Quiz M2.1

set ∅ vs set {∅}

- empty set vs singleton set - set ∅: empty set - set {∅}: single element of set is empty set ∅

sets with other sets as members

- A = {∅,{a},{b},{a,b}} - B = {x ∣ x is a subset of the set {a,b}}. - two sets are equal where A = B. - also, {a} ∈ A, but a ∉ A

A ⊆ B

- A is a subset of the set B. - A ⊆ B iff the quantification: ∀x(x ∈ A → x ∈ B) **to show that A is not a subset of B we need only find one element x ∈ A with x ∉ B. Such an x is a counterexample to the claim that x ∈ A implies x ∈ B. - use these rules to determine if one set is a subset of another

Set builder notation

- O = {x | x ia an odd positive integer less than 10} - O = {x∈Z+ ∣x is odd and x<10} - all positive rational numbers: Q+ = {x ∈ R ∣ x = p , for some positive integers p and q}

2-tuples - ordered pairs

- The ordered pairs (a, b) and (c,d) are equal iff a = c and b = d Note that (a,b) and( b,a) are not equal unless a = b

Theorem 1 Proof

- We will prove (i ) and leave the proof of (ii ) as an exercise.Let S be a set. To show that ∅ ⊆ S, we must show that ∀x(x ∈ ∅ → x ∈ S) is true. Because the empty set contains no elements, it follows that x ∈ ∅ is always false. It follows that the conditional statement x ∈ ∅ → x ∈ S is always true, because its hypothesis is always false and a conditional statement with a false hypothesis is true. Therefore, ∀x(x ∈ ∅ → x ∈ S) is true. This completes the proof of (i). Note that this is an example of a vacuous proof. - When we wish to emphasize that a set A is a subset of a set B but that A ≠ B, we write A ⊂ B and say that A is a proper subset of B. For A ⊂ B to be true, it must be the case that A ⊆ B and there must exist an element x of B that is not an element of A. That is, A is a proper subset of B if and only if ∀x(x ∈ A → x ∈ B) ∧ ∃x(x ∈ B ∧ x ∉ A) is true. - Venn diagrams can be used to illustrate that a set A is a subset of a set B. We draw the universal set U as a rectangle. Within this rectangle we draw a circle for B. Because A is a subset of B, we draw the circle for A within the circle for B. This relationship is shown in Figure 2. Recall from Definition 2 that sets are equal if they have the same elements. A useful way to show that two sets have the same elements is to show that each set is a subset of the other. In other words, we can show that if A and B are sets with A ⊆ B and B ⊆ A, then A = B. That is, A = B if and only if ∀x(x ∈ A → x ∈ B) and ∀x(x ∈ B → x ∈ A) or equivalently if and only if ∀x(x ∈ A ↔ x ∈ B), which is what it means for the A and B to be equal. Because this method of showing two sets are equal is so useful, we highlight it here.

set

- an unordered collection of distinct objects, called elements or members of the set. - A set is said to contain its elements. - We write a∈A to denote that a is an element of the set A. - The notation a ∉ A denotes that a is not an element of the set A.

domain U

- ∀xP(x) is true over the domain U if and only if the truth set of P is the set U - ∃xP(x) is true over the domain U if and only if the truth set of P is nonempty

Cartesian product A×B×C, where A={0,1}, B={1,2}, and C={0,1,2}

A × B × C = {(0, 1, 0), (0, 1, 1), (0, 1, 2), (0, 2, 0), (0, 2, 1), (0, 2, 2), (1,1,0),(1,1,1),(1,1,2),(1,2,0),(1,2,1),(1,2,2)}

A^2 if A = {1, 2}

A2 = {(1, 1), (1, 2), (2, 1), (2, 2)}

Cartesian product of A = {1,2} and B = {a,b,c}

AxB = {(1, a), (1, b), (1, c), (2, a), (2, b), (2, c)}

Identity laws

A∩U=A A∪∅=A

Domination laws

A∪U=U A∩∅=∅

Cartesian product of B = {a,b,c} and A = {1,2}

BxA = {(a,1), (b,1),(c,1),(a,2),(b,2)}

Theorem 1

For every set S, (i) ∅⊆S and (ii) S ⊆ S - show that every nonempty set S is guaranteed to have at least two subsets, the empty set and the set S itself, that is, ∅ ⊆ S and S ⊆ S.

