# The Topology Series

## Chapter 1

Prerequisites- Set Theory

### What is topology?

In simple terms, Topology is concerned with the geometric properties of objects when it undergoes physical distortions like stretching, twisting, and bending.

Now we dive a bit deeper in terms of mathematics.

**Definition 1**– A topology on a set *X* is a collection *T* of subsets of *X* having the following properties-

*Φ*and*X*are in*T*.- The union of the elements of any subcollection of
*T*is in*T*. - The intersection of the elements of any finite subcollection of
*T*is in*T*.

**Definition 2**– A set *X* for which a topology *T* has been specified is called a topological space.

A topological space is an ordered pair (*X, T*) consisting of a set *X* and a topology *T* on *X*.

If *X* is a topological space with topology *T*, we say that a subset *U* of *X* is an open set of *X* if *U* belongs to collection of *T*.

A topological space is a set *X* are both open, and such that arbitrary unions and finite intersections of open sets are open.

If *X* is any set, the collection of all subsets of *X*, it is called the discrete Topology. The collection consisting of *X*, we shall call it the indiscrete Topology, or the trivial Topology.

**Example**– Let *X* be a set, let *T _{f}* be the collection of all subsets

*U*of

*X*such that

*X-U*either is finite or is all of

*X*. Then T

_{f}is a topology on

*X*, called the finite complement topology. Both

*X*and

*Φ*are in

*T*, since

_{f}*X-X*is finite and

*X-Φ*is all of

*X*. If {

*U*} is an indexed family of non- empty elements of

_{α}*T*, to show that ∪

_{f}*U*, is in

_{α}*T*, we compute,

_{f}*X – ∪U _{α} = ∩(X – U_{α})*.

The set is finite because each set *X – U _{α}* is finite.

*U*are non-empty elements of

_{1}, U_{2},……….,U_{n}*T*, to show that

_{f}*∩U*is in

_{1}*T*, we compute,

_{f}*X – ∩U_{i} = ∪(X – U_{i})* (i = 1, to n).

**Definition 3**– Suppose that *T* and *T’* are two topologies on a given set *X*. If* T*⊂*T’*, we say that *T’* is finer than *T*. We also say *T* is coarser than *T’*, or strictly coarser, in these two respective situations. Wes say that *T* is comparable with *T’* if either *T*⊂*T’* or *T*‘⊂*T*.

**Definition 4**– If *X* is a set, a basis for a topology on *X* is a collection ** B** of subsets

*X*(called basis elements) such that-

- For each
*x***∈***X*, there is at least one basis element*B*containing*x*. - If x belongs to the intersection of two basis elements
*B*_{1}and*B*_{2}, then there is a basis element*B*_{3}containing x such that*B*_{3}⊂*B*_{1}B*∩*_{2}.

If ** B** satisfies these two conditions, then we define the topology

*T*generated by

**as follows-**

*B*A subset *U* of *X* is said to be open in *X* (that is, to be an element of *T*) if for each *x* **∈** *U*, there is a basis element B **∈** ** B** such that

*x*

**∈**

**and**

*B***⊂**

*B**U*.

Note that each basis element is itself an element of *T*.

To understand better we will go through a few lemmas.

**Lemma 1**– Let *X* be a set; let ** B** be a basis for topology

*T*on

*X*. Then

*T*equals the collection of all unions of elements of

**.**

*B*Proof- Given a collection of elements of ** B**, they are also elements of

*T*. Because

*T*is a topology, their union is in

*T*. Conversely, given

*U*

**∈**

*T*, choose for each

*x*

**∈**

*U*an element

*B*_{x}of

**such that**

*B**x*

**∈**

*⊂*

*B*_{x}*U*. Then

*U*=

*∪*

_{x ∈ U}

*B*_{x}, so

*U*equals a union of elements of

**.**

*B***Lemma 2**– Let *X* be a topological space. Suppose that * C* is a collection of open sets of

*X*such that for each open set

*U*of

*X*and each x in

*U*, there is an element C of

*such that*

**C***x*

**∈**C ⊂

*U*. Then

*is a basis for the topology of*

**C***X*.

(Proof will be available on demand.)

**Lemma 3**– Let ** B** and

**‘ be bases for the topologies**

*B**T*and

*T’*, respectively, on

*X*. Then the following are equivalent:

*T’*is finer than*T*.- For each
*x***∈***X*and each basis for the topologies and each basis element B**∈**containing*B**x*, there is a basis element B’**∈**such that‘*B**x***∈**B’ ⊂ B.

(Proof will be available on demand.)

TO BE CONTINUED IN CHAPTER 2.