mhf 3202: the cantor-schr oder-bernstein theorem
TRANSCRIPT
Proving equinumerosity
Up to this point, the main method we have for proving that twosets A and B are equinumerous is to show that there is a function
f : A→ B
that is one-to-one and onto.
In some cases, finding such a bijection can be rather difficult.
Today we will prove a theorem that will provide a new and simplermethod for showing that two sets are equinumerous.
A preliminary definition
DefinitionLet A and B be sets. We will say that A is dominated by B,denoted A - B, if there is a one-to-one function f : A→ B.
A few examples:
I If A ∼ B, then A - B.
I If A ⊆ B, then A - B.
I P(Z+) - R.
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact?
Yes!
Is - a partial order?
It is not too hard to show that - is reflexive and transitive.
Is - anti-symmetric?
That is, if A - B and B - A, does it follow that A = B?
No.
Consider A = Z+ and B = Q.
I Z+ - Q.
I Q - Z+.
I But, Z+ 6= Q.
Note however that Z+ ∼ Q.
Is this an instance of a more general fact? Yes!
Our approach
To help us understand the general strategy of the proof, we willmake use of a series of diagrams.
First, we will represent the sets A and B as follows.
Next, let
I f : A→ B be a one-to-one function witnessing A - B and
I g : B → A be a one-to-one function witnessing B - A.
Note that if either f or g is onto, it immediately follows thatA ∼ B.
So need to consider the possibility that neither f nor g are onto.
The plan of attack
To help us understand the general strategy of the proof, we willmake use of a series of diagrams.
First, we will represent the sets A and B as follows.
Next, let
I f : A→ B be a one-to-one function witnessing A - B and
I g : B → A be a one-to-one function witnessing B - A.
Note that if either f or g is onto, it immediately follows thatA ∼ B.
So need to consider the possibility that neither f nor g are onto.
The plan of attack
To help us understand the general strategy of the proof, we willmake use of a series of diagrams.
First, we will represent the sets A and B as follows.
Next, let
I f : A→ B be a one-to-one function witnessing A - B and
I g : B → A be a one-to-one function witnessing B - A.
Note that if either f or g is onto, it immediately follows thatA ∼ B.
So we need to consider the possibility that neither f nor g is onto.
The plan of attack (continued)
Our goal is to use f and g−1 to define a one-to-one and ontofunction h : A→ B.
To do so, we will
1. split A into two pieces X and Y ;
2. split B into two pieces W and Z ;
3. X will be matched up with W by f ; and
4. Y will be matched up with Z by g .
More precisely
We will let
I W = f (X ) = {f (x) | x ∈ X} and
I Y = g(Z ) = {g(z) | z ∈ Z}.
It follows that
I f : X →W is one-to-one and onto and
I g : Z → Y is one-to-one and onto.
The desired function h
Since g : Z → Y is one-to-one and onto, g−1 : Y → Z is afunction that is also one-to-one and onto.
Then we can define a one-to-one and onto function h : A→ B tobe
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z
As a first step, we want our sets X ,Y ,W and Z to satisfy thefollowing:
I We want Y ⊆ Ran(g).
I If we let A1 = A \ Ran(g), then we must have A1 ⊆ X .
I Given an arbitrary a ∈ A1, since a ∈ X , it follows thatf (a) ∈W .
I Since g is one-to-one, g(f (a)) 6= g(z) for every z ∈ Z .
I Thus g(f (a)) ∈ X .
I Since a was arbitrary, we have f (a) ∈W and g(f (a)) ∈ X forall a ∈ A1.
I That is, f (A1) ⊆W and g(f (A1)) ⊆ X .
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
A B
g(B)
W = f(X)
Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))
Xg
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))f(A3)
A4 = g(f(A3)) g
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))A4 = g(f(A3))
f(A3)f
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))A4 = g(f(A3))
f(A3)
g
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2)) ff(A3)
A4 = g(f(A3))
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))
g
f(A3)
A4 = g(f(A3))
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2)) ff(A3)
A4 = g(f(A3))
A B
g(B)Y = g(Z) Z
g(B) = Ran(g)
A1 = A \ Ran(g) f(A1)
A2 = g(f(A1))f(A2)
A3 = g(f(A2))
g
f(A3)
A4 = g(f(A3))
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
Choosing the sets X ,Y ,W and Z (continued)
Now let A2 = g(f (A1)) ⊆ X .
