The foundation of the modern mathematical language lies in the set theory. Thus we start this course with recalling basic notions of sets and maps between sets.

Recall that a set is denoted by curly brackets ${\ldots} .$ Elements of a set must by distinct. The following important sets have special reserved notation.

- $\mathbb{N}={1,2,3, \ldots}$ is the set of all positive integers. Sometimes we call $\mathbb{N}$ the set of natural numbers.
- $\mathbb{Z}={\ldots,-2,-1,0,-1,2, \ldots}$ is the set of all integers.
- $\mathbb{Q}=\left{\frac{m}{n} \mid m \in \mathbb{Z}, n \in \mathbb{N}\right}$ is the set of rational numbers.
- $\mathbb{R}=(-\infty,+\infty)$ is the set of real numbers.

There are many ways to describe a given set.

We can specify a set by listing all its elements,

$$

A={1,2,3} .

$$

Or as a set of elements satisfying certain properties:

$$

A={a \in \mathbb{Z} \mid 1 \leqslant a \leqslant 3} .

$$

In the plain English the last definition will be read as

$$

\text { set } A \text { is the set of all integers } a \text { such that } 1 \leqslant a \text { and } a \leqslant 3 \text {. }

$$

We can also give a descriptive definition

$$

B={\text { odd integers }}

$$

or equivalently write

$$

B={2 n+1 \mid n \in \mathbb{Z}}

$$

The latter will be read as

$B$ is the set of numbers of the form $2 n+1$ where $n$ ranges over all the integers

Throughout this course we will be often using quantifier notations. Please make sure you remember and understand them.

- $\forall$ (upside-down A) means ‘for all’ or ‘for any’. For example ‘ $\forall x \in \mathbb{R}$ we have $x^{2} \geqslant 0$ ‘ means ‘for all $x \in \mathbb{R}$ we have $x^{2} \geqslant 0^{\prime} .$
- $\exists$ (mirrored E) means ‘there exists’. For example ‘ $\exists C \in \mathbb{R}$ such that $\forall x \in \mathbb{R} x^{2}>C$ ‘ means ‘there exists $C \in \mathbb{R}$ such that for all $x \in \mathbb{R} x^{2}>C^{\prime}$.
- ! means ‘unique’. For example ‘ $\forall y>0 \exists ! x>0$ such that $x^{2}=y$ ‘ means ‘for all $y>0$ there exists a unique $x>0$ such that $x^{2}=y^{\prime}$.

Recall also the following notation regarding mutual relation of sets and elements. – $a \in A$ means that element $a$ belongs to the set $A$;

- $B \subset A($ or $B \subseteq A)$ means that set $B$ is a subset of the set $A$, i.e.

$$

\forall b \in B \text { we have } b \in A \text {. }

$$

Clearly $A \subset A$ and $A=B$ if and only if $A \subset B$ and $B \subset A$;

- $B \subsetneq A$ means that $B$ is a proper subset of $A$, i.e., $B \subset A$ but $B \neq A$
- $\emptyset$ is the empty set, i.e., the set with no elements in it.

Now we describe several important operations which allow to produce new sets from given sets.

The union of two sets $A$ and $B$ is the set

$$

A \cup B={x \in A \text { or } x \in B}

$$

consisting of all elements belonging to either $A$ or $B$.

The intersection of two sets $A$ and $B$ is the set

$$

A \cap B={x \in A \text { and } x \in B}

$$

consisting of all elements belonging to both $A$ and $B$.

We can define the unions of intersections of multiple sets in an obvious way:

$$

\bigcup_{i=1}^{n} A_{i}=A_{1} \cup \cdots \cup A_{n}, \quad \bigcap_{i=1}^{n} A_{i}=A_{1} \cap \cdots \cap A_{n}

$$

Difference of sets $A$ and $B$ is the set

$$

A \backslash B={x \in A \text { and } x \notin B}

$$

Note that unlike unions and intersections, the difference is not symmetric, i.e., in general $A \backslash B \neq B \backslash A$.

8. Definition 5: Cartesian Product

Given two sets $A$ and $B$ the Cartesian product of $A$ and $B$ is

$$

A \times B={(a, b) \mid a \in A, b \in B}

$$

is the set consisting of all the pairs $(a, b)$.

Let $Q$ be the square in the plane with the vertices $(0,0),(1,0),(1,1),(0,1) .$ Then $Q=[0,1] \times[0,1] .$ In general, if is helpful to think of the Cartesian product of two sets as of a square.

Informally, a map $f$ from set $A$ to set $B$ is a rule which assigns an element $b \in B$ to each element $a \in A .$ In this case we write

$$

f: A \rightarrow B, \quad f(a)=b \text {. }

$$

Under a map $f: A \rightarrow B$ every element $a \in A$ is assigned with a unique element $f(a) \in B$.

If sets $A$ and $B$ are finite and ‘small’ we could represent a map explicitly using a mapping diagram:

Often the notion function is used as a synonym of map.

Define a map $f: \mathbb{R} \rightarrow \mathbb{R}$ given by

$$

f(x)=x^{2} .

$$

For every set $A$, there is a distinguished map

$$

\mathrm{i} d_{A}: A \rightarrow A, \quad \mathrm{i} d_{A}(a)=a \forall a \in A,

$$

called identity map. It sends any element $a \in A$ to itself.

There are several important properties which a map might satisfy.

