Properties of Sets Class 11 with Proof
1.Introduction to Properties of Sets Class 11 with Proof
In this article “Properties of Sets Class 11 with Proof”, we will read sets properties with proofs viz union of sets,intersection of sets and difference of Sets.
“Before diving into these solutions,make sure you understand the foundational concept of “Types of Sets Class 11th” and have reviewed the essential “Subsets in Class 11“. This will make following today’s step-by-step solutions much easier.”
2.Properties of Sets Class 11 with Proof
\text{(i) } A \cup B = B \cup A \ (\text{commutative law}) \\ \text{proof : } (i) \ A \cup B = B \cup A \\ \text{Let } A \text{ and } B \text{ are sets, and} \\ \text{let } x \in A \cup B \Rightarrow x \in A \lor x \in B \\ \Rightarrow x \in B \lor x \in A \\ \Rightarrow x \in B \cup A \\ \Rightarrow A \cup B = B \cup A \\ (ii) \ (A \cup B) \cup C = A \cup (B \cup C) \\ \text{(Associative law)} \\ \text{proof : } (A \cup B) \cup C = A \cup (B \cup C) \\ \text{Let } A, B, C \text{ are three sets and} \\ x \in (A \cup B) \cup C \Rightarrow [x \in (A \cup B)] \lor x \in C \\ \Rightarrow [x \in A \lor x \in B] \lor x \in C \\ \Rightarrow x \in A \lor x \in B \lor x \in C \\ \Rightarrow x \in A \lor [x \in B \lor x \in C] \\ \Rightarrow x \in A \lor [x \in B \cup C] \\ \Rightarrow x \in A \cup (B \cup C) \quad \cdots (1) \\ \text{Similarly } x \in A \cup (B \cup C) \Rightarrow x \in (A \cup B) \cup C \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ x \in A \cup (B \cup C) \iff x \in (A \cup B) \cup C \\ \Rightarrow A \cup (B \cup C) = (A \cup B) \cup C \\ (iii) \ A \cup \phi = A \ (\text{Law of identity element, } \phi \text{ is the identity of U})\\ \text{proof : } A \cup \phi = A \\ \text{Let } x \in A \cup \phi \\ x \in A \cup \phi \Rightarrow x \in A \lor x \in \phi \\ \Rightarrow x \in A [\because x \notin \phi, \phi \text{ is empty set}] \\ x \in A \cup \phi \Rightarrow x \in A \\ \Rightarrow A \cup \phi \subset A \quad \cdots (1) \\ A \subset A \cup B \\ \text{If } B = \phi \text{ then} \\ A \subset A \cup \phi \\ \Rightarrow A \subset A \cup \phi \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A \cup \phi = A \\ (iv). \text{ } A \cup A = A \ (\text{Idempotent law}) \\ \text{proof : } A \cup A = A \\ B \subset A \cup B \\ \text{If } B = A \text{ then} \\ A \subset A \cup A \quad \cdots (1) \\ \text{Let } x \in A \cup A \Rightarrow x \in A \lor x \in A \\ \Rightarrow x \in A \\ A \cup A \subset A \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A \cup A = A \\ (v) \text{ } U \cup A = U \ (\text{Law of } U) \\ \text{proof : } U \cup A = U \\ \text{We know that}\\ \text{If } B = U \text{ then} \\ U \subset U \cup A \quad \cdots (1) \\ x \in U \cup A \Rightarrow x \in U \lor x \in A \\ \Rightarrow x \in U \ [\because U \text{ is universal set}] \\ U \cup A \subset U \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ \Rightarrow U \cup A = U \\ (vi) \text{ } A \subset A \cup B \text{ and } B \subset A \cup B \\ \text{proof : } A \subset A \cup B \text{ and } B \subset A \cup B \\ \text{Let } x \in A, \text{ then} \\ x \in A \Rightarrow x \in A \cup B \ [\because \text{ by def.