# General Topology — Part 1

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# What is General Topology ?

General topology is the branch of topology that deals with the basic set-theoretic definitions and constructions used in topology. It is the foundation of most other branches of topology, including differential topology and algebraic topology.

Why should I learn General Topology? Because it is the basic language of modern mathematics! Also, it is very interesting and can train you to make logical and clear thinking.

The prerequisite of learning General Topology is Set Theory. Therefore, let’s review some propositions from Set Theory that will be frequently used later before we start our journey.

# Set Theory Review

**Proposition 1**: For any universe X and subsets A, B, and C of X, the following identities hold:

Ac = X\A

A\B = A∩Bc

A\B = A\(A∩B)

A\(B∪C) = (A\B)∩(A\C)

A\(B∪C) = A∩(B∪C)c

A\(B∩C) = (A\B)∪(A\C)

(A\B)∩C = (A∩C)\(B∩C)

A∩(B\C) = (A∩B)\C

(A∩B)c =Ac ∪ Bc

(A∪B)c =Ac ∩ Bc

A∩(B∪C) = (A∩B)∪(A∩C)

A∪(B∩C) = (A∪B)∩(A∪C)

**Proposition 2**: For any two sets A and B:

A⊆B ⇒ A\B=ø

A∩B=ø ⇒ A⊆Bc

A∩B=ø ⇒ B⊆Ac

A∩B=ø ⇒ A\B = A

A⊆B ⇒Bc⊆Ac

A⊆B ⇒[x∉B ⇒x∉A]

**Proposition 3**: For any function f:A→B, f-1:B→A, A0, A1 ⊂ A, B0, B1 ⊂ B [ f-1 means inverse of function f ]:

B0⊆B1 ⇒f-1(B0)⊆f-1(B1)

f-1(B0∪B1)=f-1(B0)∪f-1(B1)

f-1(B0∩B1)=f-1(B0)∩f-1(B1)

f-1(B0\B1)=f-1(B0)\f-1(B1)

A0⊆A1 ⇒f(A0)⊆f(A1)

f(A0∪A1)=f(A0)∪f(A1)

f(A0∩A1) ⊂ f(A0)∩f(A1)

f(A0)\f(A1) ⊂ f(A0\A1)

if f is injective:

f(A0∩A1) = f(A0)∩f(A1)

f(A0)\f(A1) = f(A0\A1)

# What is Topological Space ?

A **Topological Space** is an ordered pair (X, τ), where X is a set and τ is a collection of subsets ui of X, called **Open Set**, satisfying the following axioms:

T1: ø, X ∈ τ

T2: any (finite or infinite) union of sets in τ is itself in τ

T3: any finite intersection of sets in τ is itself in τ

Pretty abstract? Don’t ask me why the axioms for a topological space are those axioms, just believe that they are useful. I will show you in the coming articles that these axioms can lead to some real life application.

Take real line R=(-∞,∞) as an example of Topological Space.

Define **Open Set** o and τ:

τ ={o|o = ∪(a,b)}

Note that union of open intervals is still open interval and intersection of open intervals can be an empty set or open interval. Let’s prove that (R,τ) is a Topological Space:

**Proof-**

T1:

ø contains no point ⇒ ø ∈ τ

∀x ∈ X, x ∈ (x-1,x+1) ⇒ ∪(x-1,x+1) ∈ τ ⇒ R ∈ τ

T2:

o[i] ∈ τ

⇒ ∪o[i] = ∪∪(a[i],b[i]) = ∪(a[j],b[j])

⇒ ∪o[i] ∈ τ

T3:

o1, o2 ∈ τ

⇒ o1 ∩ o2 = ∪(a1[i],b1[i]) ∩ ∪(a2[j],b2[j])

⇒ o1 ∩ o2 = ∪((a1[i],b1[i]) ∩ (a2[j],b2[j])) = ∪(a[k], b[k])

⇒ o1 ∩ o2 ∈ τ

∴ (R, τ) is a Topological Space

# Bases of Topological Space

Sometimes, it is not easy to specify the whole collection of open sets. One can specify a smaller collection of subsets of X and define topology using them.

Define a **Bases** B for a topology on X as a collection of subset of X:

B = {b| b ⊆ X}

Bases B has the following properties:

Ba1. ∀x ∈ X, ∃b ∈ B such that x ∈ b ⊆ X,

Ba2. For b1, b2 ∈ B and x ∈ b1∩b2, ∃b3 ∈ B such that x ∈ b3 ⊆ b1∩b2.