If we let A3 = g(f (A2)), by a similar argument, we can concludethat A3 ⊆ X .
We can continue in this way to produce a sequence of setsA1,A2,A3, . . . such that
An ⊆ X
for every n ∈ Z+.
Finally, let X = ∪n∈Z+An.
The formal proof, 1
Suppose A - B and B - A, and let f : A→ B and g : B → A beone-to-one functions that witness this.
Moreover, let R = Ran(g) ⊆ A.
Then g : B → R is one-to-one and onto.
By our theorem on the inverses of functions, g−1 : R → B is afunction.
The formal proof, 1
Suppose A - B and B - A, and let f : A→ B and g : B → A beone-to-one functions that witness this.
Moreover, let R = Ran(g) ⊆ A.
Then g : B → R is one-to-one and onto.
By our theorem on the inverses of functions, g−1 : R → B is afunction.
The formal proof, 1
Suppose A - B and B - A, and let f : A→ B and g : B → A beone-to-one functions that witness this.
Moreover, let R = Ran(g) ⊆ A.
Then g : B → R is one-to-one and onto.
By our theorem on the inverses of functions, g−1 : R → B is afunction.
The formal proof, 1
Suppose A - B and B - A, and let f : A→ B and g : B → A beone-to-one functions that witness this.
Moreover, let R = Ran(g) ⊆ A.
Then g : B → R is one-to-one and onto.
By our theorem on the inverses of functions, g−1 : R → B is afunction.
The formal proof, 1
Suppose A - B and B - A, and let f : A→ B and g : B → A beone-to-one functions that witness this.
Moreover, let R = Ran(g) ⊆ A.
Then g : B → R is one-to-one and onto.
By our theorem on the inverses of functions, g−1 : R → B is afunction.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 2
We define a sequence of sets A1,A2,A3, . . . by recursion:
I A1 = A \ R;
I An+1 = g(f (An)) = {g(f (a)) | a ∈ An} for every n ∈ Z+.
Let X = ∪n∈Z+An (that is, X = ∪{An | n ∈ Z+}), and letY = A \ X .
Clearly A = X ∪ Y and X ∩ Y = ∅.
Now we define h : A→ B to be
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
The formal proof, 3
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
We need to verify that h is a function from A to B.
In particular, we need to ensure that if a ∈ Y , then
a ∈ Dom(g−1) = Ran(g) = R.
Note that if a /∈ R, then a ∈ A \ R = A1 ⊆ X .
It follows that if a ∈ Y , we must have a ∈ R.
The formal proof, 3
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
We need to verify that h is a function from A to B.
In particular, we need to ensure that if a ∈ Y , then
a ∈ Dom(g−1) = Ran(g) = R.
Note that if a /∈ R, then a ∈ A \ R = A1 ⊆ X .
It follows that if a ∈ Y , we must have a ∈ R.
The formal proof, 3
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
We need to verify that h is a function from A to B.
In particular, we need to ensure that if a ∈ Y , then
a ∈ Dom(g−1) = Ran(g) = R.
Note that if a /∈ R, then a ∈ A \ R = A1 ⊆ X .
It follows that if a ∈ Y , we must have a ∈ R.
The formal proof, 3
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
We need to verify that h is a function from A to B.
In particular, we need to ensure that if a ∈ Y , then
a ∈ Dom(g−1) = Ran(g) = R.
Note that if a /∈ R, then a ∈ A \ R = A1 ⊆ X .
It follows that if a ∈ Y , we must have a ∈ R.
The formal proof, 3
h(a) =
{f (a) if a ∈ Xg−1(a) if a ∈ Y
.
We need to verify that h is a function from A to B.
In particular, we need to ensure that if a ∈ Y , then
a ∈ Dom(g−1) = Ran(g) = R.
Note that if a /∈ R, then a ∈ A \ R = A1 ⊆ X .
It follows that if a ∈ Y , we must have a ∈ R.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
Claim: h is one-to-one
Let a1, a2 ∈ A be arbitrary and suppose h(a1) = h(a2).
We have two cases to consider:
I Case 1: a1 ∈ X .
