VishikaClasses VishikaClasses VishikaClasses VishikaClasses VISHIKA CLASSES A free of cost educational consultant of online/offline studies for any students in the world. SET THEORY Class 11th Mathematics By Vinayak Kushwaha M.Sc. Mathematics, Physics Powered by Vishika Group of Companies vishikaclasses@gmail.com 7505630111  SET THEORY 11th Definition: A set is a collection of well-defined objects or elements. Sets are always denoted by capital letter of English alphabets and enclosed within braces { }. Well-defined: A collection of objects is said to be well-defined, if it is possible to tell beyond doubt whether a particular given object is a member of the collection or not. Example: 1. The set of vowels in English alphabet. 𝑖𝑖. 𝑒𝑒. 𝑉𝑉 = {𝑎𝑎, 𝑒𝑒, 𝑖𝑖, 𝑜𝑜, 𝑢𝑢} 2. A set of natural numbers less 7. 𝑖𝑖. 𝑒𝑒. 𝑁𝑁 = {1,2,3,4,5,6} 3. Tea-set, Dinner-set. 4. Set of Indian rivers. etc. Representation of sets: There are three types to represent a set; 1. Roster form: In roster form, all the elements are being separated by comma’s and are enclosed within braces { }. It is also known as tabular form. Example: 𝑉𝑉 = {𝑎𝑎, 𝑒𝑒, 𝑖𝑖, 𝑜𝑜, 𝑢𝑢} 2. Set-builder form: In set-builder form, all the elements of a set are possess a single common property which is not possessed by any element outside the set. Example: 𝑉𝑉 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖𝑖𝑖 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑎𝑎𝑎𝑎𝑎𝑎ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎} where ‘:’ mean ‘is such that’ 3. Venn diagram: Venn diagram is a type of geometrical representation of a set. It consists of a rectangle, represent 𝑨𝑨 𝑩𝑩 the universal set and some closed figures (like circle) represent the set of universal set. 3 1,2 4 5,6 Example: Let 𝐴𝐴 = {1,2,3,4} and 𝐵𝐵 = {3,4,5,6} then, their Venn diagram will be Types of sets: In general, there are five types of sets; 1. Empty set: A set which does not contain any element is called empty set, null set or void set. It is denoted by { } or 𝜙𝜙. Example: 1. 𝑁𝑁 = {𝑥𝑥: 1 < 𝑥𝑥 < 2, 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛} 𝑖𝑖. 𝑒𝑒. 𝑁𝑁 = 𝜙𝜙 2. 𝑃𝑃 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑡𝑡ℎ𝑎𝑎𝑎𝑎 2} 𝑖𝑖. 𝑒𝑒. 𝑃𝑃 = 𝜙𝜙 2. Singleton set: A set which contain only one element is call singleton set. Example: 1. 𝐸𝐸 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎𝑎𝑎 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛} 𝑖𝑖. 𝑒𝑒. 𝐸𝐸 = {2} 2. 𝑊𝑊 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑤𝑤ℎ𝑜𝑜𝑜𝑜𝑜𝑜 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑡𝑡ℎ𝑎𝑎𝑎𝑎 1} 𝑖𝑖. 𝑒𝑒. 𝑊𝑊 = {0} 3. Finite set: A set which is consists a definite number of elements (may be empty) is called finite set. Example: 1. 𝐴𝐴 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸ℎ 𝑎𝑎𝑎𝑎𝑎𝑎ℎ𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎} 2. 𝑁𝑁 = {𝑥𝑥: 1 < 𝑥𝑥 < 9, 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛} 4. Infinite set: A set which is consists an indefinite number of elements is called infinite set. 1 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th Example: 1. 𝑷𝑷 = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝} 2. 𝒁𝒁+ = {𝑥𝑥: 𝑥𝑥 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖} 5. Universal set: A set which is consists some sets (may be finite or infinite) is called universal set. Example: 1. 