4.11. Proving and Disproving Set Statements. 4.11.1. Proof by Download

Transcript
4.11.
Proving and Disproving Set Statements.
4.11.1.
Proof by Exhaustion
To prove set results for finite sets, the method of
exhaustion is used. That is every element in the set is tested
to ensure it satisfies the condition.
Example:
•
Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A ⊆ A ∪ B .
To prove the statement, we must show every element in A
is in A ∪ B .
Now A ∪ B = {1, 2, 3, 4}
1 ∈ A, 1 ∈ A ∪ B
2 ∈ A, 2 ∈ A ∪ B
Thus all elements in A are in A ∪ B , and so by exhaustion
A ⊆ A ∪ B.
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Exercise:
•
Let A = {1, 2} , B = {1, 2, 3, 4}. Prove A = A ∩ B .
To prove the statement, we must show every element in A
is in A ∩ B and every element in A ∩ B is in A.
Thus all elements in A are in A ∩ B and vice versa, and so
by exhaustion A = A ∩ B .
Exercise:
•
Give an example of three sets A, B and C such that
C ⊆ A∩ B.
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4.11.2. Disproof by Counterexample.
A set result can be disproven by giving a counterexample.
To find a counterexample often creating a Venn diagram
will be of benefit.
Example:
•
Disprove A ⊆ A ∩ B .
To disprove the statement, we must give a counterexample.
Let A = {1, 2} , B = {3, 4}
Now A ∩ B = 
1 ∈ A, however 1 ∉ A ∩ B = 
Thus by counterexample A⊄A ∩ B .
Exercise:
•
Disprove A ⊆ A − B .
To disprove the statement, we must show a
counterexample.
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4.11.3. Proof by Typical Element.
To prove set results for infinite sets, generalised methods
must be used. The typical element method considers a
particular but arbitrary element of the set and by applying
knows laws, rules and definitions prove the result.
It is the method for proving subset relationships.
So prove that A ⊆ B , we must show that
∀x, ( x ∈ A ⇒ x ∈ B )
Begin by letting x ∈ A , that is, we take x to be a particular
but arbitrary element of A. Using the definitions, we prove
that x ∈ B . As long as we use no special properties of the
element x, we can conclude that A ⊆ B , which is what we
wanted to prove.
This method can be used to prove set equalities. By using
the definition A = B ⇔ ( A ⊆ B ∧ B ⊆ A) and showing
A ⊆ B ∧ B ⊆ A , that is proving ∀x, ( x ∈ A ⇒ x ∈ B ) and
∀x, ( x ∈ B ⇒ x ∈ A) , the result A = B follows. Using this
definition is sometimes called a “double containment”
proof.
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Examples:
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ⊆ A ∪ B .
Need to prove ∀x, ( x ∈ A ⇒ x ∈ A ∪ B )
Let x ∈ A , then
x∈ A ⇒ x∈ A∨ x∈B
⇒ x∈ A∪ B
∴ A ⊆ A∪ B
see note
definition of ∪
Note: Appling rules of logic, we know P ⇒ P ∨ Q is a
tautology. Let P( x ) : x ∈ A, Q( x ) : x ∈ B . Thus
x ∈ A ⇒ x ∈ A ∨ x ∈ B is a tautology in the proof above.
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ⊆ B ⇔ A ∪ B = B .
Need to prove two parts:
1. A ⊆ B ⇒ A ∪ B = B
2. A ∪ B = B ⇒ A ⊆ B
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•
Proof of 1:
KNOW: A ⊆ B , that is ∀x, ( x ∈ A ⇒ x ∈ B )
K (1)
PROVE: A ∪ B = B .
Need to prove two parts:
i. A ∪ B ⊆ B
ii. B ⊆ A ∪ B
Proof of i.:
Let x ∈ A ∪ B then
x∈ A∪ B ⇒ x∈ A∨ x∈B
⇒ x∈B ∨ x∈B
⇒ x∈B
∴A∪ B ⊆ B
Proof of ii.:
definition of ∪
by (1)
Logic rule P ∨ P ≡ P
Let x ∈ B then
x∈B ⇒ x∈ A∨ x∈B
⇒ x∈ A∪ B
∴B ⊆ A∪ B
see previous example
definition of ∪
Since A ∪ B ⊆ B and B ⊆ A ∪ B , A ∪ B = B
Thus A ⊆ B ⇒ A ∪ B = B .
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Proof of 2:
KNOW: A ∪ B = B , that is
∀x , ( x ∈ A ∪ B ⇔ x ∈ B )
K( 2)
PROVE: A ⊆ B .
Let x ∈ A then
x∈ A ⇒
⇒
⇒
∴A⊆ B
x∈ A∨ x∈B
x∈ A∪ B
x∈B
see previous example
definition of ∪
by (2)
Thus A ∪ B = B ⇒ A ⊆ B
Since A ⊆ B ⇒ A ∪ B = B and A ∪ B = B ⇒ A ⊆ B it is
proven that A ⊆ B ⇔ A ∪ B = B .
Exercise:
•
Let U be a set and let A and B be elements of  (U ) .
Prove A ∩ B ⊆ A .
That is, prove ∀x, ( x ∈ A ∩ B ⇒ x ∈ A)
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4.11.4. Proof by Equivalence of Statements.
