Book of Proof - Amazon S3

134 Proofs Involving Sets
Example 8.9 Suppose A and B are sets. If P(A) ⊆ P(B), then A ⊆ B.
Proof. We use direct proof. Assume P(A) ⊆ P(B).
Based on this assumption, we must now show that A ⊆ B.
To show A ⊆ B, suppose that a ∈ A.
Then the one-element set { a } is a subset of A, so { a } ∈ P(A).
But then, since P(A) ⊆ P(B), it follows that { a } ∈ P(B).
This means that { a } ⊆ B, hence a ∈ B.
We’ve shown that a ∈ A implies a ∈ B, so therefore A ⊆ B.
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8.3 How to Prove A = B
In proofs it is often necessary to show that two sets are equal. There is a
standard way of doing this. Suppose we want to show A = B. If we show
A ⊆ B, then every element of A is also in B, but there is still a possibility
that B could have some elements that are not in A, so we can’t conclude
A = B. But if in addition we also show B ⊆ A, then B can’t contain anything
that is not in A, so A = B. This is the standard procedure for proving A = B:
Prove both A ⊆ B and B ⊆ A.
How to Prove A = B
Proof.
[Prove that A ⊆ B.]
[Prove that B ⊆ A.]
Therefore, since A ⊆ B and B ⊆ A,
it follows that A = B.
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Example 8.10 Prove that { n ∈ Z : 35| n } = { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } .
Proof. First we show { n ∈ Z : 35| n } ⊆ { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } . Suppose
a ∈ { n ∈ Z : 35| n } . This means 35| a, so a = 35c for some c ∈ Z. Thus a = 5(7c)
and a = 7(5c). From a = 5(7c) it follows that 5| a, so a ∈ { n ∈ Z : 5| n } . From
a = 7(5c) it follows that 7| a, which means a ∈ { n ∈ Z : 7| n } . As a belongs
to both { n ∈ Z : 5| n } and { n ∈ Z : 7| n } , we get a ∈ { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } .
Thus we’ve shown that { n ∈ Z : 35| n } ⊆ { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } .
Next we show { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } ⊆ { n ∈ Z : 35| n } . Suppose that
a ∈ { n ∈ Z : 5| n } ∩ { n ∈ Z : 7| n } . By definition of intersection, this means that
a ∈ { n ∈ Z : 5| n } and a ∈ { n ∈ Z : 7| n } . Therefore it follows that 5| a and 7| a.
By definition of divisibility, there are integers c and d with a = 5c and a = 7d.
Then a has both 5 and 7 as prime factors, so the prime factorization of a