𝐴𝐴 = { {𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 }, {𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐} } 2. 𝑹𝑹 = { {𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟}, {𝑆𝑆𝑆𝑆𝑆𝑆 𝑜𝑜𝑜𝑜 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖} } Some other types of sets: On comparison of two sets, there are six types of sets; 1. Subset: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵. Then, we say that 𝐴𝐴 is a subset of 𝐵𝐵 and denoted by 𝐴𝐴 ⊆ 𝐵𝐵. There are two types of subsets; i. Proper subset: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵, but there exists at least one element of 𝐵𝐵 which are not an element of 𝐴𝐴. Then, we say that 𝐴𝐴 is a proper subset of 𝐵𝐵 and denoted by 𝐴𝐴 ⊂ 𝐵𝐵. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {1,2,3,4,5,6} ii. Improper subset: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵 and every element of 𝐵𝐵 is an element of 𝐴𝐴. Then, we say that 𝐴𝐴 is an improper subset of 𝐵𝐵 and denoted by 𝐴𝐴 ⊆ 𝐵𝐵. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {4,3,2,1} 2. Super set: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵. Then, we say that 𝐵𝐵 is a super set of 𝐴𝐴 and denoted by 𝐵𝐵 ⊇ 𝐴𝐴. There are two types of super sets; i. Proper super set: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵, but there exists at least one element of 𝐵𝐵 which is not an element of 𝐴𝐴. Then, we say that 𝐵𝐵 is a proper super set of 𝐴𝐴 and denoted by 𝐵𝐵 ⊃ 𝐴𝐴. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {1,2,3,4,5,6} ii. Improper super set: Let 𝐴𝐴 and 𝐵𝐵 be two sets such that every element of 𝐴𝐴 is an element of 𝐵𝐵 and every element of 𝐵𝐵 is an element of 𝐴𝐴. Then, we say that 𝐵𝐵 is an improper super set of 𝐴𝐴 and denoted by 𝐵𝐵 ⊇ 𝐴𝐴. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {4,3,2,1} 3. Equal set: Two sets 𝐴𝐴 and 𝐵𝐵 are said to be equal if every element of 𝐴𝐴 is an element of 𝐵𝐵 and every element of 𝐵𝐵 is an element of 𝐴𝐴. It is denoted by 𝐴𝐴 = 𝐵𝐵 . Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {4,3,2,1} 4. Equivalent sets: Two sets 𝐴𝐴 and 𝐵𝐵 are said to be equivalent if the number of the elements of 𝐴𝐴 is equal to the number of the elements of 𝐵𝐵. It is denoted by 𝑛𝑛(𝐴𝐴) = 𝑛𝑛(𝐵𝐵). Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑 } 2 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th 5. Power set: The collection of all subsets of a set 𝐴𝐴 is called the power set of 𝐴𝐴. It is denoted by 𝑃𝑃(𝐴𝐴) 𝑜𝑜𝑜𝑜 2𝐴𝐴 and if 𝐴𝐴 has 𝑛𝑛 element then, the number of subsets of 𝐴𝐴 is 2𝑛𝑛 . Example: 𝐴𝐴 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐 }, then 𝑃𝑃(𝐴𝐴) = { {𝑎𝑎}, {𝑏𝑏}, {𝑐𝑐 }, {𝑎𝑎, 𝑏𝑏}, {𝑎𝑎, 𝑐𝑐 }, {𝑏𝑏, 𝑐𝑐 }, {𝑎𝑎, 𝑏𝑏, 𝑐𝑐 }, 𝜙𝜙 } 6. Disjoint sets: Two sets 𝐴𝐴 and 𝐵𝐵 are said to be disjoint if they have no common element. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑 } Subsets of set of real numbers a. Set of natural numbers 𝑵𝑵 = {1,2,3,4,5,6, … … } b. Set of whole numbers 𝑾𝑾 = {0,1,2,3,4,5,6, … … } c. Set of integers numbers 𝒁𝒁 = {0, ±1, ±2, ±3, ±4, ±5, ±6, … … } 𝑝𝑝 d. Set of rational numbers 𝑸𝑸 = {𝑥𝑥: 𝑥𝑥 = such that 𝑝𝑝, 𝑞𝑞 ∈ 𝒁𝒁 and 𝑞𝑞 ≠ 0} 𝑞𝑞 e. Set of irrational numbers 𝑰𝑰 = {𝑥𝑥: 𝑥𝑥 ∈ 𝑹𝑹 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝑸𝑸}, symbol ‘∈’ read as ‘belong to’ and symbol ‘∉’ read as ‘does not belong to’. Or 𝑵𝑵 ⊂ 𝑾𝑾 ⊂ 𝒁𝒁 ⊂ 𝑸𝑸 ⊂ 𝑹𝑹, 𝑰𝑰 ⊂ 𝑹𝑹, 𝒁𝒁 ⊄ 𝑰𝑰 and 𝑹𝑹 = {𝑸𝑸, 𝑰𝑰} Intervals as subsets of 𝑹𝑹: Let 𝑎𝑎, 𝑏𝑏 ∈ 𝑹𝑹 and 𝑎𝑎 < 𝑏𝑏. Then the set of real numbers; i. {𝑥𝑥: 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏} is called an open interval, denoted by (𝑎𝑎, 𝑏𝑏) 𝑖𝑖. 𝑒𝑒. all those points who lying between 𝑎𝑎 and 𝑏𝑏 belongs to (𝑎𝑎, 𝑏𝑏) but 𝑎𝑎 and 𝑏𝑏 does not belongs to (𝑎𝑎, 𝑏𝑏). ii. {𝑥𝑥: 𝑎𝑎 ≤ 𝑥𝑥 ≤ 𝑏𝑏} is called a closed interval, denoted by [𝑎𝑎, 𝑏𝑏] 𝑖𝑖. 𝑒𝑒. all those points who lying between 𝑎𝑎 and 𝑏𝑏 belongs to [𝑎𝑎, 𝑏𝑏] and the points 𝑎𝑎, 𝑏𝑏 also belongs to [𝑎𝑎, 𝑏𝑏]. iii. {𝑥𝑥: 𝑎𝑎 < 𝑥𝑥 ≤ 𝑏𝑏} is called left half open interval, denoted by (𝑎𝑎, 𝑏𝑏] 𝑖𝑖. 𝑒𝑒. all those points who lying between 𝑎𝑎 and 𝑏𝑏 belongs to (𝑎𝑎, 𝑏𝑏] and the point 𝑏𝑏 also belongs to (𝑎𝑎, 𝑏𝑏]. iv. {𝑥𝑥: 𝑎𝑎 ≤ 𝑥𝑥 < 𝑏𝑏} is called right half open interval, denoted by [𝑎𝑎, 𝑏𝑏) 𝑖𝑖. 𝑒𝑒. all those points who lying between 𝑎𝑎 and 𝑏𝑏 belongs to [𝑎𝑎, 𝑏𝑏) and the point 𝑎𝑎 also belongs to [𝑎𝑎, 𝑏𝑏). (𝑎𝑎, 𝑏𝑏) [𝑎𝑎, 𝑏𝑏] (𝑎𝑎, 𝑏𝑏] [𝑎𝑎, 𝑏𝑏) 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 𝑎𝑎 𝑏𝑏 Some useful results; Theorem 1: Every set is a subset of itself. Proof 1: Let 𝐴𝐴 be any set. ∵ Every element of 𝐴𝐴 is in 𝐴𝐴. ∴ 𝐴𝐴 ⊆ 𝐴𝐴 Hence, every set is a subset of itself. Theorem 2: The empty set is a subset of any set. Proof 2: Let 𝐴𝐴 be any set and let 𝜙𝜙 be the empty set. ∵ 𝜙𝜙 contains no element at all 3 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th ∴ there is no element of 𝜙𝜙 which is not contained in 𝐴𝐴. 𝑖𝑖. 𝑒𝑒., 𝜙𝜙 ⊂ 𝐴𝐴. Hence, the empty set is a subset of any set. Theorem 3: The total number of subsets of a set containing 𝑛𝑛 elements is 2𝑛𝑛 . Proof 3: Let 𝐴𝐴 be any finite set containing 𝑛𝑛 elements then, If 𝐴𝐴 containing no element, the number of subsets of 𝐴𝐴 = 𝑛𝑛𝐶𝐶0 = 1. If 𝐴𝐴 containing one element, the number of subsets of 𝐴𝐴 = 𝑛𝑛𝐶𝐶0 + 𝑛𝑛𝐶𝐶1 = 2. If 𝐴𝐴 containing two elements, the number of subsets of 𝐴𝐴 = 𝑛𝑛𝐶𝐶0 + 𝑛𝑛𝐶𝐶1 + 𝑛𝑛𝐶𝐶2 = 4. If 𝐴𝐴 containing 𝑛𝑛 elements, the number of subsets of 𝐴𝐴 = 𝑛𝑛𝐶𝐶0 + 𝑛𝑛𝐶𝐶1 + 𝑛𝑛𝐶𝐶2 + ⋯ + 𝑛𝑛𝐶𝐶𝑛𝑛 . Using binomial theorem, we have 𝑛𝑛 𝐶𝐶0 + 𝑛𝑛𝐶𝐶1 + 𝑛𝑛𝐶𝐶2 + ⋯ + 𝑛𝑛𝐶𝐶𝑛𝑛 = (1 + 1)𝑛𝑛 = 2𝑛𝑛 Hence, the total number of subsets of a set containing 𝑛𝑛 elements is 2𝑛𝑛 . Further, the number of subsets of 𝐴𝐴, each containing no element is 𝑛𝑛𝐶𝐶0 = 1. the number of subsets of 𝐴𝐴, each containing one element is 𝑛𝑛𝐶𝐶1 = 𝑛𝑛. 𝑛𝑛(𝑛𝑛 − 1) the number of subsets of 𝐴𝐴, each containing two elements is 𝑛𝑛𝐶𝐶2 = 2 𝑛𝑛 the number of subsets of 𝐴𝐴, each containing 𝑛𝑛 elements is 𝐶𝐶𝑛𝑛 = 1. Theorem 4: Let 𝐴𝐴 and 𝐵𝐵 be two sets. Then, prove that 𝐴𝐴 = 𝐵𝐵 if and only if 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴. Proof 4: Let 𝐴𝐴 = 𝐵𝐵 Then, by the definition of equal sets, every element of 𝐴𝐴 is an element of 𝐵𝐵 and every element of 𝐵𝐵 is an element of 𝐴𝐴. This implies, 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴 Thus, 𝐴𝐴 = 𝐵𝐵 ⟹ 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴 Then, by the definition of subsets, if 𝐴𝐴 ⊆ 𝐵𝐵 implies every element of 𝐴𝐴 is an element of 𝐵𝐵 and if 𝐵𝐵 ⊆ 𝐴𝐴 implies every element of 𝐵𝐵 is an element of 𝐴𝐴. We