cardinal numbers and the continuum hypothesis · finite sizes infinite sizes cardinal arithmetic...
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Cardinal Numbers and the ContinuumHypothesis
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Introduction
1. We want a standard “size” for each set, just like thenumber of elements (which is a natural number) is thestandard size for finite sets.
2. Ordinal numbers will not quite work because differentordinal numbers can have the same size.
3. Plus, once we are past that, we want to do arithmetic.4. To start, consider the arithmetic of finite set sizes.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Introduction1. We want a standard “size” for each set, just like the
number of elements (which is a natural number) is thestandard size for finite sets.
2. Ordinal numbers will not quite work because differentordinal numbers can have the same size.
3. Plus, once we are past that, we want to do arithmetic.4. To start, consider the arithmetic of finite set sizes.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Introduction1. We want a standard “size” for each set, just like the
number of elements (which is a natural number) is thestandard size for finite sets.
2. Ordinal numbers will not quite work because differentordinal numbers can have the same size.
3. Plus, once we are past that, we want to do arithmetic.4. To start, consider the arithmetic of finite set sizes.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Introduction1. We want a standard “size” for each set, just like the
number of elements (which is a natural number) is thestandard size for finite sets.
2. Ordinal numbers will not quite work because differentordinal numbers can have the same size.
3. Plus, once we are past that, we want to do arithmetic.
4. To start, consider the arithmetic of finite set sizes.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Introduction1. We want a standard “size” for each set, just like the
number of elements (which is a natural number) is thestandard size for finite sets.
2. Ordinal numbers will not quite work because differentordinal numbers can have the same size.
3. Plus, once we are past that, we want to do arithmetic.4. To start, consider the arithmetic of finite set sizes.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem.
Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets.
Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.
2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.
3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.
5. With AB denoting the set of all functions from B to A, wehave
∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A,B and C be finite sets. Then the followinghold.
1. If A∩B = /0, then |A∪B|= |A|+ |B|.2. |A∪B|= |A|+ |B|− |A∩B|.3.|A∪B∪C|= |A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.
4. |A×B|= |A| · |B|.5. With AB denoting the set of all functions from B to A, we
have∣∣AB∣∣ = |A||B|.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).
|A∪B|+ |A∩B| =∣∣A∪ (B\A)
∣∣+ |A∩B|= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B|
=∣∣A∪ (B\A)
∣∣+ |A∩B|= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|
= |A|+∣∣B\ (A∩B)
∣∣+ |A∩B|= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|
=∣∣A∪ (B∪C)
∣∣= |A|+ |B∪C|−
∣∣A∩ (B∪C)∣∣
= |A|+ |B|+ |C|− |B∩C|−∣∣(A∩B)∪ (A∩C)
∣∣= |A|+|B|+|C|−|B∩C|−
[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣
= |A|+ |B|+ |C|− |B∩C|−∣∣(A∩B)∪ (A∩C)
∣∣= |A|+|B|+|C|−|B∩C|−
[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (parts 2 and 3 only).|A∪B|+ |A∩B| =
∣∣A∪ (B\A)∣∣+ |A∩B|
= |A|+ |B\A|+ |A∩B|= |A|+
∣∣B\ (A∩B)∣∣+ |A∩B|
= |A|+ |B|
|A∪B∪C|=
∣∣A∪ (B∪C)∣∣
= |A|+ |B∪C|−∣∣A∩ (B∪C)
∣∣= |A|+ |B|+ |C|− |B∩C|−
∣∣(A∩B)∪ (A∩C)∣∣
= |A|+|B|+|C|−|B∩C|−[|A∩B|+|A∩C|−
∣∣(A∩B)∩(A∩C)∣∣]
= |A|+ |B|+ |C|− |B∩C|− |A∩B|− |A∩C|+ |A∩B∩C|
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example.
In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.
25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer.
32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse.
18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert.
12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage.
9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake.
15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake.
How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?
M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup.
S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.
C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|
= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|
= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15
= 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Example. In the restaurant “Zum Adler” 41 people are dining.25 people have ordered bone marrow dumpling soup as anappetizer. 32 people have ordered blood sausage as the maincourse. 18 people have ordered black forest cake for desert. 12people will have bone marrow dumpling soup and bloodsausage. 9 people will have bone marrow dumpling soup andblack forest cake. 15 people will have blood sausage and blackforest cake. How many people will have all three dishes?M :=set of all people who will have bone marrow dumplingsoup. S :=set of all people who will have blood sausage.C :=set of all people who will have black forest cake.
|M∩S∩C|= |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|= 41−25−32−18+12+9+15 = 2
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition.
A cardinal number is an ordinal number α so thatfor all ordinal numbers β that are equivalent to α we haveα ⊆ β .
Definition. For every infinite set S we define the cardinality |S|of S to be the unique cardinal number α that is equivalent to S.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. A cardinal number is an ordinal number α so thatfor all ordinal numbers β that are equivalent to α we haveα ⊆ β .
Definition. For every infinite set S we define the cardinality |S|of S to be the unique cardinal number α that is equivalent to S.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. A cardinal number is an ordinal number α so thatfor all ordinal numbers β that are equivalent to α we haveα ⊆ β .
Definition.
