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Home , Complement (set theory), Empty set, Naive set theory, Union (set theory)

Naïve Theory Basic Definitions Naïve is the non-axiomatic treatment of set theory.

In the axiomatic treatment, which we will only allude to at times, a set is an undefined term. For us however, a set will be thought of as a collection of some (possibly none) objects. These objects are called the members (or elements) of the set. We use the "∈" to indicate membership in a set.

Thus, if A is a set and x is one of its members, we write x ∈ A and say "x is an of A" or "x is in A" or "x is a member of A".

Note that "∈" is not the same as the Greek letter "ε" .

Basic Definitions Sets can be described notationally in many ways, but always using the set brackets "{" and "}".

If possible, one can just list the elements of the set: A = {1,3, oranges, lions, an old wad of gum} or give an indication of the elements: ℕ = {1,2,3, ... } ℤ = {..., -2,-1,0,1,2, ...} or (most frequently in ) using set-builder notation: = {x ∈ ℝ | 1 < x ≤ 7 } or {x ∈ ℝ : 1 < x ≤ 7 } which is read as "S is the set of real x, such that x is greater than 1 and less than or equal to 7".

In some sets may be denoted by special notations. For instance, in analysis S would be written (1,7]. Basic Definitions Note that sets do not contain repeated elements. An element is either in or not in a set, never "in the set 5 times" for instance. When repeated elements are needed, we refer to , but they are not sets.

Definition: Let A and B be sets. We say that A is a of B iff every element of A is also an element of B. Symbolically, A ⊆ B ⇔ (∀x)( x∈A ⇒ x∈B ).

Observe that "being a subset" is a ... a statement that is either true or false.

Ex: A = {0,2,5} If B = ℤ then A ⊆ B but if B = ℕ, A ⊈ B. Basic Definitions Notice the following distinctions:

Let A = {0,2, Elsie the Cow} Is A a subset of B if B is: B = {0,1,2,3, lions, Elsie the Cow} Yes B = {0,1,2, lions, cows} No B = {0,2, a lion, a cow} Maybe

Definition: Two sets, A and B are equal iff every element of one is also an element of the other. Symbolically, A = B ⇔ (∀x)( x∈A ⇔ x∈B ).

More useful in proofs than this definition is the logically equivalent statement: Basic Definitions (∀x)( x∈A ⇔ x∈B ) ≡ (∀x)( x∈A ⇒ x∈B ) ∧ (∀x)( x∈A ⇐ x∈B )

Which can be written in set notation (as opposed to logic notation) as: A = B iff A ⊆ B and B ⊆ A.

Definition: For sets A and B, we use the notation A ⊂ B, to mean that A is a subset of B and A ≠ B.

Note that this is not a universally accepted notation. Some authors have used this symbol to mean subset as we have defined it. The In the axiomatic development of set theory there is one (and only one) existence and it is: Axiom: The empty set exists.

Definition: Let ∅ = {x | x ≠ x }. This set contains no elements and is called an empty set.

Notice that any contradiction can be used in the defining clause. Also, the symbol "∅" is not the Greek letter "φ" phi, it is actually of Danish origin.

To emphasize the definition we have given, notice that the predicate x ∈ ∅ is always false. The Empty Set Thm: The empty set is unique.

Pf: Suppose that A and B are empty sets. Since A is an empty set, the statement x∈A is false for all x, so (∀x)( x∈A ⇒ x∈B ) is true! That is, A ⊆ B. Since B is an empty set, the statement x∈B is false for all x, so (∀x)( x∈Β ⇒ x∈Α ) is also true. Thus, B ⊆ Α. Since A ⊆ B and B ⊆ Α we have A = B. 

We have used a direct proof here ... but had we decided to do this by contradiction we would have .... The Empty Set Thm: The empty set is unique.