If A can be written as A = { x ∈ U : P( x )} and
B = { x ∈ U : Q ( x )}, the equality of specification theorem
can be used to show that A = B by showing that
P( x ) ≡ Q( x ) , that is, by showing that P( x ) ⇔ Q ( x ) is a
tautology.
Examples:
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Let A = { x ∈  : x 2 ≤ 1} and B = { x ∈  : −1 ≤ x ≤ 1} .
Prove A = B
Let P( x ) : x 2 ≤ 1 and Q( x ) : −1 ≤ x ≤ 1 . Now
x 2 ≤ 1 ⇔ −1 ≤ x ≤ 1
∴ P( x ) ⇔ Q( x )
∴A= B
•
Let U be a set and let A and B be elements of  (U ) .
Prove that ( A ∩ B ) = A ∪ B .
We need to show that the statements defining the sets
( A ∩ B ) and A ∪ B are equivalent.
( A ∩ B ) = { x ∈ U :~ ( x ∈ A ∩ B )} definition of A
A ∪ B = { x ∈ U : x ∈ A ∪ B}
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Let P( x ) :~ ( x ∈ A ∩ B ) , and Q( x ) : x ∈ A ∪ B
x∈ A∪ B ≡ x∈ A ∨ x∈B
by definition of ∪
≡ ~ ( x ∈ A) ∨ ~ ( x ∈ B ) definition of A
≡ ~ (x ∈ A ∧ x ∈ B )
≡ ~ (x ∈ A ∩ B )
by Logic De Morgan' s
by definition of ∩
∴ Q( x ) ≡ P( x )
∴ (A ∩ B) = A ∪ B
Exercise:
•
Let U be a set and let A and B be elements of  (U ) .
Prove that ( A ∪ B ) = A ∩ B .
We need to show that the statements defining the sets
( A ∪ B ) and A ∩ B are equivalent.
( A ∪ B ) = { x ∈ U :~ ( x ∈ A ∪ B )} axiom of specification
A ∩ B = { x ∈ U : x ∈ A ∩ B} axiom of specification
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•
Let U be a set and let A, B and C be elements of P(U).
Prove that ( A − B ) − C = ( A − C ) − B .
•
Let U be a set and let X and Y be elements of (U ) .
Prove that X − Y = X ∩ Y .
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4.11.5. Proof by Set Laws.
Set equalities can be proven by using known set laws
Examples:
•
Let U be a set and let A, B and C be elements of P(U).
Prove ( A − B ) − C = ( A − C ) − B
(A − B) − C = (A − B) ∩ C
set difference
= (A ∩ B ) ∩ C
set difference
= A ∩ (B ∩ C )
associativity
= A ∩ (C ∩ B )
commutativity
= (A ∩ C ) ∩ B
associativity
= (A − C ) ∩ B
= (A − C ) − B
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set difference
set difference
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4.11.6. Further Examples.
Examples:
•
Let U be a set and let A and B be elements of P(U).
Using the following:
(i) A ⊆ B ⇔ A ∪ B = B ,
(ii): A ⊆ B ⇔ A ∩ B = A ,
(iii): A ∪ B = ( A ∩ B ) .
Prove that A ⊆ B ⇔ B ⊆ A .
Proof:
A ⊆ B ⇔ A∩ B = A
•
by part (ii)
⇔ (A ∩ B) = A
by taking complements
⇔ A∪ B = A
by part (iii)
⇔B⊆A
by part (i)
Let U be a set and let A, B and C be elements of P(U).
Disprove that A − (B − C ) = ( A − B ) − C .
Let A = {1, 2, 3}, B = {2, 3}, C = {3}.
A − (B − C ) = A − {2} = {1, 3}
( A − B ) − C = {1} − C = {1} ≠ A − (B − C )
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•
Let U be a set and let X and Y be elements of  (U ) .
Use a typical element argument to prove X − Y = X ∩ Y .
Need to prove two parts:
1. X − Y ⊆ X ∩ Y
2. X ∩ Y ⊆ X − Y
Proof of 1:Let x ∈ X − Y be a typical element.
X − Y ≡ {x ∈ U : x ∈ X − Y }
⇒ {x ∈ U : x ∈ X ∧ x ∉ Y }
Axiom of Specification
Def of set difference
⇒ {x ∈ U : x ∈ X ∧ x ∈ Y }
Def of complement
⇒ {x ∈ U : x ∈ X ∩ Y }
Def of intersection
∴ ∀x ∈ U (x ∈ X − Y ⇒ x ∈ X ∩ Y )
∴
X −Y ⊆ X ∩Y
Proof 2: Let x ∈ X ∩ Y be a typical element.
X ∩ Y ≡ {x ∈ U : x ∈ X ∩ Y }
⇒ {x ∈ U : x ∈ X ∧ x ∈ Y }
Axiom of Specification
⇒ {x ∈ U : x ∈ X ∧ x ∉ Y }
Def of intersection
Def of complement
⇒ {x ∈ U : x ∈ X − Y }
Def of set difference
∴ ∀x ∈ U (x ∈ X ∩ Y ⇒ x ∈ X − Y )
∴
X ∩Y ⊆ X −Y
∴ ∀x ∈ U (x ∈ X − Y ⇔ x ∈ X ∩ Y ),
i.e. X − Y ⊆ X ∩ Y ∧ X ∩ Y ⊆ X − Y ,
∴ X −Y = X ∩Y
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Logic
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