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.
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| Show finite complement topology is a topology. |
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.