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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of Extension

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of Extension

Two sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements.

That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.

But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far.

That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

The Axiom of ExtensionTwo sets are equal iff they have the same elements. That is, if Aand B are sets, then A = B iff for all objects x we have x ∈ A iffx ∈ B.

It may seem strange that we need an axiom to define equality.But note that equality cannot be defined based on the axiomswe have so far. That means we need another axiom specificallygeared towards equality.

The model at the end of the presentation for the Axiom ofSpecification satisfies the Axiom of Extension, too.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem.

Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff

A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof.

Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.

A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B iff

for every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, iff

for every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, iff

for every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iff

A⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Two sets A and B are equal iff A⊆ B and B⊆ A.

Proof. Let A and B be sets.A = B ifffor every object x the statement x ∈ A⇔ x ∈ B is true, ifffor every object x the statement(x ∈ A⇒ x ∈ B)∧ (x ∈ B⇒ x ∈ A) is true, ifffor every object x we have that x ∈ A implies x ∈ B and thatx ∈ B implies x ∈ A, iffA⊆ B and B⊆ A.

This is one of the rare cases in which we can preserve the iffthroughout the proof.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem.

Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection.

Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets.

Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof.

Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.

“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”:

Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B.

Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B.

Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A.

Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A.

We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.

“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”:

Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A.

Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A.

Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B.

Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B.

We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.

Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Commutativity of set intersection. Let A and B besets. Then A∩B = B∩A.

Proof. Let A and B be sets.“⊆”: Let x ∈ A∩B. Then x ∈ A and x ∈ B. Hence x ∈ B andx ∈ A. Thus x ∈ B∩A. We have proved that A∩B⊆ B∩A.“⊇”: Let x ∈ B∩A. Then x ∈ B and x ∈ A. Hence x ∈ A andx ∈ B. Thus x ∈ A∩B. We have proved that B∩A⊆ A∩B.Because A∩B⊆ B∩A and B∩A⊆ A∩B we conclude thatA∩B = B∩A.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem.

Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection.

Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets.

Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof.

Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.

“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”:

Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C).

Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C.

Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C.

Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C.

Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C.

We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.

“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”:

Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C.

Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C.

Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C.

Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C.

Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C).

We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).

Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

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Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Theorem. Associativity of set intersection. Let A,B and C besets. Then A∩ (B∩C) = (A∩B)∩C.

Proof. Let A,B and C be sets.“⊆”: Let x ∈ A∩ (B∩C). Then x ∈ A and x ∈ B∩C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A∩B and x ∈ C. Thereforex ∈ (A∩B)∩C. We have proved that A∩ (B∩C)⊆ (A∩B)∩C.“⊇”: Let x ∈ (A∩B)∩C. Then x ∈ A∩B and x ∈ C. Hencex ∈ A and x ∈ B and x ∈ C. Thus x ∈ A and x ∈ B∩C. Thereforex ∈ A∩ (B∩C). We have proved that (A∩B)∩C ⊆ A∩ (B∩C).Because A∩ (B∩C)⊆ (A∩B)∩C and(A∩B)∩C ⊆ A∩ (B∩C) we conclude that(A∩B)∩C = A∩ (B∩C).

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Visualization with Venn Diagrams

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Visualization with Venn Diagrams

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Visualization with Venn Diagrams

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Visualization with Venn Diagrams

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Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition.

Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets.

Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof.

“⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”:

Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B.

First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B.

Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B

, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A.

Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A.

Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A.

Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B.

Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B.

HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B

and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.

“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”:

For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A

and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A.

BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B.

Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B.

Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Proposition. Let A and B be sets. Then A⊆ B iff A∩B = A.

Proof. “⇒”: Let A⊆ B. First, let x ∈ A∩B. Because x ∈ A∩B,we know that x ∈ A and x ∈ B, which, in particular, means thatx ∈ A. Hence A∩B⊆ A. Conversely, let x ∈ A. Because x ∈ Aand A⊆ B we infer that x ∈ B. Therefore x ∈ A∩B. HenceA⊆ A∩B and thus A∩B = A.“⇐”: For the converse, let A∩B = A and let x ∈ A. BecauseA = A∩B, we infer that x ∈ A∩B. Thus x ∈ B. Hence A⊆ B.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Equality of Sets Can Be Hard To Prove

I Let T be the set of all twin prime numbers.I Let K be the set of all presently known twin prime

numbers.I Are they equal?I I don’t know.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Equality of Sets Can Be Hard To ProveI Let T be the set of all twin prime numbers.

I Let K be the set of all presently known twin primenumbers.

I Are they equal?I I don’t know.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Equality of Sets Can Be Hard To ProveI Let T be the set of all twin prime numbers.I Let K be the set of all presently known twin prime

numbers.

I Are they equal?I I don’t know.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Equality of Sets Can Be Hard To ProveI Let T be the set of all twin prime numbers.I Let K be the set of all presently known twin prime

numbers.I Are they equal?

I I don’t know.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension

logo1

Equality of Sets Algebra for Intersections Hierarchies of Thoughts/Techniques A Warning

Equality of Sets Can Be Hard To ProveI Let T be the set of all twin prime numbers.I Let K be the set of all presently known twin prime

numbers.I Are they equal?I I don’t know.

Bernd Schroder Louisiana Tech University, College of Engineering and Science

The Axiom of Extension