Chapter 1: Topology, Hocking and Young: A theorem due G.D. Birkhoff

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 \{ O_{\alpha}\} and \{ R_{\beta} \} be two collections of subsets of a set S, both satisfying the three defining axioms O_{1}, O_{2} and O_{3} of a topology. Let S have two topologies. We will say that the topology \mathcal{T}_{1} determined by \{ O_{\alpha} \} is a finer topology than the topology \mathcal{T}_{2} determined by \{ R_{\beta} \} if every set R_{\beta} is a union of sets O_{\alpha}, that is, each R_{\beta} is open in the \mathcal{T}_{1} topology. We will denote this situation with the symbol \mathcal{T}_{1} \geq \mathcal{T}_{2}.

Axiom 1: reflexive clearly holds because \mathcal{T}_{1} \geq \mathcal{T}_{1} as set inclusion/containment is reflexive.

Axiom 2: antisymmetric also holds true because clearly \mathcal{T}_{1} \geq \mathcal{T}_{2} and \mathcal{T}_{2} \geq \mathcal{T}_{1} together imply that \mathcal{T}_{2} = \mathcal{T}_{1}. 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 \mathcal{T}_{1} \geq \mathcal{T}_{2} and \mathcal{T}_{2} \geq \mathcal{T}_{3}, then clearly \mathcal{T}_{1} \geq \mathcal{T}_{2} \geq \mathcal{T}_{3}. 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

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