# Topology theorem solution

#### Theorem statement

Let X be a non-empty set.

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

##### Proof
• φ ∈ τ

XC=X−X =φ finite

⇒X ∈ τ

• Let A1, A2, A3, —, An ∈τ

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

A1, A2, A3, —, An ∈τ

A1c, A2c, A3c, —, Anc are finite

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

A1c∪ A2c∪ A3c∪ —∪ Anc are finite

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

(A1∩ A2∩ A3∩ —∩ An) c is finite

⇒ A1∩ A2∩ A3∩ —∩ An ∈τ

The second condition for topology is satisfied

• Let A1, A2, A3, — ∈τ

⇒ A1c, A2c, A3c, — are finite

As the intersection of a finite set is finite.

A1c∩ A2c∩ A3c∩ — are finite

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

(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 AC = 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.