Power Sets

Given a set S, the power set of S is the set of all subsets of the set S. The power set of S is denoted by P(S) -- answer is set of sets in P(S)

truth set: {x ∈ D ∣ P(x)}

Given predicate p, and a domain D, we define truth set of P to be the set of elements x in D for which P(x) is true

Size of a Set

Let S be a set. If there are exactly n distinct elements in S where n is a nonnegative integer, we say that S is a finite set and that n is the cardinality of S. The cardinality of S is denoted by |S|.

complement

Let U be the universal set. The complement of the set A, denoted by Ā, is the complement of A with respect to U. Therefore, the complement of the set A is U - A

N, Z, Z+, Q, R, R+, C

N = {0,1,2,3..}, the set of all natural numbers Z = {..., -2, -1, 0, 1, 2,..}, the set of all integers Z+ = {1,2,3,..}, the set of all positive integers Q = {p/q | p ∈ Z, q ∈ Z, and q ≠ 0}, the set of all rational numbers R = the set of all real numbers R+ = the set of all positive real numbers C = the set of all complex numbers

truth sets of P(x) is "|x| = 1," Q(x) is "x^2 = 2," and R(x) is "|x| = x"

P(x): {x ∈ Z ∣ |x| = 1} is {-1,1} Q(x): {x ∈ Z ∣ x^2 = 2} is empty set b/c no integers satisfy R(x) is {N+} OR R is N, the set of nonnegative integers

Power set of the set {0, 1, 2}

P({0, 1, 2}) = {∅, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}}

power set of the empty set ∅? power set of the set {∅}?

P(∅) = {∅} P({∅}) = {∅, {∅} }

Subsets

The set A is a subset of B, and B is a superset of A, if and only if every element of A is also an element of B. We use notation A ⊆ B to indicate that A is a subset of the set B. If, instead we want to stress that B is a superset of A, we use the equivalence notation B ⊇ A. (so A ⊆ B and B ⊇ A are equivalent statements.

Showing that A is not a subset of B

To show that A⊈B, find a single x ∈ A such that x ∉ B.

Roster method

V = {a, e, i, o, u}

infinite set

a set is said to be infinite if it is not finite

empty set ∅

a set with no elements

singleton set

a set with one element

intersection of sets A and B

denoted by A ∩ B, is the set containing those elements in both A and B.

Cartesian product of the sets A1, A2, ... , An

denoted by A1 × A2 × ⋯ × An, is the set of ordered n-tuples (a1, a2, ... , an), where ai belongs to Ai for i = 1, 2, ... , n. In other words, A1 ×A2 ×⋯×An ={(a1,a2,...,an)∣ai ∈Ai fori=1,2,...,n}

two ordered n-tuples

equal iff each corr. pair of their elements is equal - (a1, a2, ... , an) = (b1, b2, ... , bn), iff ai = bi, for i = 1,2..n

∃x ∈ S(P(x)) shorthand ∃x(x ∈ S ∧ P(x))

existential quantification of P(x) over all elements in S

disjoint of two sets

if their intersection is empty

universal set U

in venn diagrams, the universal set U, which contains all the objects under consideration, is represented by a rectangle.

Union: A ∪ B = {x ∣ x ∈ A ∨ x ∈ B}

let A and B be sets. the union of the sets A and B, denoted by A ∪ B, is the set that contains those elements that are either in A or in B, or in both

Cartesian products of A and B

let A and B be sets. the cartesian product of A and B, denoted by A x B, is the set of all ordered pairs (a,b), where a ∈ A and b ∈ B. Hence, A × B = {(a, b) ∣ a ∈ A ∧ b ∈ B} - not equal unless A = ∅ or B = ∅, so that AxB = ∅ OR A = B

difference of A-B

set containing those elements in A but not B. the difference of A and B is also called complement of B with respect to A

{N, Z, Q, R}

set with four elements N, Z, Q, R

intervals

sets of all the real numbers between two numbers and a and b ex: (a,b) = {x | a < x < b}

Showing that A is a subset of B

show that if x belongs to A, then x also belongs to B

ordered n-tuple (a1, a2, ... , an)

the ordered collection that has a1 as its first element, a2 as its second element, and an as its nth element

showing two sets are equal

to show that two sets A and B are equal, show that A ⊆ B and B ⊆ A.

equal sets

two sets are equal if and only if they have the same elements. Therefore, if A and B are sets then A and B are equal if and only if ∀x(x ∈ A ↔ x ∈ B). We write A = B if A and B are equal sets.

∀x ∈ S(P(x)) shorthand ∀x(x ∈ S → P(x))

universal quantification of P(x) over all elements in the set S

union of the sets {1,3,5} and {1,2,3}

{1,2,3,5}

Cardinality example: let A be the set of odd positive integers less than 10

|A| = 5