have, 𝐴𝐴 = 𝐵𝐵 Thus, 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴 ⟹ 𝐴𝐴 = 𝐵𝐵 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴 = 𝐵𝐵 ⟺ 𝐴𝐴 ⊆ 𝐵𝐵 and 𝐵𝐵 ⊆ 𝐴𝐴 Operation on two sets: There are three types of operations; a) Union: Let 𝐴𝐴 and 𝐵𝐵 be two sets, the union of 𝐴𝐴 and 𝐵𝐵 is the set which consists of all the elements of 𝐴𝐴 and 𝐵𝐵, but the common U elements being taken only once. It is denoted by 𝐴𝐴⋃𝐵𝐵 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑎𝑎𝑎𝑎 ′𝐴𝐴 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝐵𝐵′. A B Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {3,4,5,6}, then, 𝐴𝐴⋃𝐵𝐵 = {1,2,3,4,5,6} Properties of union: Let 𝐴𝐴 and 𝐵𝐵 be two subsets of universal set 𝑈𝑈 then; i. 𝐴𝐴⋃𝐴𝐴 = 𝐴𝐴 4 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴⋃𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 (using idempotent law) ∴ 𝐴𝐴⋃𝐴𝐴 ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃𝐴𝐴 (using idempotent law) ∴ 𝐴𝐴 ⊆ 𝐴𝐴⋃𝐴𝐴 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃𝐴𝐴 = 𝐴𝐴 ii. 𝐴𝐴⋃𝑈𝑈 = 𝑈𝑈 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃𝑈𝑈. Then, 𝑥𝑥 ∈ 𝐴𝐴⋃𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝑈𝑈 (using universal law) ∴ 𝐴𝐴⋃𝑈𝑈 ⊆ 𝑈𝑈 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝑈𝑈. Then, 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃𝑈𝑈 (using universal law) ∴ 𝑈𝑈 ⊆ 𝐴𝐴⋃𝑈𝑈 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃𝑈𝑈 = 𝑈𝑈 iii. 𝐴𝐴⋃𝜙𝜙 = 𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃𝜙𝜙. Then, 𝑥𝑥 ∈ 𝐴𝐴⋃𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 (using identity law) ∴ 𝐴𝐴⋃𝜙𝜙 ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃𝜙𝜙 (using identity law) ∴ 𝐴𝐴 ⊆ 𝐴𝐴⋃𝜙𝜙 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃𝜙𝜙 = 𝐴𝐴 iv. 𝐴𝐴⋃𝐵𝐵 = 𝐵𝐵⋃𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃𝐵𝐵. Then, 𝑥𝑥 ∈ 𝐴𝐴⋃𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐵𝐵⋃𝐴𝐴 (using commutative law) ∴ 𝐴𝐴⋃𝐵𝐵 ⊆ 𝐵𝐵⋃𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐵𝐵⋃𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐵𝐵⋃𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃𝐵𝐵 (using commutative law) ∴ 𝐵𝐵⋃𝐴𝐴 ⊆ 𝐴𝐴⋃𝐵𝐵 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃𝐵𝐵 = 𝐵𝐵⋃𝐴𝐴 v. (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 = 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶. Then, 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ⇒ (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 Using associative law, we have 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 (𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐵𝐵⋃𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) 5 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th ∴ (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 ⊆ 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶). Then, 𝑥𝑥 ∈ 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐵𝐵⋃𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 (𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ) Using associative law, we have (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 ∴ 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) ⊆ (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 ……………………… (2) Hence, from