For all sets A and B, if A ⊆ B then Bc ⊆ Ac.Some of the sentences in the following scrambled list can be used to write a proof. Hence, x ∉ A, because of the definition of subset and that A ⊆ B. Suppose A and B are sets and A ⊆ B. Let x ∈ B. Hence, x ∉ A, because A ∩ B = ∅. Therefore, by definition of complement x ∈ Ac, and thus, by definition of subset, Bc ⊆ Ac. Suppose A and B are sets and A ⊆ B. Let x ∈ Bc. By definition of complement, x ∉ B. Proof: We construct a proof by selecting appropriate sentences from the list and putting them in the correct order. 1. 2. 3. 4.
For all sets A and B, if A ⊆ B then Bc ⊆ Ac.Some of the sentences in the following scrambled list can be used to write a proof. Hence, x ∉ A, because of the definition of subset and that A ⊆ B. Suppose A and B are sets and A ⊆ B. Let x ∈ B. Hence, x ∉ A, because A ∩ B = ∅. Therefore, by definition of complement x ∈ Ac, and thus, by definition of subset, Bc ⊆ Ac. Suppose A and B are sets and A ⊆ B. Let x ∈ Bc. By definition of complement, x ∉ B. Proof: We construct a proof by selecting appropriate sentences from the list and putting them in the correct order. 1. 2. 3. 4.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
For all sets A and B, if A ⊆ B then
Bc ⊆ Ac.
Some of the sentences in the following scrambled list can be used to write a proof.
Hence, x ∉ A, because of the definition of subset and that A ⊆ B.
Suppose A and B are sets and A ⊆ B. Let x ∈ B.
Hence, x ∉ A, because A ∩ B = ∅.
Therefore, by definition of complement x ∈ Ac, and thus, by definition of subset, Bc ⊆ Ac.
Suppose A and B are sets and A ⊆ B. Let x ∈ Bc.
By definition of complement, x ∉ B.
Proof:
We construct a proof by selecting appropriate sentences from the list and putting them in the correct order.
1.
2.
3.
4.
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