For every infinite set S we define the cardinality |S|of S to be the unique cardinal number α that is equivalent to S.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. A cardinal number is an ordinal number α so thatfor all ordinal numbers β that are equivalent to α we haveα ⊆ β .
Definition. For every infinite set S we define the cardinality |S|of S to be the unique cardinal number α that is equivalent to S.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
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Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem.
Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem.
Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A.
Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof.
Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A.
ThenX ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y]
, B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y]
,g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
]
,F(X) = A\g
[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X)
= A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]
⊆ A\g[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]
= F(Y).Let C :=
⋃{H ∈P(A) : H ⊆ F(H)
}and let c ∈ C. Then there
is an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C.
Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C).
Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).
Then F(C)⊆ F(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
).
By definition of C, F(C)⊆ C. ThusC = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C.
ThusC = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Cantor-Schroder-Bernstein Theorem. Let A and Bbe sets so that there is an injective function f : A → B and aninjective function g : B → A. Then there is a bijective functionh : A → B.
Proof. Define F(X) := A\g[B\ f [X]
]for all X ⊆ A. Then
X ⊆ Y implies f [X]⊆ f [Y], B\ f [X]⊇ B\ f [Y],g[B\ f [X]
]⊇ g
[B\ f [Y]
],
F(X) = A\g[B\ f [X]
]⊆ A\g
[B\ f [Y]
]= F(Y).
Let C :=⋃{
H ∈P(A) : H ⊆ F(H)}
and let c ∈ C. Then thereis an H ∈P(A) with c ∈ H ⊆ F(H)⊆ F(C). Hence C ⊆ F(C).Then F(C)⊆ F
(F(C)
). By definition of C, F(C)⊆ C. Thus
C = F(C).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.).
C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C
= F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C)
= A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]
impliesg[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C
and then B\ f [C] = g−1[A\C]. Henceg−1
∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C].
Henceg−1
∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C].
Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C.
Then h|C : C → f [C] andh|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective.
So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective.
Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y.
If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x)
= f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x)
6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6=
f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y)
= h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y).
If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x)
= g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x)
6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y)
= h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y).
Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C.
Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C].
So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Proof (concl.). C = F(C) = A\g[B\ f [C]
]implies
g[B\ f [C]
]= A\C and then B\ f [C] = g−1[A\C]. Hence
g−1∣∣A\C is bijective from A\C onto B\ f [C]. Define h : A → B
by h|C := f |C and h|A\C := g−1∣∣A\C. Then h|C : C → f [C] and
h|A\C : A\C → B\ f [C] are bijective. So h is surjective. Toprove that h is injective, let x,y ∈ A be so that x 6= y. If x,y ∈ C,then h(x) = f (x) 6= f (y) = h(y). If x,y 6∈ C, thenh(x) = g−1(x) 6= g−1(y) = h(y). Otherwise, WLOG x ∈ C andy 6∈ C. Then h(x) ∈ f [C] and h(y) ∈ B\ f [C]. So h(x) 6= h(y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem.
Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set.
Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof.
Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A.
To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A
, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.
There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y .
LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z
, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let A be an infinite set. Then A×A is equivalent toA.
Sketch of proof. Apply Zorn’s Lemma to the set F of allbijective functions f : X×X → X, where X ⊆ A. To prove that amaximal element f : Y×Y → Y of F must be a bijectivefunction from A×A to A, assume that Y is not equivalent to A.There must be an injective function from Y to A\Y . LetZ ⊆ A\Y be equivalent to Y and expand f to a function from(Y ∪Z)× (Y ∪Z) to Y ∪Z, using that Y×Z∪Z×Z∪Z×Y isequivalent to Z.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition.
Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic.
Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣
2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|
3. αβ :=
∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem.
Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite.
Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β}
and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof.
Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Definition. Cardinal arithmetic. Let α and β be cardinalnumbers and let A and B be sets with |A|= α and |B|= β .
1. α +β :=∣∣A×{0}∪B×{1}
∣∣2. αβ := |A×B|3. α
β :=∣∣AB∣∣, where AB := {f : f is a function from A to B}.
Theorem. Let α,β be cardinal numbers so that one of α and β
is infinite. Then α +β = max{α,β} and αβ = max{α,β}.
Proof. Good exercise.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem.
Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers.
Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only).
Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ .
Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C
7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C).
Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ
=∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣
=∣∣AB×AC∣∣ = α
βα
γ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣
= αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
Theorem. Let α,β and γ be cardinal numbers. Then1. αβ αγ = αβ+γ
2. αγβ γ = (αβ )γ
3.(
αβ
)γ
= αβγ
Proof (part 1 only). Let A,B,C be disjoint sets with |A|= α ,|B|= β and |C|= γ . Then AB∪C is equivalent to the set AB×AC
via f ∈ AB∪C 7→ (f |B, f |C). Henceα
β+γ =∣∣AB∪C∣∣ =
∣∣AB×AC∣∣ = αβ
αγ .
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gap
ℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number.
That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.
We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.
But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom.
The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis.
2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis
logo1
Finite Sizes Infinite Sizes Cardinal Arithmetic
A Mysterious Gapℵ0 denotes the first infinite cardinal number. That is,ℵ0 = ω = N0.We know that 2ℵ0 is not countable.But is it equal to the first uncountable ordinal ℵ1?
Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Cardinal Numbers and the Continuum Hypothesis