equation (1) and (2) we have, (𝐴𝐴⋃𝐵𝐵)⋃𝐶𝐶 = 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶) b) Intersection: Let 𝐴𝐴 and 𝐵𝐵 be two sets, the intersection of 𝐴𝐴 and U 𝐵𝐵 is the set of all elements which are common in both 𝐴𝐴 and 𝐵𝐵. It is denoted by 𝐴𝐴⋂𝐵𝐵 read as ‘𝐴𝐴 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝐵𝐵’. A B Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {3,4,5,6}, then, 𝐴𝐴⋂𝐵𝐵 = {3,4} Properties of intersection: Let 𝐴𝐴 and 𝐵𝐵 be two subsets of 𝑈𝑈 then; 𝑨𝑨⋂𝑩𝑩 i. 𝐴𝐴⋂𝐴𝐴 = 𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴⋂𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 (using idempotent law) ∴ 𝐴𝐴⋂𝐴𝐴 ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂𝐴𝐴 (using idempotent law) ∴ 𝐴𝐴 ⊆ 𝐴𝐴⋂𝐴𝐴 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋂𝐴𝐴 = 𝐴𝐴 ii. 𝐴𝐴⋂𝑈𝑈 = 𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂𝑈𝑈. Then, 𝑥𝑥 ∈ 𝐴𝐴⋂𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 (using universal law) ∴ 𝐴𝐴⋂𝑈𝑈 ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂𝑈𝑈 (using universal law) ∴ 𝐴𝐴 ⊆ 𝐴𝐴⋂𝑈𝑈 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋂𝑈𝑈 = 𝐴𝐴 iii. 𝐴𝐴⋂𝜙𝜙 = 𝜙𝜙 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂𝜙𝜙. Then, 𝑥𝑥 ∈ 𝐴𝐴⋂𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝜙𝜙 (using identity law) ∴ 𝐴𝐴⋂𝜙𝜙 ⊆ 𝜙𝜙 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝜙𝜙. Then, 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂𝜙𝜙 (using identity law) ∴ 𝜙𝜙 ⊆ 𝐴𝐴⋂𝜙𝜙 ……………………… (2) Hence, from equation (1) and (2) we have, 6 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th 𝐴𝐴⋂𝜙𝜙 = 𝜙𝜙 iv. 𝐴𝐴⋂𝐵𝐵 = 𝐵𝐵⋂𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂𝐵𝐵. Then, 𝑥𝑥 ∈ 𝐴𝐴⋂𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐵𝐵⋂𝐴𝐴 (commutative law) ∴ 𝐴𝐴⋂𝐵𝐵 ⊆ 𝐵𝐵⋂𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐵𝐵⋂𝐴𝐴. Then, 𝑥𝑥 ∈ 𝐵𝐵⋂𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂𝐵𝐵 (commutative law) ∴ 𝐵𝐵⋂𝐴𝐴 ⊆ 𝐴𝐴⋂𝐵𝐵 ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋂𝐵𝐵 = 𝐵𝐵⋂𝐴𝐴 v. (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 = 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶. Then, 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ⇒ (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 Using associative law, we have 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 (𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐵𝐵⋂𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) ∴ (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 ⊆ 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃(𝐵𝐵⋃𝐶𝐶). Then, 𝑥𝑥 ∈ 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐵𝐵⋂𝐶𝐶) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 (𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ) Using associative law, we have (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 ∴ 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) ⊆ (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 ……………………… (2) Hence, from equation (1) and (2) we have, (𝐴𝐴⋂𝐵𝐵)⋂𝐶𝐶 = 𝐴𝐴⋂(𝐵𝐵⋂𝐶𝐶) vi. 