} \\ A \cup B \text{ contains all elements of } A \text{ and } B] \\ \therefore A \subset A \cup B \quad \cdots (1) \\ \text{Similarly } B \subset A \cup B \\ \text{Hence } A \subset A \cup B \text{ and } B \subset A \cup B \\ (vii) \text{ } A \subseteq B \iff A \cup B = B \\ \text{Let } A \subseteq B \text{ then} \\ x \in A \cup B \Rightarrow x \in A \lor x \in B \\ \Rightarrow x \in B \lor x \in B \ [\because A \subseteq B] \\ A \cup B \subseteq B \quad \cdots (1) \\ \text{Again } y \in B \Rightarrow y \in A \lor y \in B \ [\because A \subseteq B] \\ \Rightarrow y \in A \cup B \\ \therefore B \subseteq A \cup B \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A \cup B = B \quad \cdots (3) \\ \text{Conversely :} \\ \text{Let } A \cup B = B \text{ then} \\ x \in A \Rightarrow x \in B \ [\because A \cup B = B] \\ A \subseteq B \quad \cdots (4) \\ x \in B \Rightarrow x \in A \cup B \ [\because A \cup B = B] \\ \Rightarrow x \in A \lor x \in B\\ \Rightarrow x \in A \\ \text{B} \subset A \quad \cdots (5) \\ \text{From } (4) \text{ and } (5), \text{ we get} \\ A \cup B = B \Rightarrow A \subseteq B \quad \cdots (6) \\ \text{From } (3) \text{ and } (6), \text{ we get} \\ A \subseteq B \iff A \cup B = B \\ (viii). \ A \subseteq C, B \subseteq C \Rightarrow A \cup B \subseteq C \\ \text{Let } A \subseteq C, B \subseteq C \text{ then} \\ x \in A \cup B \Rightarrow x \in A \lor x \in B \\ \Rightarrow x \in C \lor x \in C \ [\because A \subseteq C, B \subseteq C] \\ \Rightarrow x \in A \cup B \\ \Rightarrow C \subset A \cup B \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A \cup B \subseteq C \\ (ix) \ A \cup A' = U \\ \text{proof : } A \cup A' = U \\ x \in A \cup A' \Rightarrow x \in A \lor x \in A' \\ \Rightarrow x \in U \ [\because A \subseteq U, A' \subseteq U] \\ \Rightarrow A \cup A' \subset U \quad \cdots (1) \\ x \in U \Rightarrow x \in A \lor x \in A' \\ \ [\because A \subseteq U, A' \subseteq U] \\ \Rightarrow x \in A \cup A' \\ U \subset A \cup A' \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A \cup A' = U2. Some properties of operation of Intersection
(i) \ A \cap B = B \cap A \ (\text{Commutative law}) \\ \text{Proof : } A \cap B = B \cap A \\ \text{Let } A \text{ and } B \text{ are sets} \\ A \cap B = \{x : x \in A \text{ and } x \in B\} \\ = \{x : x \in B \text{ and } x \in A\} \\ \Rightarrow A \cap B = B \cap A \\ (ii) \ (A \cap B) \cap C = A \cap (B \cap C)\\ x \in (A \cap B) \cap C \Rightarrow x \in A \cap B \land x \in C \\ \Rightarrow (x \in A \land x \in B) \land x \in C \\ \Rightarrow x \in A \land x \in B \land x \in C \\ \Rightarrow x \in A \land (x \in B \cap C) \\ \Rightarrow x \in A \cap (B \cap C) \quad \cdots (1) \\ \text{Similarly we