Use an element argument to prove each statement in 7-22. Assume that all sets are subsets of a universal set U . For all sets A , B , and C, if A ⊆ C and B ⊆ C then A ∪ B ⊆ C .

Q: Expert Answer

To determine To prove: Find the mistake in the following “proof” that for all sets A , B and C , if A ⊆ B and B ⊆ C then A ⊆ C . Suppose A , B and C are sets such that A ⊆ B and B ⊆ C . Since A ⊆ B , there is an element x such that x ∈ A and x ∈ B . Since B ⊆ C , there is an element x such that x ∈ A and x ∈ C . Hence there is an element x such that x ∈ A and x ∈ C and so A ⊆ C . Explanation Given information: Let A , B and C be 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 Calculation: Let A , B and C be are any sets. Prove that. If A ⊆ B and B ⊆ C then A ⊆ C . According to the definition of subset, the statement A ⊆ B means that for every element in A is in the set B . But the proof given in me text book, dealt with a single element x in the set A . So the statement “Since A ⊆ B , there exists an element x such that x ∈ A and x ∈ B is false. Also the statement ‘since B ⊆ C , there exists an element x such that x ∈ B and x ∈ C is false, The other error in the proof ¡s occurs in the last step

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Answer 1

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Discrete Mathematics With Applicat...

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Publisher: Cengage Learning,

ISBN: 9781337694193

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Answer 3

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Answer 4

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Answer 5

To determine

To prove:

Find the mistake in the following “proof” that for all sets A,B and C, if A⊆B and B⊆C then A⊆C.

Suppose A,B and C are sets such that A⊆B and B⊆C. Since A⊆B, there is an element x such that x∈A and x∈B. Since B⊆C, there is an element x such that x∈A and x∈C. Hence there is an element x such that x∈A and x∈C and so A⊆C.

Explanation of Solution

Given information:

Let A,B and C be any sets.

Concept used:

∪:Union of sets∩:Intersection of sets

⊆: subset of set

Calculation:

Let A,B and C be are any sets.

Prove that. If A⊆B and B⊆C then A⊆C.

According to the definition of subset, the statement A⊆B means that for every element in A is in the set B.

But the proof given in me text book, dealt with a single element x in the set A.

So the statement “Since A⊆B, there exists an element x such that x∈A and x∈B is false.

Also the statement ‘since B⊆C, there exists an element x such that x∈B and x∈C is false,

The other error in the proof ¡s occurs in the last step

Use an element argument to prove each statement in 7-22. Assume that all sets are subsets of a universal set U . For all sets A , B , and C, if A ⊆ C and B ⊆ C then A ∪ B ⊆ C .

Discrete Mathematics With Applicat...

Solutions

Discrete Mathematics With Applicat...