Chapter 6 Set Theory

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dont forget the set of all

If S is a set and P(x) is a property that elements of S may or may not satisfy, then a set A may be defined by writing A = {x ∈ S | P(x)}, (left curly brace) "the set of all" (the vertical bar) "such that" which is read "the set of all x in S such that P of x."

Operations on Sets

Let A and B be subsets of a universal set U. 1. The union of A and B, denoted A ∪ B, is the set of all elements that are in at least one of A or B. 2. The intersection of A and B, denoted A ∩ B, is the set of all elements that are common to both A and B. 3. The difference of B minus A (or relative complement of A in B), denoted B − A, is the set of all elements that are in B and not A. 4. The complement of A, denoted Ac, is the set of all elements in U that are not in A. Symbolically: A ∪ B = {x ∈ U | x ∈ A or x ∈ B}, A ∩ B = {x ∈ U | x ∈ A and x ∈ B}, B − A = {x ∈ U | x ∈ B and x / ∈ A}, Ac = {x ∈ U | x / ∈ A}.

Element Argument: The Basic Method for Proving That One Set Is a Subset of Another

Let sets X and Y be given. To prove that X ⊆ Y , 1. suppose that x is a particular but arbitrarily chosen element of X, 2. show that x is an element of Y .

mutually disjoint

Sets A1, A2, A3 . . . are mutually disjoint (or pairwise disjoint or nonoverlapping) if, and only if, no two sets Ai and Aj with distinct subscripts have any elements in common. More precisely, for all i, j = 1, 2, 3, . . . Ai ∩ Aj = ∅ whenever i does not equal j.

disjoint set

Two sets are called disjoint if, and only if, they have no elements in common. Symbolically: A and B are disjoint ⇔ A ∩ B = ∅.

To disprove that a set X is a subset of a set Y , you show that there is _____.

an element in X that is not in Y

The empty set is a set with _____.

no elements

The notation A ⊆ B is read "_____" and means that _____.

the set A is a subset of the set B; for all x, if x ∈ A then x ∈ B (Or: every element of A is also an element of B)

Given sets A1, A2, . . . , An , the Cartesian product A1 × A2 × . . . × An is _____.

the set of all ordered n-tuples (a1, a2, . . . , an), where ai is in Ai for all i = 1, 2, . . . , n

The power set of a set A is _____.

the set of all subsets of A

To use an element argument for proving that a set X is a subset of a set Y , you suppose that _____ and show that _____.

x is any [particular but arbitrarily chosen] element of X; x is an element of Y

An element x is in A ∩ B if, and only if, _____.

x is in A and x is in B (Or: x is in both A and B)

An element x is in A ∪ B if, and only if, _____.

x is in A or x is in B (Or: x is in at least one of the sets A and B)

An element x is in B − A if, and only if, _____.

x is in B and x is not in A

An element x is in A^c if, and only if, _____.

x is in the universal set and is not in A

partition

A finite or infinite collection of nonempty sets {A1, A2, A3 . . .} is a partition of a set A if, and only if, 1. A is the union of all the Ai 2. The sets A1, A2, A3, . . . are mutually disjoint.

A collection of nonempty sets A1, A2, A3, . . . is a partition of a set A if, and only if, _____.

A is the union of all the sets A1, A2, A3, . . . and Ai ∩ Aj = ∅ whenever i != j.

Sets A and B are disjoint if, and only if, _____.

A ∩ B = ∅ (Or: A and B have no elements in common)

subsets

A ⊆ B ⇔ ∀x, if x ∈ A then x ∈ B. A(not a subset)B ⇔ ∃x such that x ∈ A and x (not)∈ B. A is a proper subset of B ⇔ (1) A⊆B, and (2) there is at least one element in B that is not in A.

power set

Given a set A, the power set of A, denoted P (A), is the set of all subsets of A.

Notation

Given real numbers a and b with a ≤ b: (a, b) = {x ∈ R | a < x < b} [a, b] = {x ∈ R | a ≤ x ≤ b} (a, b] = {x ∈ R | a < x ≤ b} [a, b) = {x ∈ R | a ≤ x < b}. The symbols∞and −∞ are used to indicate intervals that are unbounded either on the right or on the left: (a,∞)={x ∈ R | x > a} [a,∞) ={x ∈ R | x ≥ a} (−∞, b)={x ∈ R | x < b} [−∞, b)={x ∈ R | x ≤ b}.

Set Equality

Given sets A and B, A equals B, written A = B, if, and only if, every element of A is in B and every element of B is in A. Symbolically: A = B ⇔ A ⊆ B and B ⊆ A.


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