𝑓 −1 𝐵 = 𝐴. 3. 𝑓 𝐶 ⊆ 𝑆 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝐶 ⊆ 𝑓 −1 𝑆. 4. 𝐼𝑓 𝐶 ⊆ 𝐷, 𝑡ℎ𝑒𝑛 𝑓 𝐶 ⊆ 𝑓 𝐷. 5. 𝐼𝑓 𝑆 ⊆ 𝑇, 𝑡ℎ𝑒𝑛 𝑓 −1 𝑆 ⊆ 𝑓 −1 𝑇. 6. 𝑓 ∪𝑖𝜖𝐼 𝑈𝑖 = ∪𝑖∈𝐼 𝑓 𝑈𝑖. 7. 𝑓 ∩𝑖∈𝐼 𝑈𝑖 ⊆ ∩𝑖∈𝐼 𝑓 𝑈𝑖. 8.𝑓 −1 ∪𝑘∈𝐾 𝑉𝑘 =∪𝑘∈𝐾 𝑓 −1 𝑉𝑘.9.𝑓 −1 ∩𝑘∈𝐾 𝑉𝑘 =∩𝑘∈𝐾 𝑓 −1 𝑉𝑘  4.3 Composition and Inverse Functions 𝐿𝑒𝑡 𝑓: ℝ → ℝ 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓 𝑥 = 𝑥2 + 3 𝑇ℎ𝑖𝑠 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑏𝑢𝑖𝑙𝑡 𝑢𝑝 𝑜𝑢𝑡 𝑜𝑓 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 “𝑖𝑛𝑠𝑖𝑑𝑒” 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛, 𝑎𝑛𝑑 𝑤𝑒 𝑎𝑝𝑝𝑙𝑦 𝑡ℎ𝑒 𝐶ℎ𝑎𝑖𝑛 𝑅𝑢𝑙𝑒 𝑡𝑜 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 Definition 4.3.1. 𝐿𝑒𝑡 𝐴, 𝐵 𝑎𝑛𝑑 𝐶 𝑏𝑒 𝑠𝑒𝑡𝑠, 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑓: 𝐴 → 𝐵 𝑎𝑛𝑑 𝑔: 𝐵 → 𝐶 𝑏𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. 𝑡ℎ𝑒 𝒄𝒐𝒎𝒑𝒐𝒔𝒊𝒕𝒊𝒐𝒏 𝑜𝑓 𝑓 𝑎𝑛𝑑 𝑔 𝑖𝑠 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔 ⃘𝑓: 𝐴 → 𝐶 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑔 ⃘𝑓 𝑥 = 𝑔 𝑓 𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐴  Function Composition ƒ g a ƒ(a) g(ƒ(a)) = gf(a) g f A B C Functions 25 Observe also that for the composition of two functions to be defined, the codomain of the first function must equal the domain of the second function. Example 4.3.2. 𝐿𝑒𝑡 𝑓, 𝑔: ℝ → ℝ 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 𝑓 𝑥 = 𝑥 2 𝑎𝑛𝑑 𝑔 𝑥 = 𝑥 + 3 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ ℝ. Are f ο g and g ο f defined? Yes Because Codomain of f = Domain of g Codomain of g = Domain of f 𝑓 ⃘𝑔 𝑥 = 𝑥 + 3 2 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ ℝ. 𝑔 ⃘𝑓 𝑥 = 𝑥 2 + 3 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ ℝ. Example 4.3.2. Let k : R → R be defined by k(x) = sinx for all x ϵ R, and let h: (0;∞) →R be defined by h(x) = ln x for all x ϵ (0;∞) Are k ο h and h ο k defined? k ο h is defined but h ο k is not defined Because Codomain of h = Domain of k Codomain of k ≠ Domain of h 𝑘 ⃘ℎ 𝑥 = sin 𝑙𝑛𝑥 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 0, ∞. ln 𝑠𝑖𝑛𝑥 𝑖𝑠 𝑛𝑜𝑡 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ ℝ.  𝐼𝑓 𝑓: 𝐴 → 𝐵 𝑎𝑛𝑑 𝑔: 𝐵 → 𝐶 𝑎𝑟𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠, 𝑡ℎ𝑒𝑛 𝑤𝑒 𝑐𝑎𝑛 𝑓𝑜𝑟𝑚 𝑔 ⃘𝑓: 𝐴 → 𝐶 , by “commutative diagrams.” Lemma 4.3.3. 𝐿𝑒𝑡 𝐴, 𝐵, 𝐶 𝑎𝑛𝑑 𝐷 𝑏𝑒 𝑠𝑒𝑡𝑠 , 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑓: 𝐴 → 𝐵 𝑎𝑛𝑑 𝑔: 𝐵 → 𝐶 𝑎𝑛𝑑 ℎ: 𝐶 → 𝐷 𝑏𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. 1. ℎ ⃘𝑔 ⃘𝑓 = ℎ ⃘ 𝑔 ⃘𝑓 𝐴𝑠𝑠𝑜𝑐𝑖𝑎𝑡𝑖𝑣𝑒 𝐿𝑎𝑤. 2. 𝑓 ⃘1𝐴 = 𝑓 𝑎𝑛𝑑 1𝐵 ⃘𝑓 = 𝑓 𝐼𝑑𝑒𝑛𝑡𝑖𝑡𝑦 𝐿𝑎𝑤. Definition 4.3.4. 𝐿𝑒𝑡 𝐴 𝑎𝑛𝑑 𝐵 𝑏𝑒 𝑠𝑒𝑡𝑠 , 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑓: 𝐴 → 𝐵 𝑎𝑛𝑑 𝑔: 𝐵 → 𝐴 𝑏𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠. 1. 𝑇ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔 𝑖𝑠 𝑎 𝒓𝒊𝒈𝒉𝒕 𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝑓𝑜𝑟 𝑓 𝑖𝑓 𝑓 ⃘𝑔 = 1𝐵. 2. 𝑇ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔 𝑖𝑠 𝑎 𝒍𝒆𝒇𝒕 𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝑓𝑜𝑟 𝑓 𝑖𝑓 𝑔 ⃘𝑓 = 1𝐴. 3. 𝑇ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑔 𝑖𝑠 𝑎𝑛 𝒊𝒏𝒗𝒆𝒓𝒔𝒆 𝑓𝑜𝑟 𝑓 𝑖𝑓 𝑖𝑡 𝑖𝑠 𝑏𝑜𝑡ℎ 𝑎 𝑟𝑖𝑔ℎ𝑡 𝑖𝑛𝑣𝑒𝑟𝑠𝑒 𝑎𝑛𝑑 𝑎 𝑙𝑒𝑓𝑡 𝑖𝑛𝑣𝑒𝑟𝑠𝑒. Example (Group work) 𝑬𝒙𝒆𝒓𝒄𝒊𝒔𝒆 𝟒. 𝟑. 𝟗. 𝐿𝑒𝑡 ℎ, 𝑘: ℝ → ℝ 𝑏𝑒 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑏𝑦 4𝑥 + 1, 𝑥≥0 3𝑥, 𝑥≥0 ℎ 𝑥 =ቊ 𝑘 𝑥 =ቊ 𝑥, 𝑥 < 0, 𝑥 + 3, 𝑥