I Case 2: a1 ∈ Y .
In Case 1, we will argue that if a1 ∈ X and a2 ∈ Y we have acontradiction.
Thus we must have a1 ∈ X and a2 ∈ X , and then we will arguethat a1 = a2.
Case 2 will proceed similarly.
h is one-to-one, Case 1
Case 1: Suppose that a1 ∈ X and suppose for the sake ofcontradiction that a2 ∈ Y .
Then by definition of h, we have
I h(a1) = f (a1)
I h(a2) = g−1(a2).
Since h(a1) = h(a2), it follows that
f (a1) = g−1(a2).
Applying g to both sides yields
g(f (a1)) = g(g−1(a2)) = a2.
h is one-to-one, Case 1
Case 1: Suppose that a1 ∈ X and suppose for the sake ofcontradiction that a2 ∈ Y .
Then by definition of h, we have
I h(a1) = f (a1)
I h(a2) = g−1(a2).
Since h(a1) = h(a2), it follows that
f (a1) = g−1(a2).
Applying g to both sides yields
g(f (a1)) = g(g−1(a2)) = a2.
h is one-to-one, Case 1
Case 1: Suppose that a1 ∈ X and suppose for the sake ofcontradiction that a2 ∈ Y .
Then by definition of h, we have
I h(a1) = f (a1)
I h(a2) = g−1(a2).
Since h(a1) = h(a2), it follows that
f (a1) = g−1(a2).
Applying g to both sides yields
g(f (a1)) = g(g−1(a2)) = a2.
h is one-to-one, Case 1
Case 1: Suppose that a1 ∈ X and suppose for the sake ofcontradiction that a2 ∈ Y .
Then by definition of h, we have
I h(a1) = f (a1)
I h(a2) = g−1(a2).
Since h(a1) = h(a2), it follows that
f (a1) = g−1(a2).
Applying g to both sides yields
g(f (a1)) = g(g−1(a2)) = a2.
h is one-to-one, Case 1
Case 1: Suppose that a1 ∈ X and suppose for the sake ofcontradiction that a2 ∈ Y .
Then by definition of h, we have
I h(a1) = f (a1)
I h(a2) = g−1(a2).
Since h(a1) = h(a2), it follows that
f (a1) = g−1(a2).
Applying g to both sides yields
g(f (a1)) = g(g−1(a2)) = a2.
h is one-to-one, Case 1 (continued)
From the previous slide we now have g(f (a1)) = a2.
Since a1 ∈ X = ∪n∈Z+An, we can choose some n ∈ Z+ such thata1 ∈ An. Then we have
a2 = g(f (a1)) ∈ g(f (An)) = An+1 ⊆ X ,
and hence a2 ∈ X , contradicting our assumption that a2 ∈ Y .
Thus, both a1, a2 ∈ X , so
f (a1) = h(a1) = h(a2) = f (a2).
But f is one-to-one, and thus a1 = a2.
h is one-to-one, Case 1 (continued)
From the previous slide we now have g(f (a1)) = a2.
Since a1 ∈ X = ∪n∈Z+An, we can choose some n ∈ Z+ such thata1 ∈ An. Then we have
a2 = g(f (a1)) ∈ g(f (An)) = An+1 ⊆ X ,
and hence a2 ∈ X , contradicting our assumption that a2 ∈ Y .
Thus, both a1, a2 ∈ X , so
f (a1) = h(a1) = h(a2) = f (a2).
But f is one-to-one, and thus a1 = a2.
h is one-to-one, Case 1 (continued)
From the previous slide we now have g(f (a1)) = a2.
Since a1 ∈ X = ∪n∈Z+An, we can choose some n ∈ Z+ such thata1 ∈ An. Then we have
a2 = g(f (a1)) ∈ g(f (An)) = An+1 ⊆ X ,
and hence a2 ∈ X , contradicting our assumption that a2 ∈ Y .
Thus, both a1, a2 ∈ X , so
f (a1) = h(a1) = h(a2) = f (a2).
But f is one-to-one, and thus a1 = a2.
h is one-to-one, Case 1 (continued)
From the previous slide we now have g(f (a1)) = a2.
