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De Morgan's Law

Question):- 1...

Question

Question):- 1 (i)- Let X be a non empty set and P(X) be the power set of X. Let the binary operation on P(X) be defined by A * B = (A - B) $\cup$ (B - A) for all A, B $\in$ P(X). Show that the empty set $ϕ$ is the identity element for the operation * and all the elements A of P(X) are invertible with $A^{- 1} = A$ .

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Solution

$* P (X) \times P (X) \to P (X) A * B = (A - B) \cup (B - A) A * B = (A - B) \cup (B - A) = (B - A) \cup (A - B) = B * A * i s c o m m u t a t i v e L e t E b e t h e i d e n t i t y e l e m e n t t h e n A * E = E * A = (A - E) \cup (E - A) = A [F r o m s e t t h e o r y w e k n o w (A - B) \cup (B - A) = (A \cup B) - (A \cap B)] (A \cup E) - (A \cap E) = A(1) A l l t e r m s o f A a r e c o n t a i n e d i n A \cup E, a n d w e a r e r e m o v i n g t e r m s f r o m A \cup E t h a t a r e c o m m o n t o A a n d E, i . e . w e a r e r e m o v i n g s o m e t e m s o f A f r o m A \cup E a n d s t i l l i t g i v e s u s A, t h i s m e a n s t h e s e t w e a r e s u b t r a c t i n g m u s t b e e m p t y s e t A \cap E = ϕ_____________(2) P u t i n e q u a t i o n 1 (A \cup E) - ϕ = A (A \cup E) = A t a k i n g i n t e r s e c t i o n w i t h E b o t h s i d e s (A \cup E) \cap E = A \cap E (A \cap E) \cup (E \cap E) = A \cap E ϕ \cup E = ϕ E = ϕ L e t C t h e i n v e r s e o f A A * C = C * A = (A - C) \cup (C - A) = (A \cup C) - (A \cap C) = E = ϕ (A \cup C) - (A \cap C) = ϕ D I F F E R E N C E O F T W O S E T S I S A N E M P T Y S E T, T H E R E F O R E, T H E T W O S E T S M U S T B E E Q U A L A \cup C = A \cap C L e t a b e a n y e l e m e n t o f s e t A, i . e . a \in A a \in A \cup C a \in A \cap C a \in A a n d a \in C a \in C T h u s e v e r y e l e m e n t o f A b e l o n g s t o C A \subset C L e t c b e a n y e l e m e n t o f s e t C, i . e . c \in C c \in A \cup C c \in A \cap C c \in A a n d c \in C c \in A T h u s e v e r y e l e m e n t o f C b e l o n g s t o A C \subset A C \subset A a n d A \subset C, t h e r e f o r e A = C C = A^{- 1} = A$

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X

be a non- empty set and

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Given a non-empty set X, let *: P(X) × P(X) → P(X) be defined as A * B = (A − B) ∪ (B − A), &mnForE; A, B ∈ P(X). Show that the empty set Φ is the identity for the operation * and all the elements A of P(X) are invertible with A⁻¹ = A. (Hint: (A − Φ) ∪ (Φ − A) = A and (A − A) ∪ (A − A) = A * A = Φ).

Given non-empty set X, consider the binary operation $* : P (X) \times P (X) \to P (X)$ given by $A * B = A \cap B \forall A, B$ in P(X), where P(X)is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation.

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