Summary of set theory related chapters in [1]

Contents Index

## Set Theory

### Zermelo–Fraenkel

**Axiom (Sets are objects)**. If $A$ is a set, then $A$ is also an object. In particular, given two sets $A$ and $B$, it is meaningful to ask whether $A$ is also an element of $B$.

**Axiom (Axiom of extensionality, equality of set)**. Two sets $A$ and $B$ are equal, $A=B$, iff every element of $A$ is an element of $B$ and vice versa.

Comment: The “is an element of ” relation $\in$ obeys the axiom of substitution ( equality axiom)

**Axiom (Empty set).** There exists a set $\emptyset$, known as the empty set, which contains no elements. i.e. for every object $x$, we have $x \notin \emptyset$

**Lemma (Single Choice) **Let $A$ be a non-empty set. Then there exists an object $x$ such that $x \in A$

**Axiom (Singleton sets and pair sets)** If $a$ is an object, then there exists a set $\{ a \}$ (singleton). If $a$ and $b$ are objects, then there exists a set $\{ a, b \}$ (pair).

**Axiom (Pairwise union):** Given any two sets $A, B$, there exists an union set $A \cup B$

**Lemma** Union operation is commutative and associative

**Definition (Subsets)** Let $A,B$ be sets. We say that $A$ is a subset of $B$, denoted $A \subseteq B$ iff every element of $A$ is also an element of $B$.

**Proposition (Sets are partially ordered by set inclusion)** Let $A,B,C$ be sets. If $A \subseteq B$ and $B \subseteq C$, then $A \subseteq C$. If $A \subseteq B$ and $B \subseteq A$ then $A=B$

**Axiom (Axiom of specification, axiom of separation)** Let $A$ be a set, and for each $x \in A$, let $P(x)$ be a property pertaining to $x$ (either true or false). Then there exists a set, called $\{ x \in A | P(X) \}$

**Definition (Intersections)**. The intersection $S_1 \cap S_2$ of two sets is defined to be the set

$$ S_1 \cap S_2 = \{ x \in S_1 | x \in S_2 \} $$

**Definition (Difference sets)** Given two sets $A$ and $B$, we define the set $A-B$ or $A \setminus B$

$$ A \setminus B := \{ x \in A | x \not\in B \}$$

**Proposition (Sets form a boolean algebra) **Let $A,B,C$ be sets, and let $X$ be a set containing $A,B,C$ as subsets

- (minimum element) $A \cup \emptyset = A$ and $\emptyset \cap A = \emptyset$
- (maximum element) $A \cup X = X$ and $A \cap X = A$
- (identity) $A \cap A = A, A \cup A = A$
- (commutativity) $A \cup B = B \cup A$ and $A \cap B = B \cap A$
- (associativity) $(A \cup B) \cup C = A \cup (B \cup C), (A \cap B) \cap C = A \cap (B \cap C)$
- (distributivity) $A \cap (B \cup C) = (A \cap B) \cup (A \cap C), A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$
- (Partition) $A \cup (X \setminus A) = X, A \cap (X \setminus A) = \emptyset$
- (De morgan laws) $X \setminus (A \cup B) = (X \setminus A) \cap (X \setminus B), X \setminus (A \cap B) = (X \setminus A) \cup (X \setminus B) $

**Axiom (Replacement) **Let $A$ be a set. For ay object $x \in A$, and any object y, suppose we have a statement $P(x,y)$ pertaining to $x$ and $y$, such that for each $x \in A$ there is at most one $y$ for which $P(x,y)$ is true. Then there exists a set $\{ y | P(x,y) \text{ is true for some } x \in A \}$

Remark: this can be used to apply function to element of a set to create another set

**Axiom (Infinity)** There exists a set $\mathcal{N}$, whose elements are called natural numbers such that Peano axioms hold

**Axiom (Axiom of regularity, axiom of foundation)** If $A$ is a non-empty set, then there is at least one element $x$ of $A$ which is either not a set, or is disjoint from $A$.

### Russel Paradox

Hardly anything more unfortunate can befall a scientific writer than to have one of the foundations of his edifice shaken after the work is finished 🙁

Gottlob Frege

**Paradox Axiom (Universal specification, Axiom of comprehension)**: Suppose for every object $x$, we have a property $P(x)$ pertaining to $x$. Then there exists a set $\{ x | P(X) \text{ is true } \}$

Remark: Allowing axiom of comprehension could allow the following contradictory set

$$ A = \{X | X \not\in X \} $$

Remark: allowing this axiom can easily derive other previous axioms. But it is too good to be true.

### Function

**Definition (Functions) **Let $X,Y$ be sets, and let $P(x,y)$ be a property pertaining to an object $x \in X$ and an object $y \in Y$, such that for every $x \in X$ there is exactly one $y \in Y$ for which $P(x,y)$ is true. Then we define the function $f: X \rightarrow Y$ defined by $P$ to be the object which, given any input $x \in X$, assigns an output $f(x) \in Y$ for which $P(x, f(x))$ is true.

Comment: One common way to define a function is simply to specify its domain, its range, and how one generates the output $f(x)$ from each input; this is known as an explicit definition of a function. In other cases, we only define a function by specifying what property $P(x,y)$ links the input $x$ with the output $f(x)$; this is an implicit definition of a function.

