Topology 01: Set Theory and Logic

Lingerois Shanghai Jiao Tong University

Update: January 10, 2020

1 Preliminarys of Set Theory

We skip naive set theory, one could refer to the book for a review.

We introduce a term for set theory. We say set A and B intersects if A ∩ B , ∅. Note Vacuously True: If p is false, for any q, p ⇒ q is always true. We say this kind of proposition is vacuously true. One can comprehend the truth of p ⇒ q by using the notion of set theory. Let A = {x|p(x) = 1}, and B = {x|q(x) = 1}, then p(x) ⇒ q(x) is saying that if x ∈ A then x ∈ B, which is equal to A ⊂ B. For x < A (equal to p(x) = 0), of course x < B (equal to q(x) = 0). Proposition 1.1 p ⇒ q is equal to ¬q ⇒ p.

Proof. The only case that p ⇒ q is false is when p is true and q is false, which implies ¬q ⇒ ¬p is false. A similar argument can prove ¬q ⇒ ¬p is true then p ⇒ q is true. Theorem 1.2 DeMorgan’s Law

A − (B ∪ C) = (A − B) ∪ (A − C) A − (B ∩ C) = (A − B) ∩ (A − C)

The set of sets is called collection, denoted by A, B, ··· . The collection of all the subsets of A is called the power set, denoted by P(A). We define the the union and intersection of all sets in A.

For given A, the union of elements in A is defined as Ø A = {x|∃A ∈ A, x ∈ A} A∈A And the intersection of A’s elements is defined as Ù A = {x|x} A∈A A problem here is what if A = ∅? One can prove that the union is still well defined if A =. But for the intersection, "∀A ∈ A, x ∈ A" is vacuously true for any X. Therefore we don’t define the intersection of empty A.

We skip the discuss of naive Cartesian product.

2 Function

We define the function f : A → B as a subset of C × D. We define it like this to circumvent the vague description of "corresponding". Definition 2.1 A rule of assignment is a subset r of C × D, where for any (a, b),(a, b0) ∈ r, we have b = b0.

The definition is saying that r can only contain at most one element (a, b) for any a ∈ C. We say

C is the domain of r;

2 D is the codomain of r; {b|∃(a, b) ∈ r} ⊂ D is the range of r, which is also called the image set of r. Remark We don’t use the term "range" for B of f : A → B, otherwise, we use "codomain" to keep in line with other notes. Definition 2.2 A function f is a rule of assignment r, together with a domain B that contains the image set of r. The domain of r is called the domain of f , the image set of r is called the image set of f . B is called the codomain of f . Definition 2.3 For functions f : A → B and g : B → C, the composition of f and g, denoted by f ◦ g, is a function f ◦ g : A → C, defined as ( f ◦ g)(a) = f (g(a)).

We skip the definition of injection, surjection and bijection, by shown them by fig.1.

(a) injection (b) surjection

(c) bijection (d) not an injection, nor a surjection

Figure 1: Compare of Functions

Remark We use "injective", "surjective" and "bijective" for adjectives. We never use ambiguous words like "one-to-one" and "onto".

Some notions are used frequently. For function f : X → Y, let A ⊂ X and B ⊂ Y, we define

The image of A as f (A) = { f (x)|x ∈ A}; The preimage of B as f −1(B) = {x| f (x) ∈ B}.

3 By defining them, we have Proposition 2.4 We have the following holds:

1. x ∈ f −1(B) ⇐⇒ f (x) ∈ B; 2. f −1( f (A)) ⊂ A; 3. f ( f −1)(B) ⊃ B; 4. f (A ∩ B) ⊂ f (A) ∩ f (B).

We illustrate the possible non-equality of 2. and 3. by fig.2.

(a) f −1(f (A)) ( A (b) f (f −1)(B) ) B

Figure 2: Compare of Functions

3 Relation

Definition 3.1 A relation on a set A is a subset C of A × A.

3.1 Equivalence Relation

We skip the definition of equivalence relation and partition. Note When we defines equivalence class and partition, we never says the equivalence classes C of some relation C on A is an partition of A. This fact is true, but one has to prove it. To prove this, one has to show that (1) every x ∈ A lies in some C ∈ C; (2) any x ∈ A never lies in two distinct classes C1,C2 ∈ C.