can prove that} \\ x \in A \cap (B \cap C) \Rightarrow x \in (A \cap B) \cap C \quad \cdots (2) \\ \text{From } (1) \text{ and } (2) \text{, we get} \\ x \in (A \cap B) \cap C \iff x \in (A \cap B) \cap C \\ \Rightarrow (A \cap B) \cap C = (A \cap B) \cap C \\ (iii) \text{ } x \in A \Rightarrow x \notin \phi \\ x \in A, x \notin \phi \Rightarrow \phi \cap A = \phi \\ \text{Similarly} \\ x \in A \Rightarrow x \in U \ [\because A \subseteq U] \\ x \in A, x \in U \Rightarrow U \cap A = A \\ (iv) \text{ } A \cap A = A \\ \text{proof : } A \cap A = A \\ A \cap B \subset B \\ \text{If } B = A \text{ then} \\ A \cap A \subset A \\ \text{If } x \in A \Rightarrow x \in A \land x \in A \\ \Rightarrow x \in (A \cap A) \\ A \subset A \cap A \\ A \cap A \subset A, A \subset A \cap A \\ \Rightarrow A = A \cap A \\ (v) \text{ } A \cap (B \cup C) = (A \cup B) \cup (A \cap C) \\ \text{proof : } x \in A \cap (B \cup C) = (A \cup B) \cup (A \cap C) \\ x \in A \cap (B \cup C) \\ \Rightarrow x \in A \land x \in B \cup C\\ \Rightarrow (x \in A \land x \in A) \land (x \in B \lor x \in C) \\ \Rightarrow (x \in A \land x \in E) \lor (x \in A \land x \in C) \\ \Rightarrow x \in A \cap B \lor x \in A \cap C \\ \Rightarrow x \in (A \cap B) \cup (A \cap C) \\ \Rightarrow A \cap (B \cup C) \subset (A \cap B) \cup (A \cap C) \quad \cdots (1) \\ \text{Again } y \in (A \cap B) \cup (A \cap C) \\ \Rightarrow y \in (A \cap B) \lor y \in (A \cap C) \\ \Rightarrow (y \in A \land y \in B) \lor (y \in A \land y \in C) \\ \Rightarrow (y \in A \lor y \in A) \land (A \in B \lor y \in C) \\ \Rightarrow y \in A \land y \in B \cup C \\ \Rightarrow y \in A \cap (B \cup C) \\ \therefore (A \cap B) \cup (A \cap C) \subset A \cap (B \cup C) \quad \cdots (2) \\ \text{From } (1) \text{ and } (2) \text{, we get} \\ A \cap (B \cup C) = (A \cap B) \cup (A \cap C) \\ (vii) \ A \subset B \iff A \cap B = A \\ \text{proof : } A \subset B \iff A \cap B = A \\ \text{Let } A \subset B \text{ then} \\ x \in A \cap B \Rightarrow x \in A \land x \in B \\ \Rightarrow x \in A \land \ x \in A \ [\because A \subset B] \\ \Rightarrow x \in A \\ A \cup B \subset A \quad \cdots (1) \\ \text{Again } y \in A \Rightarrow y \in A \land y \in B \ [\because A \subset B] \\ \Rightarrow y \in A \cap B \\ \therefore A \subset A \cap B \quad \cdots (2) \\ \text{From } (1) \text{ and } (2) \text{, we get} \\ A \subset B \dots \Rightarrow A \cap B = A \quad \cdots (3) \\ \text{Conversely :} \\ \text{Let } A \cup B = A \\ \text{then } x \in A \Rightarrow x \notin A' \Rightarrow x \in B' \\ \therefore A \cap B = A \Rightarrow A \subset B \quad \cdots (4) \\ \text{Now from } (3) \text{ and } (4) ; \\ A \subset B \iff A \cap B = A \\ (viii) \text{ } A \subset C, B \subset C \Rightarrow A \cap B \subset C \\ \text{proof : } A \subset C, B \subset C \Rightarrow A \cap B \subset C \\ \text{Let } A \subset C, B \subset C \\ \text{Then } x \in A \cap B \Rightarrow x \in A \land x \in B \\ \Rightarrow x \in C \land x \in C \ [\because A \subset C, B \subset C] \\ \Rightarrow x \in C \\ A \cap B \subset C \\ A \subset C, B \subset C \Rightarrow A \cap B \subset C \\ (ix) \text{ } A \cap A' = \phi \\ x \in A \cap A' \Rightarrow x \in A \land x \notin A' \\ x \in A, x \notin A' \Rightarrow A \cap A' = \phi
