Topology theorem solution

Table of Contents

• Theorem statement
  • Proof
  • Example
  • solution

Theorem statement

Let X be a non-empty set.

τ= {φ, X, A⊆X such that A^C = X−A is finite} prove that τ is a topology on X.

Proof

• φ ∈ τ

           X^C=X−X =φ finite

           ⇒X ∈ τ

• Let A₁, A₂, A₃, —, A_n ∈τ

We want to show that A₁∩ A₂∩ A₃∩ —∩ A_n ∈τ

A₁, A₂, A₃, —, A_n ∈τ   

A₁^c, A₂^c, A₃^c, —, A_n^c are finite

As the union of a finite number of a finite set is finite.

A₁^c∪ A₂^c∪ A₃^c∪ —∪ A_n^c are finite

              By De Morgan’s law                                                    (A∩B)^C=A^C∪B^C                                      

         (A₁∩ A₂∩ A₃∩ —∩ A_n) ^c is finite

         ⇒ A₁∩ A₂∩ A₃∩ —∩ A_n ∈τ

The second condition for topology is satisfied

• Let A₁, A₂, A₃, — ∈τ    

⇒ A₁^c, A₂^c, A₃^c, — are finite

As the intersection of a finite set is finite.

A₁^c∩ A₂^c∩ A₃^c∩ — are finite

              By De Morgan’s law                                                     (A∪B)^C=A^C∩B^C

         (A₁∪ A₂∪ A₃∪ —) ^c is finite

         ⇒ A₁∪ A₂∪ A₃∪ — ∈τ

The third condition for topology is satisfied

Hence τ is a topology on X.

Example

Let X be a non-empty set.

τ= {φ, X, A⊆X such that A^C = X−A is finite} prove that τ is a topology on X.

solution

• X, φ ∈τ

This shows that X and φ belong to τ.

So the first condition for topology is satisfied

This shows that the union of any number of numbers of τ belongs to τ.

The second condition for topology is satisfied.

This shows that the intersection of any number of numbers of τ belongs to τ.

The third condition for topology is satisfied.

This shows that τ is the topology on X.