Pf: BWOC suppose that A and B are distinct empty sets. Since A is an empty set, the statement x∈A is false for all x, so (∀x)( x∈A ⇒ x∈B ) is true! That is, A ⊆ B. Since B is an empty set, the statement x∈B is false for all x, so (∀x)( x∈Β ⇒ x∈Α ) is also true. Thus, B ⊆ Α. Since A ⊆ B and B ⊆ Α we have A = B. →← 

The contradiction method in this case does not add anything to the , it just puts an unnecessary layer over the basic direct method ... this is not wrong, its just in bad taste!! The Universal Subset Thm: The empty set is a subset of every set.*

Pf: Let A be a set. Since x ∈ ∅ is false for all x, (∀x)( x ∈ ∅ ⇒ x∈Α ) is true. Thus, ∅ ⊆ A. Since A was arbitrary, ∅ is a subset of every set.  ∗ Observe that the statement is a quantified statement: (∀A, A a set)( ∅ ⊆ A) To prove a "for all" statement, one starts with an arbitrary element of for the variable ... assuming no properties other than membership, and show that the statement is true for it. Warnings Some minor points concerning the empty set:

1. ∅ is the empty set, {∅} is NOT the empty set. {∅} is a set that contains 1 element (i.e., the empty set), if you want to use set brackets to describe the empty set, use "{}".

2. While ∅ is a subset of every set, it is rarely ever a member of a set (to be so, it must be in the list of elements). Thus, ∅ ∈ {1, Elsie, ∅} as well as ∅ ⊆ {1, Elsie, ∅} but ∅ ∉ {1, Elsie, {∅}} Text Problems

T T T T F T F F F F T T 2.1(b)

Thm: For every set A, A ⊆ A.

Pf: Let A be an arbitrary set. Since x∈A ⇒ x∈A for all x (whether x∈A is true or not) (∀x)(x∈A ⇒ x∈A ) is true. So, A ⊆ A for all sets A.  Theorem 2.2

Thm: Let A, B and C be sets. If A ⊆ B and B ⊆ C, then A ⊆ C.

Pf: Let x ∈ A. Since A ⊆ Β, x ∈ B. Since B ⊆ C, x ∈ C. Thus, for all x, x ∈ A ⇒ x ∈ C. So, A ⊆ C. 

We refer to this result as the "transitivity of subset inclusion". Proper Definition: For any set A, ∅ and A are called improper subsets of A. All other subsets are called proper subsets.

Example: Let S = {1, 2, Elsie}

The proper subsets of S are: {1}, {2}, {Elsie} {1, 2}, {1, Elsie}, {2, Elsie}

The improper subsets of S are: ∅, {1,2, Elsie} Power Sets Definition: Given a set S, the of S, denoted P (S), is the set consisting of all the subsets of S. P (S) = {B | B ⊆ S}. Important! The elements of the power set of a set are themselves sets, both proper and improper.

If S = {0,2,5} then {5} ∈ P (S) but 5 ∉ P (S). On the other hand, 5 ∈ S and {5} ∉ S.

Example: If S = {1, 2, Elsie} then

P (S) = {∅, {1}, {2}, {Elsie},{1, 2}, {1, Elsie}, {2, Elsie}, S} Power Sets Thm: If a set S has n elements then P (S) has 2n elements.

We will provide at least two proofs of this important result in later sections when we talk about the techniques used.

Example: If S = ∅ then P (∅) = {∅} which is not the empty set!

S contains 0 elements, and P (∅) has 20 = 1 element. Power Sets Thm: Let A and B be sets. Then A ⊆ B iff P (A) ⊆ P (B). Pf: (⇒ Sufficiency) Let C ∈ P (A), then C ⊆ A. Since A ⊆ Β we have by Thm 2.2 that C ⊆ B. Thus, C ∈  (B). Since C was arbitrary, P (A) ⊆ P (B).

(⇐ Necessity) Let x ∈ A. Then {x} ⊆ A. Since P (A) ⊆ P (B), {x} ∈ P (B), so {x} ⊆ B. Thus x ∈ B, and since x was arbitrary, A ⊆ B.  Both parts of this proof illustrate the method of "element chasing" used frequently in proofs involving sets. Power Sets Thm: Let A and B be sets. Then A ⊆ B iff P (A) ⊆ P (B). An alternate proof the the necessity of this theorem can be given which does not involve element chasing: (⇐ Necessity) Since A ∈ P (A) and P (A) ⊆ P (B), we have that A ∈ P (B). Thus, A ⊆ B. 

Compare this with the previous proof: (⇐ Necessity) Let x ∈ A. Then {x} ⊆ A. Since P (A) ⊆ P (B), {x} ∈ P (B), so {x} ⊆ B. Thus x ∈ B, and since x was arbitrary, A ⊆ B.  Text Problems

T T T T T T T F T T T F

P ({∅, {∅}}) = { ∅, {∅, {∅}}, {∅}, {{∅}} }. Set Operations

Definitions: Let A and B be sets.