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) = (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶) Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶). Then, 𝑥𝑥 ∈ 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐵𝐵⋂𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 { 𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 } Using distributive law, we have (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶) ∴ 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) ⊆ (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶) ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶). Then, 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐴𝐴⋃𝐶𝐶 ) ⇒ (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵) 𝑎𝑎𝑎𝑎𝑎𝑎 (𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 ) Using distributive law, we have 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 { 𝑥𝑥 ∈ 𝐵𝐵 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 } ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐵𝐵⋂𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) ∴ (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶) ⊆ 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃(𝐵𝐵⋂𝐶𝐶 ) = (𝐴𝐴⋃𝐵𝐵)⋂(𝐴𝐴⋃𝐶𝐶) vii. 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) = (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶) Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶). Then, 𝑥𝑥 ∈ 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐵𝐵⋃𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 { 𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 } 7 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th Using distributive law, we have (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵) 𝑜𝑜𝑜𝑜 (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶) ∴ 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) ⊆ (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶) ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶). Then, 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵) 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ (𝐴𝐴⋂𝐶𝐶 ) ⇒ (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵) 𝑜𝑜𝑜𝑜 (𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐶𝐶 ) Using distributive law, we have 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 { 𝑥𝑥 ∈ 𝐵𝐵 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐶𝐶 } ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ (𝐵𝐵⋂𝐶𝐶 ) ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) ∴ (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶) ⊆ 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋂(𝐵𝐵⋃𝐶𝐶 ) = (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴⋂𝐶𝐶) c) Difference: Let 𝐴𝐴 and 𝐵𝐵 be two sets, the difference of 𝐴𝐴 and 𝐵𝐵 is the set of all elements which are a member of 𝐴𝐴 but not a member of 𝐵𝐵. It is denoted by 𝐴𝐴 − 𝐵𝐵. Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {3,4,5,6}, then, 𝐴𝐴 − 𝐵𝐵 = {1,2} and A B 𝐵𝐵 − 𝐴𝐴 = {5,6}. 𝑨𝑨 − 𝑩𝑩 Properties of difference: Let 𝐴𝐴 and 𝐵𝐵 be two subsets of 𝑈𝑈 then; i. 𝐴𝐴 − 𝐵𝐵 ≠ 𝐵𝐵 − 𝐴𝐴 ii. (𝐴𝐴 − 𝐵𝐵) − 𝐶𝐶 ≠ 𝐴𝐴 − (𝐵𝐵 − 𝐶𝐶) iii. 𝐴𝐴 − 𝜙𝜙 = 𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴 − 𝜙𝜙). Then, 𝑥𝑥 ∈ (𝐴𝐴 − 𝜙𝜙) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 ∴ (𝐴𝐴 − 𝜙𝜙) ⊆ 𝐴𝐴 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ (𝐴𝐴 − 𝜙𝜙) ∴ 𝐴𝐴 ⊆ (𝐴𝐴 − 𝜙𝜙) ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴 − 𝜙𝜙 = 𝐴𝐴 iv. 