3. Some properties of Difference operation
(i) \text{ } A - A = \phi \\ \text{proof : } A - A = \phi \\ A - \phi = A, \phi - A = \phi \\ x \in A - A \Rightarrow x \in A \land x \notin A \\ \Rightarrow x \in \phi \\ A - A \subset \phi \quad \cdots (1) \\ \text{But } \phi \subset A - A \quad \cdots (2) \\ \therefore \text{From } (1) \text{ and } (2), \text{ we get} \\ A - A = \phi \\ A - \phi = \{x : x \in A \land x \notin \phi\} \\ \Rightarrow x \in A \\ \Rightarrow A - \phi = A \\ \phi - A = \{x : x \in \phi \land x \notin A\} \\ \Rightarrow \phi - A = \phi \ [\because x = \phi] \\ (ii) \text{ } A - B \neq B - A \ (\text{when } A \neq B) \\ \text{proof : } A - B \neq B - A \\ A - B = \{x : x \in A \land x \notin B\} \\ B - A = \{x : x \in B \land x \notin A\}\\ \text{ i.e. } A - B \text{ has those element} \\ \text{which is not in } B, \text{ but is in } A. \\ \text{Similarly } B - A \text{ has those} \\ \text{element which is not in } A \text{ but is in } B. \\ A \neq B \\ \therefore A - B \text{ and } B - A \text{ has not} \\ \text{common element.} \\ \therefore (A - B) \cap (B - A) = \phi \\ \Rightarrow A - B \neq B - A \\ (iv) \ (A - B) \cap B = \phi \\ \text{proof : } (A - B) \cap B = \phi \\ \text{If } x \in B \Rightarrow x \notin A - B \ (\text{By def.}) \\ x \in B, x \notin A - B \Rightarrow (A - B) \cap B = \phi \\ (v) \ (A - B) \cup (A \cap B) = A \\ \text{proof : } (A - B) \cup (A \cap B) = A \\ \text{Let } A - B = P \\ (A - B) \cup (A \cap B) = P \cup (A \cap B) \\ = (P \cup A) \cap (P \cup B) \ [\text{By distributive law}] \\ = [(A - B) \cup A] \cap [(A - B) \cup B] \\ = [(A \cap B') \cup A] \cap [(A \cap B') \cup B] \\ \ [\because A - B = A \cap B'] \\ = [(A \cup A) \cap (B' \cup A)] \cap [(A \cup B) \cap (B' \cup B)] \\ = [A \cap (B' \cup A)] \cap [(A \cup B) \cap U] \\ = [A \cap (A \cup B')] \cap [(A \cup B) \cap U] \ [\text{By distributive law}] \\ = [A \cap (A \cup B') \cap (B' \cup B)] \cap [(A \cup A') \cap (B' \cup A)] \quad \text{(By distributive law)} \\ = [(A \cup B) \cap (B' \cup B)] \cap [(A \cup A') \cap (B' \cup A)] \quad \text{(By distributive law)} \\ = [(A \cup B) \cap U] \cap [U \cap (B' \cup A)] \ [\because A \cup A' = U] \\ = (A \cup B) \cap (A' \cup B') \ [\because A \cap U = U \cap A = A] \\ = (A \cup B) \cap (A \cap B)' \ [\because A' \cup B' = (A \cap B)'] \\ = (A \cup B) - (A \cap B) \ [\because A \cap B' = A - B] \\ \Rightarrow (A - B) \cup (B - A) = (A \cup B) - (A \cap B) \\ (vi) \ (A - B) \cup (B - A) = (A \cup B) - (A \cap B) \\ \text{proof : } (A - B) \cup (B - A) = (A \cup B) - (A \cap B) \\ \therefore A - B = A \cap B' \land B - A = B \cap