Define **Open Set** o and τ

τ ={o|o ⊆ X, ∀x ∈ o ∃b ∈ B s.t. x ∈ b ⊆ o}

Let’s prove that (*X, *τ) is a Topological Space generated by Bases B:

**Proof**-

T1:

ø contains no point ⇒ ø ∈ τ

∀x ∈ X ∃b ∈ B s.t. x ∈ b ⊆ X **[By Ba1]**

⇒ X ∈ τ

T2:

o[i] ∈ τ

⇒ ∪o[i] ⊆ X and ∀x ∈ ∪o[i] ∃o[α] s.t. x ∈o[α]

⇒ ∪o[i] ⊆ X and ∀x ∈ ∪o[i] ∃b ∈ B s.t. x ∈ b ⊆ o[α] ⊆ ∪o[i] **[By defintion of τ]**

⇒ ∪o[i] ∈ τ

T3:

o1, o2 ∈ τ

⇒ o1 ∩ o2 ⊆ X and o1 ∩ o2 = {x|∃b1, b2 ∈ B s.t. x ∈ b1 ⊆ o1 and x ∈ b2 ⊆ o2}

⇒ o1 ∩ o2 ⊆ X and o1 ∩ o2 = {x|∃b1, b2 ∈ B s.t. x ∈ b1 ∩ b2 ⊆ o1 ∩ o2}

⇒ o1 ∩ o2 ⊆ X and o1 ∩ o2 = {x|∃b3 ∈ B s.t. x ∈ b3 ⊆ b1 ∩ b2 ⊆ o1 ∩ o2} **[By Ba2]**

⇒ o1 ∩ o2 ∈ τ

∴(X,τ) is a Topological Space generated by Bases B

With the help of bases, we can express the open set as a union of bases.

Define **Open Set** o and τ’

τ’ ={o|o ⊆ X, o = ∪b[i], b[i] ∈ B}

Let’s prove that τ = τ’:

**Proof-**

∀o ∈ τ’

⇒ o = ∪b[i] ⊆ X

⇒ ∀x ∈ o ∃b[α] ∈ B s.t. x ∈ b[α] and b[α] ⊆ ∪b[i]

⇒ ∀x ∈ o ∃b[α] ∈ B s.t. x ∈ b[α] ⊆ o **[ By o = ∪b[i] ]**

⇒ o ∈ τ

⇒ τ’ ⊆ τ

∀o ∈ τ

⇒ ∀x ∈ o ∃bx ∈ B s.t. x ∈ bx and bx ⊆ o

⇒ o = ∪bx[i]

⇒ o ∈ τ’

⇒ τ ⊆ τ’

Take real line R=(-∞,∞) as an example. Remember that we define Open Set o = ∪(a,b), the bases will be open interval (a,b) in this case.

# Sub-bases of Topological Space

Define a **Sub-bases** S for a topology on X as a collection of subset of X:

S = {s| s ⊆ X} ⇒ ∪S = ∪{s|s ∈ S} = X

Define a collection B’ for a topology on X as a collection of subset of X:

B’ = {b| b = ∩s[i], s[i] ∈ S} ⇒ S ⊂ B’

Let’s prove that B’ is a Bases of topology on X:

**Proof-**

Ba1:

X = ∪S

⇒ ∀x ∈ X, ∃s ∈ S s.t. x ∈ s ⊆ X

⇒ ∀x ∈ X, ∃s ∈ B’ s.t. x ∈ s ⊆ X **[ By S ⊂ B’ ]**

Ba2:

b1, b2 ∈ B’ and x ∈ b1 ∩ b2

⇒ b1=∩s[i], b2=∩s[j], x ∈ ∩s[i] ∩ ∩s[j]

⇒ ∃b3 ∈ B’ s.t. x ∈ b3 where b3 = b1∩b2 = ∩s[i] ∩ ∩s[j] = ∩s[k]

Therefore, **Open Set** can be defined in the following 2 ways:

1. Every **Open Set** is a union of **Bases** elements.

2. Every **Open Set** is a union of finite intersections of **Sub-bases** elements.

Take real line R=(-∞,∞) again as an example. The sub-bases will be half open interval (-∞,b) and (a,∞) in this case. Intersection of (-∞,b) and (a,∞) will be (a,b) when a < b.