4 3.2 Order

Definition 3.2 An order relation (or simple order, linear order) C of set A is a relation that satiesfy the follows:

1. For any x, y ∈ A such that x , y, we have x ∼ y or y ∼ x; 2. There exists no x ∈ A such that x ∼ x; 3. If x ∼ y and y ∼ z, then x ∼ z.

The order relation of a set is saying that one can line up the elements of A by a thread, in the order of the order relation. We often use "<" instead of "∼" to denote the relations in an order relation. Note There’s no parlance of "strict simple order" or "non-strict simple order". A simple order never includes x ∼ x for any x ∈ A. The notion x ≤ y means one of the following two statements holds:

x and y are different elements in A and x < y; x and y are different notions of the same elements in A. Definition 3.3 For a set X, let < be the simple order on X. For a < b, we denote

{x|a < x < b} as (a, b), and we say it is a open interval. If (a, b) is empty, we say b is the immediate successor of a.

Definition 3.4 For set A and B has simple order order type if there exists an bijection f : A → B such that

a1

As for A, B with simple order, we can define a simple order for A × B.

Definition 3.5 For set A and B has simple order

(a1, b1) < (a2, b2)

We say it is the dictionary order relation.

5 3.3 Least Upper Bound Property

We now discuss least upper bound property, which is a amazing as well as useful property used frequently in analysis. Let’s define something first.

Definition 3.6 Let A be a set with simple order <. Let A0 be a subset of A. We say b is the largest element in A0 if b0 ∈ A0 and for any x ∈ A0 it holds that x ≤ b0. Similarly, we say b is the smallest element in A0 if for any x ∈ A0 it holds that b ≤ x. Remark In fact the names with "largest" and "smallest" are quite straightforward since we use "maximum" and "minimal" more often.

Definition 3.7 We say A0 ⊂ A is bounded above if there exists b ∈ A such that for all x ∈ A0 it holds that x ≤ b; We say b is a upper bound of A0. If there’s a smallest element in the collection of the upper bounds of A, we say it is the least upper bound (or supremum) of A0, denoted by sup A0.

Similarly, one could define bounded below, lower bound, greatest lower bound. Definition 3.8 We say a set A with simple order has least upper bound property if for all its nonempty subset A0, if A0 has an upper bound, it has least upper bound. Similarly, we cound define greatesst lower bound property.

Notice that discussing all (from the start of section 3.3) these concepts is significant only if the order is specified. The simple order can be omitted if there’s no ambiguity.

An amazing of the previous definition is that the least upper bound property and the greatest lower bound property is exactly the same property. Theorem 3.9 Ordered set A has least upper bound property if and only if it has greatest lower bound property.

We postpone the proof of theorem 3.9 in the notes of Mathematical Analysis.

6 4 Integers and Real Numbers

We skip the definition of binary operations. Definition 4.1 A set R we called real numbers, is a field (with addition + and multiplication ·), with a simple order < and the following properties (assuming x, y ∈ R):

1. If x > y, then x + z > y + z; 2. If x > y and z > 0, then x · z > y · z; 3. Simple order > has least upper bound property; 4. If x < y, there exists z ∈ R such that x < z < y.

We say a field is an ordered field if it satisfies properties 1. and 2.. We say a set with simple order is an linear continuum if it satisfies properties 3. and 4..

Note we don’t prove such R exists, since it is too complex. A proof can be found in James R. Munkres’ topology book’s appendix. Now we move on to define positive integers. Definition 4.2 We say a subset of real numbers is inductive if it includes 1 and x + 1 ∈ A if x ∈ A. Let A be the collection of all the indutive subset of R. Define positive integer as

Z+ = ∩A∈A A

Note We skipped the definition of field. 1 is the unity in the field R. The definition of Z+ will use the definition of 1.

Let n ∈ Z+, we denote Sn = {i ∈ Z+|i < n}. We say Sn is a section of Z+. Similarly, we can define a section for every set with simple order.

We define integers as the set of Z+ ∩ {0} ∩ (−Z+).

Theorem 4.3 [Well-Ordering Property] Every nonempty subset of Z+ has a smallest element.

Proof. (Sketch) Prove by induction. Let A ⊂ Z+ with the above property, then 1 ∈ A. One could prove A is inductive.