**Definition (Equality of functions)** Two functions $f: X \rightarrow Y, g: X \rightarrow Y$ with the same domain and range are said to be equal $f=g$ iff $f(x) = g(x)$ for all $x \in X$

**Definition (Composition)** Let $f: X \rightarrow Y$ and $g: Y \rightarrow Z$ be two functions, such that the range of $f$ is the same set as the domain of $g$. We then define the composition $g \circ f: X \rightarrow Z$ of the two functions $g$ and $f$ to be the function defined explicitly by the formula

$$ (g \circ f)(x) := g(f(x))$$

**Lemma (Composition is associative)** Let $f: Z \rightarrow W, g: Y \rightarrow Z$ and $h: X \rightarrow Y$ be functions. Then $f \circ (g \circ ) = (f \circ g) \circ h$

**Definition (injective)** A function $f$ is one-to-one (or injective) if different elements map to different elements

$$ x \neq x’ \implies f(x) \neq f(x’) $$

**Definition (surjective)** A function $f$ is onto (or surjective) if every element in $Y$ comes from applying $f$ to some element in $X$: for every $y \in Y$, there exists $x \in X$ such that $f(x) = y$

**Definition (bijective) **A function is called bijective or invertible if it is both injective and surjective

### Images and Inverse Images

**Definition (Images of Sets)** If $f: X \rightarrow Y$ is a function from $X$ to $Y$, and $S$ is a set in $X$, we define $f(S)$ to be the set

$$ f(S) := \{ f(x) | x \in S \} $$

**Definition (Inverse images)** If $U$ is a subset of $Y$, we define the set $f^{-1}(U)$ to be the set

$$ f^{-1}(U) := \{ x \in X | f(x) \in U \} $$

**Axiom (Power set axiom)** Let $X,Y$ be sets. Then there exists a set, denoted $Y^{X}$, which consists of all the functions from $X$ to $Y$

Remark: this is to basically create larger set such as $\mathbf{R}$

**Lemma (existence of power set)** Let $X$ be a set, then following set exists

$$ \{ Y | Y \subseteq X \} $$

**Axiom (Union) **Let $A$ be a set, all of whose elements are themselves sets. Then there exists a set $\cup A$ whose elements are precisely those objects which are elements of the elements of $A$.

### Cartesian Products

**Definition (Ordered Pair)** If $x$ and $y$ are any objects, we define the ordered pair $(x,y)$ to be a new object, consisting of $x$ as its first component and $y$ as its second component. Two ordered pairs $(x,y)$ and $(x’, y’)$ are considered equal iff both their components match

Remark: pair can be implemented with set as $(x,y) = \{ \{ x \}, \{ x, y \} \}$

**Definition (Cartesian Product)** If $X$ and $Y$ are sets, then we define the Cartesian product $X \times Y $ to be the collection of ordered pairs

$$ X \times Y = \{ (x,y) | x \in X, y \in Y \} $$

**Definition (Ordered n-tuple and n-fold Cartesian product) **Let $n$ be a natural number. An ordered n-tuple $(x_i)_{1 \leq i \leq n}$ (also denoted $(x_1, …, x_n)$ is a collection of objects $x_i$. If $(X_i)_{1 \leq i \leq n}$ is an ordered n-tuple of sets, we define their Cartesian product $\prod_{1 \leq i \leq n} X_i$ (also denoted $X_1 \times … \times X_n$) by

$$ \prod_{1 \leq i \leq n} := \{ (x_i)_{1 \leq i \leq n} | x_i \in X_i \text{ for all } 1 \leq i \leq n \}$$

**Definition (Finite choice).** Let $n \geq 1$ be a natural number and for each natural number $ 1 \leq i \leq n$, let $X_i$ be a non-empty set. Then there exists an n-tuple $(x_i)_{1 \leq i \leq n}$ such that $x_i \in X_i$ for all $1 \leq i \leq n$

Remark: infinite number of choices requires the axiom of choice

### Cardinality

**Definition (Equal cardinality)** $X$ and $Y$ have equal cardinality iff there exists a bijection $f: X \rightarrow Y$

The notion of having equal cardinality is an equivalence relation

**Definition **Let $n$ be a natural number. A set $X$ is said to have cardinality $n$ iff it has equal cardinality with $\{ i \in N | i < n \} $. We also say that $X$ has $n$ elements iff it has cardinality of $n$

**Proposition (Uniqueness of cardinality)** Let $X$ be a set of some cardinality $n$. Then $X$ cannot have any other cardinality.

**Lemma** Suppose that $n \geq 1$ and $X$ has cardinality $n$. Then $X$ is non-empty, and if $x$ is any element of $X$, then the set $X – \{ x \}$ has cardinality $n-1$.

**Definition (Finite sets)** A set is infinite iff it has cardinality $n$ for some natural number $n$; otherwise, the set is called infinite. If $X$ is a finite set, we use $\#(X)$ to denote the cardinality of $X$

## Logic

## Reference

[1] Tao, Terence. *Analysis*. Vol. 1. Hindustan Book Agency, 2006.