# Subspace Topology

Define a **Subspace** **Topology** τY where Y is a subset of X

τY ={oY | oY= Y∩o, o ∈ τ, Y ⊆ X}

With this topology, Y is called a **Subspace** of X. Its open sets consist of intersections of open sets of X with Y. Let’s prove that (Y,τY) is a Topological Space:

**Proof-**

T1:

ø ∈ τ ⇒ ø=Y∩ø ∈ τY

X ∈ τ ⇒Y=Y∩X ∈ τY

T2:

oY[i] ∈ τY

⇒ ∪oY[i] = ∪{Y∩o[i]| o[i] ∈ τ}

⇒ ∪oY[i] = Y∩(∪o[i]), ∪o[i] ∈ τ

⇒ ∪oY[i] ∈ τY

T3:

oY1, oY2 ∈ τY

⇒ oY1 ∩ oY2 = (Y∩o1) ∩ (Y∩o2), o1 ∈ τ, o2 ∈ τ

⇒ oY1 ∩ oY2 = Y∩(o1 ∩ o2), o1 ∩ o2 ∈ τ

⇒ oY1 ∩ oY2 ∈ τY

∴ (Y, τY) is a Topological Space

# Bases of **Subspace** **Topology**

Consider the collection BY for **Subspace** **Topology** τY:

BY = {Y∩b| b ∈ B, Y ⊆ X}

Let’s prove that BY is a **Bases** of **Subspace** **Topology** τY:

**Proof-**

Ba1.

∀x ∈ Y ⊆ X

⇒ ∃b ∈ B s.t. x ∈ b ⊆ X **[By Ba1]**

⇒ ∃Y∩b ∈ BY s.t. x ∈ Y∩b ⊆ Y

Ba2.

bY1, bY2 ∈ BY and x ∈ bY1 ∩ bY2

⇒ bY1=Y∩b1, bY2=Y∩b2, x ∈ Y∩b1 ∩ Y∩b2

⇒ ∃b3 ∈ B s.t. x ∈ b3 ⊆ b1∩b2, x ∈ Y **[By Ba2]**

⇒ ∃bY3 ∈ BY s.t. x ∈ bY3 where bY3 = Y∩b3 ⊆ Y ∩ b1∩b2

⇒ ∃bY3 ∈ BY s.t. x ∈ bY3 ⊆ Y ∩ b1∩Y∩b2

⇒ ∃bY3 ∈ BY s.t. x ∈ bY3 ⊆ bY1∩ bY2

Take closed interval [0, 1] as an example, it is the subspace of R. The bases will be [0, 1] ∩ (a, b) = {[0,b), (a,1], (a,b), ø, [0,1]}.

# Sub-bases of Subspace Topology

Define the collection SY for **Subspace Topology** τY with **Sub-bases** S for a topology on X:

SY = {s’| s’=Y∩s, s ∈ S} ⇒

∪SY = ∪{s’|s’ ∈ SY} = ∪{Y∩s|s ∈ S} = Y∩(∪S) = Y∩X = Y

Define a collection BY’ for **Subspace** **Topology** τY:

BY’ = {b| b = ∩s’[i], s’[i] ∈ SY} ⇒ SY ⊂ BY’

Let’s prove that BY’ is a Bases of **Subspace** **Topology** τY:

**Proof-**

Ba1:

Y = ∪SY

⇒ ∀x ∈ Y, ∃s’ ∈ SY s.t. x ∈ s’ ⊆ Y

⇒ ∀x ∈ Y, ∃s’ ∈ BY’ s.t. x ∈ s’ ⊆ Y **[ By SY **⊂** BY’ ]**

Ba2:

b1, b2 ∈ BY’ and x ∈ b1 ∩ b2

⇒ b1=∩s’[i], b2=∩s’[j], x ∈ ∩s’[i] ∩ ∩s’[j]

⇒ ∃b3 ∈ BY’ s.t. x ∈ b3 where b3 = b1'∩b2' = ∩s’[i] ∩ ∩s’[j] = ∩s’[k]

Thus, SY is a **Sub-bases** of **Subspace** **Topology** τY.

Take closed interval [0, 1] again as an example, it is the subspace of R. The sub-bases will be [0, 1] ∩ (-∞,b) = [0, b) and [0, 1] ∩ (a,∞)= (a,1] when 0 < a< b < 1.

# Conclusion

In this post, we have studied the basic idea of the Topological Space, Open Set, Subspace Topology, Bases and Sub-Bases.

In the next article, I will show you how to use the language of General Topology to define continuous function on a more general space instead of Euclidean space.

# Your feedback is highly appreciated and will help me to continue my articles.

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Reference:

“https://en.wikipedia.org/wiki/Topological_space”

“https://en.wikipedia.org/wiki/Open_set”

“https://en.wikipedia.org/wiki/Base_(topology)”

“https://en.wikipedia.org/wiki/Subbase”

“https://en.wikipedia.org/wiki/Subspace_topology”

“Topology (2nd edition) by James Munkres”