If A,B,C are finite sets whose elements are from the same universal set U and n(A) denotes the number of element in the set A (a) Show by means of venn diagram that n(A ∪ B) = n(A) + n(B) − n(A ∩ B) (b) Using the fact that (A ∪ B ∪ C) = (A∪B)∪C =A∪(B∪C) deduce an expression for (A∪B∪C) (c) If n(A∪B)= n(A∩B), what can be said about A and B? How did you reach your conclusion. Thank you in advance


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Question Number 203545 by Mastermind last updated on 21/Jan/24

$If A, B, C are finite sets whose elements are$ $from the same universal set U and n (A)$ $denotes the number of element in the set A$ $(a) Show by means of venn diagram that$ $n (A \cup B) = n (A) + n (B) - n (A \cap B)$ $(b) Using the fact that (A \cup B \cup C) = (A \cup B) \cup C$ $= A \cup (B \cup C) deduce an expression for$ $(A \cup B \cup C)$ $(c) If n (A \cup B) = n (A \cap B), what can be said$ $about A and B ? How did you reach your conclusion .$ $Thank you in advance$
	Answered by deleteduser1 last updated on 21/Jan/24
	$b) n∣(A∪B)∪C∣=n[P∪C] where P=A∪B n(P∪C)=n(P)+n(C)−n(P∩C) =n(A)+n(B)+n(C)−n(A∩B)−n(P∩C) P∩C=(A∩B)∩C=(A∩C)∪(B∩C) n(PnC)=n(A∩C)+n(B∩C)−n[(A∩C)∩(B∩C)] ⇒n(A∪B∪C)=n(A)+n(B)+(C)−n(A∩B) −n(A∩C)−n(B∩C)+n(A∩B∩C) c) Suppose one of the sets ,say A, is not a subset of the other,B. Then,∃ a_1 ∈A s.t. a_1 ∉B ⇒n(A∪B)>n(A∩B)⇒A⊆B⇒n(A∩B)=n(A) Suppose B\A ≠∅⇒n(B\A)=k≥1 then n(A∪B)=n(A)+n(B)−n(A∩B)=n(B) =n(B\A)+n(A)=k+n(A)>n(A∩B) ⇒n(B\A)=∅⇒A=B$
	$b) n ∣ (A \cup B) \cup C ∣= n [P \cup C] w h e r e P = A \cup B$ $n (P \cup C) = n (P) + n (C) - n (P \cap C)$ $= n (A) + n (B) + n (C) - n (A \cap B) - n (P \cap C)$ $P \cap C = (A \cap B) \cap C = (A \cap C) \cup (B \cap C)$ $n (P n C) = n (A \cap C) + n (B \cap C) - n [(A \cap C) \cap (B \cap C)]$ $\Rightarrow n (A \cup B \cup C) = n (A) + n (B) + (C) - n (A \cap B)$ $- n (A \cap C) - n (B \cap C) + n (A \cap B \cap C)$ $c) S u p p o s e o n e o f t h e s e t s, s a y A, i s n o t a s u b s e t$ $o f t h e o t h e r, B . T h e n, \exists a_{1} \in A s . t . a_{1} \notin B$ $\Rightarrow n (A \cup B) > n (A \cap B) \Rightarrow A \subseteq B \Rightarrow n (A \cap B) = n (A)$ $S u p p o s e B ∖ A \neq \emptyset \Rightarrow n (B ∖ A) = k ⩾ 1$ $t h e n n (A \cup B) = n (A) + n (B) - n (A \cap B) = n (B)$ $= n (B ∖ A) + n (A) = k + n (A) > n (A \cap B)$ $\Rightarrow n (B ∖ A) = \emptyset \Rightarrow A = B$
	Commented by Mastermind last updated on 21/Jan/24

	$Thank you but whats the meaning of s . t . ?$
	Commented by Mastermind last updated on 21/Jan/24

	$besides i could not identify solution a and solution b$
	Commented by deleteduser1 last updated on 21/Jan/24

	$s u c h t h a t$
	Commented by Mastermind last updated on 21/Jan/24

	$Thank you but help me separate the solution$ $separately .$ $Thank you$
	Commented by Mastermind last updated on 21/Jan/24

	$Where is the solution A ?$
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