One can prove theorem 4.4 by using theorem 4.3.

7 Theorem 4.4 [Strong Induction Principle] Let A ⊂ Z+. If for arbitrary n ∈ Z+, there is Sn ⊂ A, then A = Z+.

theorem 4.4 provide us with a new type of proof by induction.

Another property of Z+ is that it has no upper bound in R. We call this the Archimedean ordering property of real line.

5 Cartesian Set

In this section, we are going to define Cartesian set for countable collection. Definition 5.1 Let A be a nonempty collection. A indexing function of A is a surjection f : J → A, where J is called the index set. Collecion A together with f is called the indexed family of sets. Given α ∈ J, set f (α) is denoted by Aα. The indexed familiy of set itself is denoted by {Aα}α∈J or {Aα} for simplicity.

Indexing functions are surjection, but we don’t acquire them to be injections. Example 5.2 For A × B, we usually say "the 1st coordinate" and "the 2nd coordinate". We are actually doing f : {0,1} → {A, B} and 1 7→ A,2 7→ B.

Indexing function gives us a new notion for the union and intersection of sets, like Ø Aα = {x|existsα ∈ J : xinAα} aα∈J Ù Aα = {x|∀α ∈ J : xinAα} aα∈J

Definition 5.3 Let m ∈ Z+. Given set X, a m-tuple of X is defined as a function:

x : {1, ··· , m} → X

Similarly, a ω-tuple of X is defined a function

x : Z+ → X

We denote x(i) as xi for both m-tuple and ω-tuple.

8 For both tuples defined in theorem 5.3, one can think that there are m or countable different locations in a array. And the tuple x pointed out which element of X lies in every specific location.

We are going to define Cartesian product for countable family of set, by using m-tuple or ω-tuple. Ð Definition 5.4 For J = [m], let A = {Aα}α∈J , Let X = J ∈α Aα, then we define the Cartesian set for A as m Y Ai or A1 × · · · × Am i=1 be all the m-tuple in X that satisfy for each i ∈ [m], xi ∈ Ai. Similarly, one can de fine the Cartesian set Y Ai

i∈Z+

Why are we making things so complex? You might think its easy to define the Cartesian product of finite family of sets by induction. But define by tuple makes it easier to define such Cartesian set for infinite or even uncountable family of sets.

6 Finite Set

Definition 6.1 We say a set A is finite, if there’s some section Sn of Z+ and a bijection f : A → Sn. Otherwise, we say A is infinite

The definition works well for ∅, since there’s a bijection f : ∅ → S1 = ∅. We say a finite set with bijection between Sn has cardinality n − 1. We say ∅ has cardinality 0. One might notice that the cardinality of a finite set is telling us how may elements does the set contains. However, we haven’t prove that the cardinality of a finite set is unique.

Lemma 6.2 Let n ∈ Z+, A be a set, a0 ∈ A. There exists a bijection f : A → [n] iff there exists a bijection g : A − {a0} → [n − 1].

The proof of this lemma is easy. One can just give a transformation between such f and g in theorem 6.2.

9 Definition 6.3 Let A be a set. Presume for some n ∈ Z+, there exists a bijection f : A → [n]. Let B ( A, then there exists no bijection g : B → [n], but there exists some h : B → [m] for some m < n.

Proof. (Sketch) Prove can be done by indcuction. Let C ⊂ Z+ of all the n that makes the theorem holds. Then 1 ∈ C obviously. One can prove that for any n ∈ C, there is n +1 ∈ C, then C = Z+.

Corollary 6.4 If A is a finite set, there exists no bijection between A and any of its proper subset.

Corollary 6.5 Z+ is not a finite set.

Proof. There’s a bijection f : Z+ → Z+ − {0}, such that f (n) = n + 1.

We can now prove that every finite set has a unique cardinality. Corollary 6.6 The cardinality of finite set A is determined by A.

Proof. (Sketch) If A has two distinct cardinality m and n, one could counstruct a bijection f : [m] → [n]. This is impossible due to theorem 6.4.

Corollary 6.7 The union or Cartesian product of finite number of sets is finite.

