Prove that the collection of all topologies on a given set S constitutes a lattice under the partial ordering of finer/coarser topology of a set.

Proof: NB. This theorem has already been proven by eminent mathematician G. D. Birkhoff.

PS: I am making my own attempt here. Your comments are most welcome !

Part 1: TST: Coarser/finer is a partial ordering of topologies.

To prove axiom 1: this order is reflexive.

Let and be two collections of subsets of a set S, both satisfying the three defining axioms , and of a topology. Let S have two topologies. We will say that the topology determined by is a finer topology than the topology determined by if every set is a union of sets , that is, each is open in the topology. We will denote this situation with the symbol .

Axiom 1: reflexive clearly holds because as set inclusion/containment is reflexive.

Axiom 2: antisymmetric also holds true because clearly and together imply that . In other words, these two topologies are equivalent, or they give rise to same basis.

Axiom 3: Transitivity holds because set inclusion/containment is transitive. If and , then clearly . In other words, there can be a chain of finer/coarser topologies for a given set.

Part 2:

TST: Under this definition of partial ordering of topologies, the partial ordering forms a lattice.

From axiom 3 proof, we know that there can exist a chain of finer/coarser topologies. So supremum and infimum can exist in a partial ordering of topologies. Hence, such a partial ordering forms a lattice.

QED.

Kindly let me know your comments and oblige about my above attempt.

Regards,

Nalin Pithwa