The of A and B, denoted A ∪ B, is the set A ∪ B = {x | x ∈ A ∨ x ∈ B}.

The of A and B, denoted A ∩ B, is the set A ∩ B = {x | x ∈ A ∧ x ∈ B}.

The relative of B in A (difference) denoted by A – B, is the set A – B = {x | x ∈ A ∧ x ∉ B}. Venn Diagrams Frequently, one sees diagrams used to represent the relationships between sets, called Venn diagrams. For instance:

Venn diagrams should be used with care, and you should never an argument on what such a picture looks like, since one diagram can not represent all possible relationships. Complements An unstated assumption that we have made is that lurking in the background is a set that contains all the objects that may be considered in the context of a specific problem. This is known as the (or Universe of Discourse) for the problem. It is a set of convenience, and changes with different problems.

Definition: The complement of a set A, which we denote by A~ (or Ac), is the relative difference of A in the Universal set U. That is, A~ = U – A = {x ∈ U | x ∉ A}.

The reason we need to introduce the universal set here is that if it didn't exist the complement of A would not be a set!! Russell's Definition: A set is said to be ordinary if it is not a member of itself.

Ex: A = {1,2,3,A} is not an ordinary set. {abstract ideas} is not an ordinary set.

Now consider the set P = {all ordinary sets}.

Is P ordinary? If P were an ordinary set then P ∈ P, so P would not be ordinary. →← So, P is not ordinary .... but then P ∉ P, so P is an ordinary set. →← !!!!!!! The only way out of this paradox .... P is not a set! The Halting Problem Suppose there is a program P which looks at a program S and its data set and decides whether or not S will go into an infinite loop with this data set.

Modify the program P to P*, by adding code to P which will go into an infinite loop if P is about to declare that S will not go into an infinite loop and stop normally otherwise (declaring S will go into an infinite loop).

Now, feed program P* into itself.

If P* stops (halts), it is because P* goes into an infinite loop !! If P* goes into an infinite loop, it is because P* halts. !!!!!!!!!! So, P* and therefore P can not exist. Prop: a) ∅ ∩ A = ∅ and ∅ ∪ A = A

b) A ∩ B ⊆ A

c) A ⊆ A ∪ B

d) A ∪ B = B ∪ A and A ∩ B = B ∩ A

e) A ∪ (B ∪ C) = (A ∪ B) ∪ C and A ∩ ( B ∩ C) = ( A ∩ B) ∩ C

f) A ∪ A = A = A ∩ A

g) If A ⊆ B then A ∪ C ⊆ B ∪ C and A ∩ C ⊆ B ∩ C. ∅ ∩ A = ∅ Pf: To prove set we need to show that ∅ ∩ A ⊆ ∅, and ∅ ⊆ ∅ ∩ A. However, we have already proved that this second containment is true, so we only need to prove the first. To prove ∅ ∩ A ⊆ ∅, we have to show that (∀x ∈ U)( x ∈ ∅ ∩ A ⇒ x ∈ ∅). ∅ ∩ Α = {x | x ∈∅ ∧ x ∈ A}, but since x ∈∅ is false for all x, x ∈∅ ∧ x ∈ A is always false. Thus, x ∈ ∅ ∩ A is always false, so x ∈ ∅ ∩ A ⇒ x ∈ ∅ is true for all x. 

Α better proof: BWOC suppose ∅ ∩ A ≠ ∅. Then ∃x (x ∈ ∅ ∩ A). Which implies, by the definition of intersection, that ∃x (x ∈ ∅) →←  ∅ ∪ A = A

Pf: We first show that ∅ ∪ A ⊆ A. Let x ∈ ∅ ∪ A. By the definition of union, x ∈ ∅ ∨ x ∈ A . Since x ∈ ∅ is always false, we must have x ∈ A is true. Thus, ∀x (x ∈ ∅ ∪ A ⇒ x ∈ A) and so, ∅ ∪ A ⊆ A. STOP!!!! ... we are being sloppy. This statement is supposed to be true for all sets A. What if A is = ∅? We have used the statement "x ∈ A is true" in this "proof", but if A is the empty set, that is false!!