7 Countable Set and Uncountable Set

Definition 7.1 A infinite set is countably infinite if there’s a bijection

f : A → Z+

Example 7.2 Z+ × Z+ is countably infinite. We can map (x, y) ∈ Z+ × Z+toZ+ by doing 1 (x, y) 7→ (x0 = x + y − 1, y0 = y) 7→ (x0 − 1)x0 + y0 2 This map first maps the "squre" to the "lower triangle", then to the "line". Definition 7.3 A set is countable if it is finite or countably infinite. Otherwise, we say it is uncountable.

10 Theorem 7.4 Let B be a nonempty set, the following conditions are equal:

1. B is countable;

2. There exists a surjection f : Z+ → B;

3. There exists an injection g : B → Z+. Remark We have skipped a similar version of theorem 7.4 for finite set since its straightforward.

One don’t have to care to much about the construction of bijections from the countably infinite set will introduce later to Z+. But one principle used in the construction is quite important, refer to theorem 7.5 and theorem 7.6.

Definition 7.5 Given the infinite subset C ⊂ Z+, there exists a unique function h : Z+ → C that satisfy: the smallest element of C, for i = 1 h(i) =  (1) the smallest element of [C − h({1, ··· ,i − 1})], for all i > 1  eq. (1) is called the recursion formula. The definition given by it is called the recursive definition. Theorem 7.6 Let A be a set. Given an formula: define h(1) as the unique element in A. For i > 1, define h(i) as the unique such element, only related to the value of h on positive integers less than i. This formula gives us a unique h : Z+ → A.

Theorem 7.6 can be proved by using theorem 7.5. We are not going to prove it.

Lemma 7.7 Let C ⊂ Z+ be an infinite set, then C is a countably infinite set.

Proof. (Sketch) Prove by construct a bijection h : Z+ → C by using theorem 7.6.

Corollary 7.8 Any subset of countable set is countable.

Corollary 7.9 Set Z+ × Z+ is countably infinite.

It’s a little bit interesting that one can use the number theoretic method to prove theorem 7.9.

Proof. (Sketch) Define the injection f : Z+ × Z+ → Z+ as

f (n, m) = 2n3m is just fine.

11 Corollary 7.10 Rational numbers Q is countable. Theorem 7.11 The Cartesian product of finite numbers of countable set is countable.

Still, I like to prove theorem 7.11 by using the number theoretic method.

Sadly, countable Cartesian product of countable sets is not always countable. Theorem 7.12 Let X = {0,1}. Xω is uncountable.

Theorem 7.12 is quite important for computer science. The method of proving it is called diagonalization. Theorem 7.13 Let A be a set. There exists no injection f : P(A) → A as well as surjection g : A → P(A).

Proof. We prove by diagonalization again, in another form. We just have to prove the second part of theorem 7.13.

Leet g : A → P(A) be an arbitrary functon, we prove g is not surjection. For any a ∈ A, a can be in g(a) or not. Let B ⊂ A be the set of all a < g(a), namely

B = {a|a ∈ A − g(a)}

Then there’s no element b ∈ B such that g(b) = B, due to the definition of B. Therefore, g is not a surjection.

8 Principle of Recursive Definition

We are skipping this section temporarily. Maybe I’ll write this section after reading it with a food for thought in an spare afternoon.

12 9 Infinite Set and Axiom of Choice

In this section, we are going to talk about a wired axiom and an inconceivable result comes from it. We starts with a theorem. Theorem 9.1 Let A be a set. The following conditions of A are equivalent:

1. There exists an injection f : Z+ → A; 2. There exists a bijection between A and its proper subset; 3. A is an infinite set.

Proof. (1.⇒2.) Define f (a) = an, where f is the injection in 1., let B be the image set of f . Define g as

g(an) = an+1, for an ∈ B, g(x) = x, for x ∈ A − B

Then g : A → A − {a1} is a bijection. g can be illustrated as fig.3.

Figure 3: the construction of g

(2.⇒3.) This is the contrapositive of theorem 6.4.

(3.⇒1.) We prove by using "recursive definition". We choose a1 ∈ A to be f (1). For n ∈ Z+, we choose an ∈ A − {a1, ··· , an−1} and let f (n) = an. Then f is a injection from Z+ to A.

The (3.⇒1.) in the previous proof is morbid. How can we make sure we could choose an element from a set? The axiom of choice is going to solve this.