A' \ [\because (viii)] \\ \text{L.H.S. } (A - B) \cup (B - A) \\ = (A \cap B') \cup (B \cap A')\\ = [(A \cap B') \cup B] \cap [(A \cap B') \cup A] \\ \text{(By distributive law)} \\ = [(A \cup B) \cap (B' \cup B)] \cap [(A \cup A) \cap (B' \cup A)] \\ \text{(By distributive law)} \\ = [(A \cup B) \cap U] \cap [U \cap (B' \cup A)] \ [\because A \cup A' = U] \\ = (A \cup B) \cap (A' \cup B') \ [\because A \cap U = U \cap A = A] \\ = (A \cup B) \cap (A \cap B)' \ [\because A' \cup B' = (A \cap B)'] \\ = (A \cup B) - (A \cap B) \ [\because A \cap B' = A - B] \\ \Rightarrow (A - B) \cup (B - A) = (A \cup B) - (A \cap B) \\ (vii) \text{ } A \subset B \Rightarrow A - B = \phi \\ \text{proof : } A \subset B \Rightarrow A - B = \phi \\ \text{Let } A \subset B \text{ then} \\ x \in A - B \Rightarrow x \in A \land x \notin B \\ \Rightarrow x \in B \land x \notin B \ [A \subset B] \\ \Rightarrow x \in \phi \\ \therefore A - B \subset \phi \\ \phi \subset A - B \ [\phi \text{ is subset of every set}] \\ \therefore A - B = \phi \\ (viii) \text{ } A - B = A \cap B', B - A = A' \cap B \\ \text{proof : } A - B = A \cap B', B - A = A' \cap B \\ \text{Let } x \in A - B \text{ then} \\ x \in A - B \Rightarrow x \in A \land x \notin B \\ \Rightarrow x \in A \land x \in B' \\ \Rightarrow x \in A \cap B' \\ \Rightarrow A - B \subset A \cap B' \quad \cdots (1) \\ \text{Again } y \in A \cap B' \text{ then} \\ y \in A \cap B' \Rightarrow y \in A \land y \in B' \\ \Rightarrow y \in A \land y \notin B \\ \Rightarrow y \in A - B \\ \therefore A \cap B' \subset A - B \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A - B = A \cap B' \\ \text{Similarly we can prove that} \\ B - A = A' \cap B4. De Morgan’s Law
(1) \text{ } (A \cup B)' = A' \cap B' \\ \text{proof : } (A \cup B)' = A' \cap B' \\ x \in (A \cup B)' \Rightarrow x \notin A \cup B \\ \Rightarrow x \notin A \land x \notin B \\ \Rightarrow x \in A' \land x \in B' \\ \Rightarrow x \in A' \cap B' \\ (A \cup B)' \subset A' \cap B' \quad \cdots (1) \\ \text{Conversely : } x \in A' \cap B' \Rightarrow x \in A' \land x \in B' \\ \Rightarrow x \notin A \land x \notin B \\ \Rightarrow x \notin A \cup B \\ \Rightarrow x \in (A \cup B)' \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ A' \cap B' = (A \cup B)' \\ (2) \text{ } x \in (A \cap B)' \Rightarrow x \notin A \cap B \\ \text{proof : } \Rightarrow x \notin A \lor x \notin B \\ \Rightarrow x \in A' \lor x \in B' \\ \Rightarrow x \in A' \cup B' \\ (A \cup B)' \subset A' \cup B' \quad \cdots (3) \\ \text{Conversely : } x \in A' \cup B' \Rightarrow x \in A' \lor x \in B' \\ \Rightarrow x \notin A \lor x \notin B \\ \Rightarrow x \notin A \cap B \\ \Rightarrow x \in (A \cap B)' \\ A \cup B' \subset (A \cap B)' \quad \cdots (4) \\ \text{From } (3) \text{ and } (4), \text{ we get} \\ A' \cup B' = (A \cap B)'(5.) Distributive Law
