Prove each statement in 39-44. For every positive integer n , if A and B 1 , B 2 , B 3 , .... are any set, then ∩ i = 1 n ( A × B i ) = A × ( ∩ i = 1 n B i ) .

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To determine To Prove: For all integers n ≥ 1 if A and B 1 , B 2 , B 3 , ...... are any sets, then ∩ n = 1 n ( A × B i ) = A × ( ∩ i = 1 n B i ) Explanation Given information: Let A and B 1 , B 2 , B 3 , .......... , B n ( n ≥ 1 ) are any sets. Concept used: ∪ : U n i o n of sets ∩ : I n t e r sec t i o n of sets ⊆ : subset of set ∅ : Null set Calculation: This prove ∩ n = 1 n ( A × B i ) = A × ( ∩ i = 1 n B i ) is done in following ways. Suppose A and B 1 , B 2 , B 3 , .......... , B n are any sets. Proof that ∩ n = 1 n ( A × B i ) ⊆ A × ( ∩ i = 1 n B i ) . Suppose ( x , y ) is any element in ∩ n = 1 n ( A × B i ) . By definition of intersection ( x , y ) ∈ A × B i For Cartesian product x ∈ A ( 2 ) and y ∈ B 1 for some Implies that y ∈ ∩ i = 1 n B i . Thus x ∈ A and y ∈ ∩ i = 1 n B i and so by definition of Cartesian product

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Discrete Mathematics With Applicat...

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EPP + 1 other

Publisher: Cengage Learning,

ISBN: 9781337694193

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Answer 5

To determine

To Prove:

For all integers n≥1 if A and B1,B2,B3,...... are any sets, then

∩n=1n(A×Bi)=A×(∩i=1nBi)

Explanation of Solution

Given information:

Let A and B1,B2,B3,..........,Bn(n≥1) are any sets.

Concept used:

∪:Union of sets∩:Intersection of sets

⊆: subset of set

∅: Null set

Calculation:

This prove ∩n=1n(A×Bi)=A×(∩i=1nBi) is done in following ways.

Suppose A and B1,B2,B3,..........,Bn are any sets.

Proof that ∩n=1n(A×Bi)⊆A×(∩i=1nBi).

Suppose (x,y) is any element in ∩n=1n(A×Bi).

By definition of intersection

(x,y)∈A×Bi For Cartesian product

x∈A (2) and y∈B1

for some

y∈∩i=1nBi

Thus x∈A and y∈∩i=1nBi and so by definition of Cartesian product

Prove each statement in 39-44. For every positive integer n , if A and B 1 , B 2 , B 3 , .... are any set, then ∩ i = 1 n ( A × B i ) = A × ( ∩ i = 1 n B i ) .

Discrete Mathematics With Applicat...

Solutions

Discrete Mathematics With Applicat...