13 Axiom 9.2 [Axiom of Choice] Given any family A of nonempty pairwise non-intersect sets, there exists a set C s.t. every set in A has exactly one common element with C.

C is just like choices of sets in A. C chooses exactly one element from each set in A. We can get theorem 9.3 instantly from theorem 9.2. Lemma 9.3 [Existence of a choice function] Given any nonempty collection B of sets. There exists a function Ø c : B → B B∈B s.t. for any B ∈ B, c(B) is a element of B.

Use theorem 9.3, one can choose a element in the proof of theorem 9.1. The proof is finally a proof. Remark We skipped the prove of theorem 9.3. One can consider what is B in theorem 9.3 in the proof of theorem 9.1.

10 Well-Ordered Set

Definition 10.1 We say a set A with simple order is well-ordered, if every nonempty subset of A has a smallest element.

When talking about well-ordering, one has to specific the simple order. The simple order can be omitted when defaulted.

Theorem 10.2 Every nonempty finite set has the order type of section Sn of Z+, therefore is well-ordered.

It is not surprising that every finite set is well-ordered, what does surprise us is that every set is well-ordered. Theorem 10.3 Let A be a set, there exists a simple order on A that makes A well-ordered.

Theorem 10.3 has been proved by Ernst F. F. Zermelo in 1904. Let’s see a example of an uncountable set that is well-ordered.

14 Definition 10.4 Let X be a simple ordered set. Given a ∈ X, define Sα as

Sα = {x|x ∈ X and x < α}

We say Sα is a section of X at α.

Lemma 10.5 There exists a well-ordered set A with Ω as its largest element s.t. SΩ is uncountable, but for every α ∈ A such that α , Ω, Sα is countable.

We can prove this counterintuitive lemma by using the counterintuitive theorem 10.3.

Proof. Let B be a uncountable well-ordered set. Let C = B × 1,2 be the well-ordered set with dictionary order. Note the smallest element of any nonempty subset D of C can be chosen by: (1)

If D has some element (b,1), then choose (b1,1) where b1 is the smallest element of {b|(b,1) ∈ D};

(2) Otherwise choose (b2,2) where b2 is the smallest element of {b|(b,2) ∈ D}.

C has some uncountable section. Let Ω being the smallest element such that SΩ of C is uncountable. Then A = SΩ ∪ {Ω}.

Such SΩ in theorem 10.5 is a quite important example in topology. We prove a property of it.

Theorem 10.6 Let SΩ be such set in theorem 10.5. If E is a countable set in SΩ, E has upper bound in SΩ.

Proof. Let A be a countable subset of SΩ. Then for any a ∈ A, Sa is countable. Therefore Ð B = a∈A Sa is countable which implies B , SΩ. Let x be an arbitrary element in SΩ − B, x is a supper bound of A.

11 Maximum Principle

Maximum principle is equivalent to axiom of choice. We are goint to talk about some of it. Definition 11.1 Given set A, we say a relation ≺ is an strict partial order, if the following properties hold:

15 1. (Irreflexivity) For any a ∈ A, a ≺ a does not hold; 2. Transitivity If a ≺ b, b ≺ c then a ≺ c.

Compared to simple order, strict partial order doesn’t require any two distinct elements can be compared. One can not line up elements by the strict partial order. The strict partial order looks more like a net. Example 11.2 The prowerset P(A) of any set A has a strict partial order ⊂.

Example 11.3 . Let A be all the interior of circles on R2, the ⊆ relation of A is a strict partial order. See fig.4

(a) A ≺ B ≺ C (b) A ≺ C, B ≺ C but A, B are in- comparable

Figure 4: Strict Partial Order of Circles in R2

Theorem 11.4 [Maximum Principle] Let A be a set, ≺ be a strict partial order on A. There exists a maximum subset B with simple order.

One can prove maximum principle as a practice using axiom of choice. Definition 11.5 Let A be a set, ≺ be a strict partial order. Let B be a subset of A, we say c ∈ A is a upper bound of B if for every b ∈ B it holds that b = c or b ≺ c. We say m ∈ A is a maximum element in A if for any a ∈ A, m ≺ a is false. Theorem 11.6 [Zorn’s Lemma] Let A be a strict partial ordered set. If every simple ordered subset of A has a supper bound, there’s a maximum element in A.

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