# 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.