13. Definition 6: Bijection, injection and surjection

A map $f: A \rightarrow B$ is called injective if it does not $g l u e$ elements together, i.e.,

$$

\forall x, y \in A: \quad \text { If } f(x)=f(y), \text { then } x=y .

$$

A map $f: A \rightarrow B$ is called surjective if its image is the whole $B$, i.e.,

$$

\forall b \in B, \exists a \in A \text { such that } f(a)=b \text {. }

$$

A map which is injective and surjective is called bijective. Problem 1: Change exactly one arrow in the Figure 1 to make the map bijective. In how many ways you can do it?

Function $f(x)=x^{2}$ defines a map $f: \mathbb{R} \rightarrow \mathbb{R}$. This map not injective, since $f(1)=f(-1)=1$. It is also not surjective, since $f(x)$ never takes the value $-1$.

The $\operatorname{map} g: \mathbb{R} \rightarrow \mathbb{R}$ defined by the function $g(x)=x^{3}$ is on the contrary bijective.

$$

f: A \rightarrow B, \quad g: B \rightarrow C,

$$

the we can form a new map $(g \circ f)$ called the composition of $f$ and $g$ :

$$

(g \circ f): A \rightarrow C, \quad(g \circ f)(a):=g(f(a))

$$

NB: Note that maps in a composition $g \circ f$ are applied from right to left – order in which you take composition is important!

Map $g: B \rightarrow A$ is the inverse of the map $f: A \rightarrow B$ if

$$

(g \circ f)=\mathrm{i} d_{A}, \text { and }(f \circ g)=\mathrm{i} d_{B}

$$

The inverse map is denoted by $f^{-1}$.

Map

$$

g:[0,+\infty) \rightarrow[0,+\infty), \quad g(x)=\sqrt{x}

$$

is the inverse of the map

$$

f:[0,+\infty) \rightarrow[0,+\infty), \quad f(x)=x^{2}

$$

When does a map admits an inverse? The following theorem answers this question.

Map $f: A \rightarrow B$ admits an inverse if and only if $f$ is bijective. In this case the inverse is unique.

19. NB: Whenever we encounter if and only if claim, a proof in two directions is required!

20. Proof. Direction $\Rightarrow$.

Assume that $f$ admits an inverse $g: B \rightarrow A .$ We are about to prove that $f$ is injective and bijective.

- $(f$ is injective). Indeed, assume that $f(x)=f(y)$ for some $x, y \in A .$ Apply $g$ to both sides of this identity, then

$$

x=\mathrm{i} d_{A}(x)=(g \circ f)(x)=g(f(x))=g(f(y))=(g \circ f)(y)=\mathrm{i} d_{A}(y)=y .

$$

So $f$ is injective

- ( $f$ is surjective). Take any $b \in B$. Then

$$

f(g(b))=(f \circ g)(b)=\mathrm{i} d_{B}(b)=b \text {, }

$$

so $b$ is in the range of $f$, and therefore $f$ is surjective.

Let us define function $g: B \rightarrow A$. Given any $b \in B$ we can find $a \in A$ such that $f(a)=b$, since $f$ is surjective. Then we define

$$

g(b)=a

$$

It remains to check that $g$ is indeed the inverse map.

- Check that $(f \circ g)=\mathrm{i} d_{B} .$ Take any $b \in B .$ Then by definition of $g$, we have $f(g(b))=b$. Therefore indeed $(f \circ g)(b)=b$.
- Check that $(g \circ f)=\mathrm{i} d_{A} .$ Take any $a \in A .$ Then we want to check that

$$

(g \circ f)(a)=a

$$

We already know that

$$

f(g(f(a)))=f(a)

$$

since $f \circ g=\mathrm{i} d_{B}$. Now, as $f$ is injective the latter implies

$$

g(f(a))=a,

$$

as required.

23. Definition 8: Equivalence relation

Equivalence relation on a set $A$ is a subset $R \subset A \times A$ satisfying the following properties

- (reflexive) for any $a \in A$ we have $(a, a) \in R$
- (symmetric) if $(a, b) \in R$, then $(b, a) \in R$.
- (transitive) if $(a, b) \in R$ and $(b, c) \in R$, then $(a, c) \in R .$

If we have $(a, b) \in R$ we will write ‘ $a \sim b$ ‘ and say ‘ $a$ is equivalent to $b$ ‘.

- $R={(a, a) \mid a \in A} .$ In other words, any element $a$ is equivalent only to itself.
- $R={(a, b) \mid a, b \in A} .$ In other words, any element $a \in A$ is equivalent to any other element $b \in A$.
- Take $R=\mathbb{Z}$ and define $R={(m, n) \mid m-n$ is even $} \subset \mathbb{Z} \times \mathbb{Z}$.

If $a \in A$ and $\sim$ is an equivalence relation, we can form an equivalence class

$$

[a]={b \in A \mid b \sim a} .

$$

Let $\sim$ be an equivalence relation on the set $A$. Then any two equivalence classes $[a],[b]$ either coincide element-wise, or do not intersect.

The above exercise allows to define a new set consisting of the equivalence classes:

The set of equivalence classes in A modulo an equivalence relation $\sim$ is defined as

$$

A / \sim={[a] \mid a \in A} .

$$

There is a natural map

$$

A \rightarrow A / \sim, \quad a \mapsto[a]

$$

sending each element to its equivalence class. Problem 2: When is the natural map $A \rightarrow A / \sim$ surjective? injective? bijective?

Going back to Example 6, we see that

If $R={(a, a) \mid a \in A}$ then $A / \sim$ is the same as $A$

If $R={(a, b) \mid a, b \in A}$ then $A / \sim$ has only one element

Finally, in the last example $Z / \sim$ consists of two elements: class of odd integers and class of even integers.