A \cup (B \cap C) = (A \cup B) \cap (A \cup C) \\ x \in A \cup (B \cap C) \\ \Rightarrow x \in A \lor x \in B \cap C \\ \Rightarrow (x \in A \lor x \in A) \lor (x \in B \land x \in C)\\ \Rightarrow (x \in A \lor x \in B) \land (x \in A \lor x \in C) \\ \Rightarrow x \in A \cup B \land x \in A \cup C \\ \Rightarrow x \in (A \cup B) \cap (A \cup C) \\ A \cup (B \cap C) \subset (A \cup B) \cap (A \cup C) \quad \cdots (1) \\ y \in (A \cup B) \cap (A \cup C) \\ \Rightarrow y \in (A \cup B) \land y \in (A \cup C) \\ \Rightarrow (y \in A \lor y \in B) \land (y \in A \lor y \in C) \\ \Rightarrow (y \in A \land y \in A) \lor (y \in B \lor y \in C) \\ \Rightarrow y \in A \lor y \in B \cap C \\ \Rightarrow y \in A \cup (B \cap C) \\ \therefore (A \cup B) \cap (A \cup C) \subset A \cup (B \cap C) \quad \cdots (2) \\ \text{From } (1) \text{ and } (2), \text{ we get} \\ (A \cup B) \cap (A \cup C) = A \cup (B \cap C)
3.Practice Question for Students
To prove that
(1)A - (B \cup C) = (A - B) \cap (A - C)
(2)A - (B \cap C) = (A - B) \cup (A - C)
By solving the above questions,one can understand the Properties of Sets Class 11 with Proof well because the concept is well understood when you solve the questions practically.
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Also Read This Article:- Set theory
4.Frequently Asked Questions Related to Properties of Sets Class 11 with Proof
Q:1.What is meant by union of two sets?
Ans:The union of two sets A and B is the set C which consists of all those elements which are either in A or in B (including those which are in both).In symbols,we write
A \cup B =\{x : x \in A \text{ or } x \in B \}
Q:2.What is meant by intersection of two sets?
Ans:The intersection of two sets A and B is the set of all those elements which belong to both A and B.Symbolically,we write
A \cap B==\{x : x \in A \text{ or } x \in B \}
Q:3.Define Difference between two sets
Ans:The difference of the two sets A and B in this order in the set of elements which belong to A but not B.Symbolically,we write A-B and read as “A minus B”.
By answering the above questions,you can know about the primary terms of Properties of Sets Class 11 with Proof.
5.छात्र-छात्राओं से आज का सवाल (Today’s Question to Students):
🎯 विनर्स कॉर्नर:क्या आप इस सवाल का सही जवाब दे सकते हैं?अपने नाम के साथ नीचे कमेंट करें!सही जवाब देने वाले Top Students के नाम हमारी आगे post/is article ke update में Photo ya Special Mention के साथ publish की जाएगी।अपना जवाब अभी दर्ज करें!👇
*एक धावक ने 5 पारी वाली दौड़ में से केवल 1 \frac{1}{4} पारियाँ दौड़ी।अब बताइए कि उसे दौड़ का कितना भाग दौड़ना शेष रह गया।*
(A runner only ran 1 \frac{1}{4} innings out of 5 innings of running.Now tell me how much of the race he has left to run.)
*पिछली प्रश्नोत्तरी का उत्तर*
\frac{5}{7}=0.71 ,\frac{3}{5}=0.60, \frac{9}{12}=0.75
आरोही क्रम=\frac{3}{5},\frac{5}{7}, \frac{9}{12}
*Previous Quiz Solution*
\frac{5}{7}=0.71 ,\frac{3}{5}=0.60, \frac{9}{12}=0.75
Ascending Order=\frac{3}{5},\frac{5}{7}, \frac{9}{12}
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