Since a1 ∈ X = ∪n∈Z+An, we can choose some n ∈ Z+ such thata1 ∈ An. Then we have
a2 = g(f (a1)) ∈ g(f (An)) = An+1 ⊆ X ,
and hence a2 ∈ X , contradicting our assumption that a2 ∈ Y .
Thus, both a1, a2 ∈ X , so
f (a1) = h(a1) = h(a2) = f (a2).
But f is one-to-one, and thus a1 = a2.
h is one-to-one, Case 1 (continued)
From the previous slide we now have g(f (a1)) = a2.
Since a1 ∈ X = ∪n∈Z+An, we can choose some n ∈ Z+ such thata1 ∈ An. Then we have
a2 = g(f (a1)) ∈ g(f (An)) = An+1 ⊆ X ,
and hence a2 ∈ X , contradicting our assumption that a2 ∈ Y .
Thus, both a1, a2 ∈ X , so
f (a1) = h(a1) = h(a2) = f (a2).
But f is one-to-one, and thus a1 = a2.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
h is one-to-one, Case 2
Case 2: a1 ∈ Y .
We can argue as in Case 1 that a2 ∈ Y .
Thus h(a1) = h(a2) implies that
g−1(a1) = g−1(a2).
Applying g to both sides yields
a1 = g(g−1(a1)) = g(g−1(a2)) = a2.
In either case, we have a1 = a2.
Since a1 and a2 were arbitrary, it follows that h is one-to-one.
Claim: h is onto
Let b ∈ B be arbitrary.
Since g(b) ∈ A, we have two cases to consider:
I Case 1: g(b) ∈ X ;
I Case 2: g(b) ∈ Y .
Claim: h is onto
Let b ∈ B be arbitrary.
Since g(b) ∈ A, we have two cases to consider:
I Case 1: g(b) ∈ X ;
I Case 2: g(b) ∈ Y .
Claim: h is onto
Let b ∈ B be arbitrary.
Since g(b) ∈ A, we have two cases to consider:
I Case 1: g(b) ∈ X ;
I Case 2: g(b) ∈ Y .
Claim: h is onto
Let b ∈ B be arbitrary.
Since g(b) ∈ A, we have two cases to consider:
I Case 1: g(b) ∈ X ;
I Case 2: g(b) ∈ Y .
Claim: h is onto
Let b ∈ B be arbitrary.
Since g(b) ∈ A, we have two cases to consider:
I Case 1: g(b) ∈ X ;
I Case 2: g(b) ∈ Y .
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 1: g(b) ∈ X
Choose n ∈ Z+ such that g(b) ∈ An.
Since g(b) ∈ Ran(g) = R and A1 = A \ R, it follows thatg(b) /∈ A1.
Thus, g(b) ∈ An for some n > 1, so An = g(f (An−1)).
Then we can choose some a ∈ An−1 such that g(f (a)) = g(b).
Since g is one-to-one, it follows that f (a) = b.
Since a ∈ An−1, we know a ∈ X and hence h(a) = f (a) = b.
Thus b ∈ Ran(h).
h is onto, Case 2: g(b) ∈ Y
In this case, we apply h to g(b) to get
h(g(b)) = g−1(g(b)) = b.
Thus b ∈ Ran(h).
In either case, b ∈ Ran(h).
Since b was arbitrary, h is onto.
h is onto, Case 2: g(b) ∈ Y
In this case, we apply h to g(b) to get
h(g(b)) = g−1(g(b)) = b.
Thus b ∈ Ran(h).
In either case, b ∈ Ran(h).
Since b was arbitrary, h is onto.
h is onto, Case 2: g(b) ∈ Y
In this case, we apply h to g(b) to get
h(g(b)) = g−1(g(b)) = b.
Thus b ∈ Ran(h).
In either case, b ∈ Ran(h).
Since b was arbitrary, h is onto.
h is onto, Case 2: g(b) ∈ Y
In this case, we apply h to g(b) to get
h(g(b)) = g−1(g(b)) = b.
Thus b ∈ Ran(h).
In either case, b ∈ Ran(h).
Since b was arbitrary, h is onto.
h is onto, Case 2: g(b) ∈ Y
In this case, we apply h to g(b) to get
h(g(b)) = g−1(g(b)) = b.
Thus b ∈ Ran(h).
In either case, b ∈ Ran(h).
Since b was arbitrary, h is onto.