# Proof;

## To Show finite complement topology is a topology. OR Let X be a non empty set and denote the collection of subsets of X whose complements are finite, together with empty set. Then T is topology on X. OR Let T = (X, τ) be a finite complement space. Then τ is a topology on T.

Show finite complement topology is a topology. |

**Related Theorems;**

Show that the intersection of two topologies on X is a topology.

If A is a subset of a topological space, then ∂(A)⊆Cl(A).

We have to show that T satisfied the definition of Topology on X.

1) The Union of any number of members of T belong to T.

Let A1, A2, ..., An,... ∈ T

⇒ A’1, A’2, A’3, ... , A’n,...

All are finite sets.

⇒ A’1 ∩ A’2 ∩ A’3 ,..., ∩ A’n ∩ ,... is a finite set.

By DeMorgan’s Law

A’1 ∩ A’2 ∩ A’3 ,..., ∩ A’n ∩ ,... = ( A1 ∪ A2 ∪ A3 ,..., ∪ An ∪ ,... )’

⇒ A1 ∪ A2 ∪ A3 ,..., An ∪ ,.. ∈ T

2) The intersection of any finite number of members of T belong to T.

Let A1, A2, ..., An ∈ T

⇒ A’1, A’2, A’3, ... , A’n

All are finite sets.

⇒ A’1 ∪ A’2 ∪ A’3 ∪ ,..., ∪ A’n is also a finite set.

By DeMorgan’s Law

A’1 ∪ A’2 ∪ A’3 ∪ ,..., ∪ A’n = ( A1 ∩ A2 ∩ A3 ∩ ,..., ∩ An )'

⇒ A1 ∩ A2 ∩ A3 ∩ ,..., ∩ An ∈ T

3) The empty set and X belong to T.

By Definition , Φ ∈ T

As we know that X’ = Φ is a finite set

⇒ X ∈ T

#### Hence proved that finite complement topology is, in fact, a topology.