𝐴𝐴 − 𝐴𝐴 = 𝜙𝜙 (Do yourself) Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴 − 𝐴𝐴). Then, 𝑥𝑥 ∈ (𝐴𝐴 − 𝐴𝐴) ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝜙𝜙 ∴ (𝐴𝐴 − 𝐴𝐴) ⊆ 𝜙𝜙 ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ (𝐴𝐴 − 𝐴𝐴) ∴ 𝜙𝜙 ⊆ (𝐴𝐴 − 𝐴𝐴) ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴 − 𝐴𝐴 = 𝜙𝜙 v. 𝑈𝑈 − 𝐴𝐴 = 𝐴𝐴′ (Do yourself) 8 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th Let 𝑥𝑥 be an arbitrary element of (𝑈𝑈 − 𝐴𝐴). Then, 𝑥𝑥 ∈ (𝑈𝑈 − 𝐴𝐴) ⇒ 𝑥𝑥 ∈ 𝑈𝑈 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ 𝑈𝑈 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝑈𝑈⋂𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴′ ∴ (𝑈𝑈 − 𝐴𝐴) ⊆ 𝐴𝐴′ ……………………… (1) Conversely, let 𝑥𝑥 be an arbitrary element of 𝐴𝐴′ 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝑈𝑈⋂𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝑈𝑈 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝑈𝑈 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ (𝑈𝑈 − 𝐴𝐴) ∴ 𝐴𝐴′ ⊆ (𝑈𝑈 − 𝐴𝐴) ……………………… (2) Hence, from equation (1) and (2) we have, 𝑈𝑈 − 𝐴𝐴 = 𝐴𝐴′ Symmetric difference: The symmetric difference of two sets 𝐴𝐴 and 𝐵𝐵 is defined as, 𝐴𝐴∆𝐵𝐵 = (𝐴𝐴 − 𝐵𝐵)⋃(𝐵𝐵 − 𝐴𝐴) Example: 𝐴𝐴 = {1,2,3,4} , 𝐵𝐵 = {3,4,5,6}, then, 𝐴𝐴∆𝐵𝐵 = {1,2,5,6} Complement of a set: Let 𝑈𝑈 be a universal set and 𝐴𝐴 ⊆ 𝑈𝑈, then 𝐴𝐴’ is said to be the complement of the set 𝐴𝐴, if 𝐴𝐴′ ⊆ 𝑈𝑈 but 𝐴𝐴’ is consists A’ all the elements which are not an element of 𝐴𝐴. A Example: 𝑈𝑈 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐, 𝑑𝑑, 𝑒𝑒, 𝑓𝑓} , 𝐴𝐴 = {𝑎𝑎, 𝑏𝑏, 𝑐𝑐 } , 𝐴𝐴′ = {𝑑𝑑, 𝑒𝑒, 𝑓𝑓} Properties of complement: Let 𝐴𝐴 and 𝐴𝐴′ be two subsets of 𝑈𝑈 such that 𝐴𝐴′ is the complement of the set 𝐴𝐴 then; i. 𝐴𝐴⋂𝐴𝐴′ = 𝜙𝜙 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋂𝐴𝐴′ . Then, 𝑥𝑥 ∈ 𝐴𝐴⋂𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝜙𝜙 ∴ 𝐴𝐴⋂𝐴𝐴′ ⊆ 𝜙𝜙 ……………………… (1) Conversely, let 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋂𝐴𝐴′ ∴ 𝜙𝜙 ⊆ 𝐴𝐴⋂𝐴𝐴′ ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋂𝐴𝐴′ = 𝜙𝜙 ii. 𝐴𝐴⋃𝐴𝐴′ = 𝑈𝑈 Let 𝑥𝑥 be an arbitrary element of 𝐴𝐴⋃𝐴𝐴′ . Then, 𝑥𝑥 ∈ 𝐴𝐴⋃𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝑈𝑈 ∴ 𝐴𝐴⋃𝐴𝐴′ ⊆ 𝑈𝑈 ……………………… (1) Conversely, let 𝑥𝑥 ∈ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴⋃𝐴𝐴′ ∴ 𝑈𝑈 ⊆ 𝐴𝐴⋂𝐴𝐴′ ……………………… (2) Hence, from equation (1) and (2) we have, 𝐴𝐴⋃𝐴𝐴′ = 𝑈𝑈 iii. (𝐴𝐴′ )′ = 𝐴𝐴 Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴′ )′.Then, 𝑥𝑥 ∈ (𝐴𝐴′ )′ ⇒ 𝑥𝑥 ∉ 𝐴𝐴′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴 ∴ (𝐴𝐴′ )′ ⊆ 𝐴𝐴 ……………………… (1) 9 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th ′ ′ )′ Conversely, let 𝑥𝑥 ∈ 𝐴𝐴 ⇒ 𝑥𝑥 ∉ 𝐴𝐴 ⇒ 𝑥𝑥 ∈ (𝐴𝐴 ∴ 𝐴𝐴 ⊆ (𝐴𝐴′ )′ ……………………… (2) Hence, from equation (1) and (2) we have, (𝐴𝐴′ )′ = 𝐴𝐴 iv. 𝑈𝑈′ = 𝜙𝜙 and 𝜙𝜙′ = 𝑈𝑈 Let 𝑥𝑥 be an arbitrary element of 𝑈𝑈′.Then, 𝑥𝑥 ∈ 𝑈𝑈′. ⇒ 𝑥𝑥 ∉ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝜙𝜙 ∴ 𝑈𝑈′ ⊆ 𝜙𝜙 ……………………… (1) Conversely, let 𝑥𝑥 ∈ 𝜙𝜙 ⇒ 𝑥𝑥 ∉ 𝑈𝑈 ⇒ 𝑥𝑥 ∈ 𝑈𝑈 ′ ∴ 𝜙𝜙 ⊆ 𝑈𝑈′ ……………………… (2) Hence, from equation (1) and (2) we have, 𝑈𝑈′ = 𝜙𝜙 v. (𝐴𝐴⋃𝐵𝐵)′ = 𝐴𝐴′⋂𝐵𝐵′ Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋃𝐵𝐵)′ . Then, 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)′ ⇒ 𝑥𝑥 ∉ 𝐴𝐴⋃𝐵𝐵 ⇒ 𝑥𝑥 ∉ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∉ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴′ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴′⋂𝐵𝐵′ ∴ (𝐴𝐴⋃𝐵𝐵)′ ⊆ 𝐴𝐴′⋂𝐵𝐵′ ……………………… (1) ′ ′ Conversely, let 𝑥𝑥 ∈ 𝐴𝐴 ⋂𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴′ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∈ 𝐵𝐵′ ⇒ 𝑥𝑥 ∉ 𝐴𝐴 𝑜𝑜𝑜𝑜 𝑥𝑥 ∉ 𝐵𝐵 ⇒ 𝑥𝑥 ∉ 𝐴𝐴⋃𝐵𝐵 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋃𝐵𝐵)′ ∴ 𝐴𝐴′⋂𝐵𝐵′ ⊆ (𝐴𝐴⋃𝐵𝐵)′ ……………………… (2) Hence, from equation (1) and (2) we have, (𝐴𝐴⋃𝐵𝐵)′ = 𝐴𝐴′⋂𝐵𝐵′ vi. (𝐴𝐴⋂𝐵𝐵)′ = 𝐴𝐴′⋃𝐵𝐵′ Let 𝑥𝑥 be an arbitrary element of (𝐴𝐴⋂𝐵𝐵)′ . Then, 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)′ ⇒ 𝑥𝑥 ∉ 𝐴𝐴⋂𝐵𝐵 ⇒ 𝑥𝑥 ∉ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐵𝐵 ⇒ 𝑥𝑥 ∈ 𝐴𝐴′ 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴′⋃𝐵𝐵′ ∴ (𝐴𝐴⋂𝐵𝐵)′ ⊆ 𝐴𝐴′⋃𝐵𝐵′ ……………………… (1) Conversely, let 𝑥𝑥 ∈ 𝐴𝐴′⋃𝐵𝐵′ ⇒ 𝑥𝑥 ∈ 𝐴𝐴′ 𝑜𝑜𝑜𝑜 𝑥𝑥 ∈ 𝐵𝐵′ ⇒ 𝑥𝑥 ∉ 𝐴𝐴 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 ∉ 𝐵𝐵 ⇒ 𝑥𝑥 ∉ 𝐴𝐴⋂𝐵𝐵 ⇒ 𝑥𝑥 ∈ (𝐴𝐴⋂𝐵𝐵)′ ∴ 𝐴𝐴′⋃𝐵𝐵′ ⊆ (𝐴𝐴⋂𝐵𝐵)′ ……………………… (2) Hence, from equation (1) and (2) we have, (𝐴𝐴⋂𝐵𝐵)′ = 𝐴𝐴′ ⋃𝐵𝐵′ Some useful results; for any set 𝐴𝐴, 𝐵𝐵 and 𝐶𝐶, we have 𝐴𝐴 ⊆ 𝐵𝐵 = 𝐵𝐵′ ⊆ 𝐴𝐴′ 𝐴𝐴⋃(𝐴𝐴⋂𝐵𝐵) = 𝐴𝐴 𝐴𝐴 − 𝐵𝐵 = 𝐴𝐴⋂𝐵𝐵′ (𝐴𝐴 − 𝐵𝐵)⋂𝐵𝐵 = 𝜙𝜙 𝐴𝐴⋂(𝐴𝐴⋃𝐵𝐵) = 𝐴𝐴 𝐵𝐵 − 𝐴𝐴 = 𝐵𝐵⋂𝐴𝐴′ (𝐴𝐴 − 𝐵𝐵) = 𝐴𝐴 ⟺ (𝐴𝐴⋂𝐵𝐵) = 𝜙𝜙 (𝐴𝐴⋂𝐵𝐵)⋃(𝐴𝐴 − 𝐵𝐵) = 𝐴𝐴 𝑃𝑃(𝐴𝐴⋂𝐵𝐵) = 𝑃𝑃(𝐴𝐴)⋂ 𝑃𝑃(𝐵𝐵) (𝐴𝐴⋃𝐵𝐵) = (𝐴𝐴⋂𝐵𝐵) ⟺ 𝐴𝐴 = 𝐵𝐵 𝐴𝐴⋃(𝐵𝐵 − 𝐴𝐴) = (𝐴𝐴⋃𝐵𝐵) 𝑃𝑃(𝐴𝐴)⋃𝑃𝑃(𝐵𝐵) ⊂ 𝑃𝑃(𝐴𝐴⋃𝐵𝐵) 𝐴𝐴 − (𝐵𝐵⋃𝐶𝐶 ) = (𝐴𝐴 − 𝐵𝐵)⋂(𝐴𝐴 − 𝐶𝐶 ) 𝐴𝐴 − (𝐵𝐵⋂𝐶𝐶 ) = (𝐴𝐴 − 𝐵𝐵)⋃(𝐴𝐴 − 𝐶𝐶 ) 10 ©Copyright Vishika Group of Companies. All Rights Reserved.  SET THEORY 11th (𝐴𝐴⋃𝐵𝐵) − 𝐶𝐶 = (𝐴𝐴 − 𝐶𝐶 )⋃(𝐵𝐵 − 𝐶𝐶 ) (𝐴𝐴⋂𝐵𝐵) − 𝐶𝐶 = (𝐴𝐴 − 𝐶𝐶 )⋂(𝐵𝐵 − 𝐶𝐶 ) (𝐴𝐴 − 𝐵𝐵)⋃(𝐵𝐵 − 𝐴𝐴) = (𝐴𝐴⋃𝐵𝐵) − (𝐴𝐴⋂𝐵𝐵) (𝐴𝐴 − 𝐵𝐵)⋂(𝐵𝐵 − 𝐴𝐴) = (𝐴𝐴⋂𝐵𝐵) − (𝐴𝐴⋃𝐵𝐵) Application of sets: There are two applications of sets; a) In general, let 𝐴𝐴 and 𝐵𝐵 be two finite sets, then 𝑛𝑛(𝐴𝐴⋃𝐵𝐵) = 𝑛𝑛(𝐴𝐴) + 𝑛𝑛(𝐵𝐵) − 𝑛𝑛(𝐴𝐴⋂𝐵𝐵) Example: In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis? Solution: Let 𝐴𝐴 be the set of people who like cricket and let 𝐵𝐵 be the set of people who like tennis. Then, the set of people who like both cricket and tennis will be 𝐴𝐴⋂𝐵𝐵 and the set of people who like tennis only and not cricket will be 𝑈𝑈 − 𝐴𝐴. Therefore, 𝑛𝑛(𝐴𝐴) = 40, 𝑛𝑛(𝐴𝐴⋃𝐵𝐵) = 65, 𝑛𝑛(𝐴𝐴⋂𝐵𝐵) = 10, 𝑛𝑛(𝐵𝐵) =? 𝑛𝑛(𝑈𝑈 − 𝐴𝐴) = 𝑛𝑛(𝐴𝐴⋃𝐵𝐵) − 𝑛𝑛(𝐴𝐴) = 65 − 40 = 25 Now, 𝑛𝑛(𝐴𝐴⋃𝐵𝐵) = 𝑛𝑛(𝐴𝐴) + 𝑛𝑛(𝐵𝐵) − 𝑛𝑛(𝐴𝐴⋂𝐵𝐵) 65 = 40 + 𝑛𝑛(𝐵𝐵) − 10 𝑛𝑛(𝐵𝐵) = 35 b) In general, let 𝐴𝐴, 𝐵𝐵 and 𝐶𝐶 be three finite sets, then 𝑛𝑛(𝐴𝐴⋃𝐵𝐵⋃𝐶𝐶 ) = 𝑛𝑛(𝐴𝐴) + 𝑛𝑛(𝐵𝐵) + 𝑛𝑛(𝐶𝐶 ) − 𝑛𝑛(𝐴𝐴⋂𝐵𝐵) − 𝑛𝑛(𝐵𝐵⋂𝐶𝐶 ) − 𝑛𝑛(𝐴𝐴⋂𝐶𝐶 ) + 𝑛𝑛(𝐴𝐴⋂𝐵𝐵⋂𝐶𝐶) Example: In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find the number of people who read at least one of the newspapers and the number of people who read exactly one newspaper. Solution: Let us consider that 𝑈𝑈 be the universal set then, 𝑛𝑛(𝑈𝑈) = 60, 𝑛𝑛(𝐻𝐻) = 25, 𝑛𝑛(𝑇𝑇) = 26, 𝑛𝑛(𝐼𝐼 ) = 26, 𝑛𝑛(𝐻𝐻⋂𝑇𝑇) = 11, 𝑛𝑛(𝑇𝑇⋂𝐼𝐼 ) = 8, 𝑛𝑛(𝐻𝐻⋂𝐼𝐼 ) = 9, 𝑛𝑛(𝐻𝐻⋂𝑇𝑇⋂𝐼𝐼 ) = 3 ∴ 𝑛𝑛(𝐻𝐻⋃𝑇𝑇⋃𝐼𝐼 ) = 𝑛𝑛(𝐻𝐻) + 𝑛𝑛(𝑇𝑇) + 𝑛𝑛(𝐼𝐼 ) − 𝑛𝑛(𝐻𝐻⋂𝑇𝑇) − 𝑛𝑛(𝑇𝑇⋂𝐼𝐼 ) − 𝑛𝑛(𝐻𝐻⋂𝐼𝐼 ) + 𝑛𝑛(𝐻𝐻⋂𝑇𝑇⋂𝐼𝐼) = 25 + 26 + 26 − 11 − 9 − 8 + 3 𝑛𝑛(𝐻𝐻⋃𝑇𝑇⋃𝐼𝐼 ) = 52 The number of people who read exactly one newspaper is = 𝑛𝑛(𝐻𝐻⋃𝑇𝑇⋃𝐼𝐼 ) − {𝑛𝑛(𝐻𝐻⋂𝑇𝑇) + 𝑛𝑛(𝑇𝑇⋂𝐼𝐼 ) + 𝑛𝑛(𝐻𝐻⋂𝐼𝐼 ) − 2𝑛𝑛(𝐻𝐻⋂𝑇𝑇⋂𝐼𝐼 )} = 52 − {11 + 9 + 8 − 6} = 30 11 ©Copyright Vishika Group of Companies. All Rights Reserved.