To fix this problem, we deal with the special case of A = ∅ first. ∅ ∪ A = A (again) Pf: If A = ∅, then ∅ ∪ ∅ = {x | x ∈ ∅ ∨ x ∈ ∅ } = ∅, since the condition is a contradiction. Thus, ∅ ∪ A = A is true in this case. We may therefore assume that A ≠ ∅. We first show that ∅ ∪ A ⊆ A. Suppose x ∈ ∅ ∪ A. By the definition of union, x ∈ ∅ ∨ x ∈ A . Since x ∈ ∅ is always false, we must have x ∈ A is true. Thus, ∀x (x ∈ ∅ ∪ A ⇒ x ∈ A) and so, ∅ ∪ A ⊆ A. Now we show that A ⊆ ∅ ∪ A. Let x ∈ A. By the definition of union, x ∈ ∅ ∪ A. Thus, ∀x (x ∈ A ⇒ x ∈ ∅ ∪ A) and so, A ⊆ ∅ ∪ A. ❏ A ∩ B ⊆ A

Pf: If A ∩ B = ∅, then the statement is true for any set A. Thus we may assume that A ∩ B ≠ ∅. Let x ∈ A ∩ B. By the definition of intersection, x ∈ A and x ∈ B. In particular, x ∈ A. So, (∀x)( x ∈ A ∩ B ⇒ x ∈ A) and we have A ∩ B ⊆ A. ❏ If A ⊆ B then A ∪ C ⊆ B ∪ C Pf: If A ∪ C = ∅ then the statement is true, so we may assume that A ∪ C ≠ ∅. Let x ∈ A ∪ C. Case I: x ∈ A. Since A ⊆ B, x ∈ A ⇒ x ∈ B. x ∈ B ⇒ x ∈ B ∪ C. Case II: x ∈ C. x ∈ C ⇒ x ∈ B ∪ C.

Thus, in either case, x ∈ A ∪ C ⇒ x ∈ B ∪ C. So, (∀x)(x ∈ A ∪ C ⇒ x ∈ B ∪ C) and we have A ∪ C ⊆ B ∪ C. ❏ Propositions Prop: a) (A ∪ B)~ = A~ ∩ B~ and (A ∩ B)~ = A~ ∪ B~

b) A~~ = A

c) A - B = A ∩ B~

d) A ⊆ B iff B~ ⊆ A~. (A ∪ B)~ = A~ ∩ B~

Pf: (A ∪ B)~ = U - (A ∪ B) = {x | x ∈ U ∧ x ∉ Α ∪ Β} = {x | x ∈ U ∧ ~ (x ∈ Α ∪ Β)} = {x | x ∈ U ∧ ~ (x ∈ Α ∨ x ∈ Β)} = {x | x ∈ U ∧ (~x ∈ Α ∧ ∼x ∈ Β)} = {x | x ∈ U ∧ (x ∉ Α ∧ x ∉ Β)} = {x | (x ∈ U ∧ x ∉ Α) ∧ (x ∈ U ∧ x ∉ Β)} = {x | (x ∈ U - Α) ∧ (x ∈ U - Β)} = {x | (x ∈ A~) ∧ (x ∈ B~)}

= Α∼ ∩ Β∼ ❏ A - B = A ∩ B~ Pf: If A-B = ∅ then (∀ x ∈ A)(~(x ∉ B)) thus A ∩ B~ = {x ∈ U| x ∈ A ∧ x ∉ B} = ∅, and the statement is true. We may therefore assume that A-B ≠ ∅. Let x ∈ A-B. Then x ∈ A and x ∉ B. x ∈ U ∧ x ∉ B ⇒ x ∈ B~, i.e. x ∈ A ∩ B~. So, A - B ⊆ A ∩ B~. If A ∩ B~ = ∅ then (∀ x ∈ A)(~(x ∉ B)), so A-B = ∅ and the statement is true. We may therefore assume that A ∩ B~ ≠ ∅. Let y ∈ A ∩ B~ . Thus, y ∈ A and y ∈ B~. y ∈ B~ ⇒ y ∈ U ∧ y ∉ B. So y ∈ A and y ∉ B, i.e. y ∈ A – B. Hence, A ∩ B~ ⊆ A – B and we have A - B = A ∩ B~. ❏ A ⊆ B iff B~ ⊆ A~

Pf: A ⊆ B iff (∀x ∈ U)(x ∈ A ⇒ x ∈ B) iff (∀x ∈ U)(~(x ∈ B) ⇒ ~(x ∈ A)) iff (∀x ∈ U)(x ∈ B~ ⇒ x ∈ A~) iff B~ ⊆ A~ ❏