## Franklin's Notes

### Finite topological space

A finite topological space is exactly what it sounds like - topological space $(X,\mathcal T)$ in which the set of points $X$ is finite. The study of finite topological spaces is interesting from a combinatorial/algebraic perspective, as well as a good source of counterexamples. Here's a list of some "nice" facts about finite topological spaces, which are proven in detail below.

• A finite topological space always has an irreducible basis.

• Finite topological spaces correspond to finite distributive lattices in a convenient way.

• Connectedness of a finite topological space corresponds to primality of its lattice.

• Finite topological spaces can have nontrivial homotopy, see e.g. the pseudocircle .

Proposition 1. Every finite topological space has an irreducible basis .

Proof. Let $(X,\mathcal T)$ be a finite topological space. For each point $x\in X$, we may construct a minimal open set containing $x$ by intersecting all of the open sets of $\mathcal T$ that contain $x$ (since there are only finitely many of them, making their intersection open as well). Let us denote this minimal open set by $B_x$.

The collection of minimal open sets $B_x$ comprises an irreducible basis for the topology $\mathcal T$. Firstly, we know that it is a basis for $\mathcal T$ because every open set $O\in\mathcal T$ can be written in the form since $B_x\subset O$ for each $x\in O$. Finally, we know that this basis is irreducible because if $B_x$ were a union of other basis elements, then one of these other basis elements $B_y$ would have to contain $x$, but this would imply that $B_x$ was a subset of that basis element, and therefore either $B_x=B_y$ or the union of basis elements is strictly larger than $B_x$, which is a contradiction in either case. $\blacksquare$

Proposition 2. For every finite topological space, the set of closed sets comprises another topology on the same set, called the cotopology. If the original topology is denoted $\mathcal T$, the cotopology is usually denoted $\mathcal T^\ast$.

Proof. In a finite topological space, there is no distinction between arbitrary unions/intersections and finite unions/intersections. Thus, since the open sets are closed under arbitrary unions, their complements (the closed sets) are closed under arbitrary intersections; and since the open sets are closed under finite i.e. arbitrary intersections, the closed sets are closed under arbitrary unions. $\blacksquare$

There is a nice relationship between finite topological spaces and finite distributive lattices :

Proposition 3. If $L$ is the lattice of open sets of a topological space $(X,\mathcal T)$, then it is distributive. Conversely, if $L$ is a finite distributive lattice, then it is the lattice of open sets of some finite topological space.

Proof. DeMorgan's laws are sufficient to prove that the lattice of open sets of any topological space (not necessarily finite) is distributive:

Now, on to the converse, which is quite a bit more difficult. We will show that given a distributive lattice $L$, we may express $L$ as being isomorphic to a topology on $W\subset L$, the set of join-irreducible elements of $L$, with the open sets being the sets of the form $W_\ell$, i.e. the lower sets of all join-irreducible elements beneath some element $\ell\in L$ of the lattice, as described here .

Let $\ell_1,\ell_2\in L$ and let $W_1,W_2$ be the sets of join-irreducible elements of $L$ beneath $\ell_1$ and $\ell_2$ respectively. As we have seen previously, the join of the elements of $W_1$ is equal to $\ell_1$, and similarly for $W_2$ and $\ell_2$. That is, Consequently, we have that and $W_1\cup W_2$ is downward closed as well. Additionally, we have that by the distributivity of $L$. Notice that although $w\land w'$ may not be join-irreducible, it can always be expressed as a join of the elements of a downward-closed set of join-irreducible elements of $L$, each of which will be $\leq$ both $w$ and $w'$, meaning that they will be in both $W_1\cap W_2$, or in $W_1\cap W_2$. Thus $\ell_1\land \ell_2$ can be expressed as a join of elements in both $W_1$ and $W_2$, meaning that However, each element $w''\in W_1\cap W_2$ has $(w'',w'')\in W_1\times W_2$, meaning that Hence, we have that Notice that $W_1\cap W_2$ is downward closed as well.

Hence, if we let $\mathcal W$ be the set of lower sets of join-irreducible elements $\subset W$, we may define a function $f:\mathcal W\to L$ sending each such set to the join of its elements in $L$, and the above argument shows that and additionally, because each $\ell\in L$ has a representation as the join of all join-irreducible elements beneath it, we have that $f$ is surjective, and by Birkhoff's Theorem , $f$ is injective. Hence, $f$ establishes an isomorphism between the lattice of sets $\mathcal W$ and the lattice $L$. Additionally, $(W,\mathcal W)$ comprises a topological space, and we have that $L$ is the lattice of open sets of this space, proving the theorem. $\blacksquare$

This correspondence between finite distributive lattices and finite topological spaces gives us a very convenient way of describing connected finite topological spaces in terms of their lattices of open sets.

Proposition 4. A finite topological space is connected iff its lattice of open sets is prime .

Proof. We will prove the equivalent assertion that a finite topological space is disconnected iff its lattice of open sets is composite.

If a finite topological space $(X,\mathcal T)$ is disconnected, then it can be written in the form $X = X_1\cup X_2$, where $X_1$ and $X_2$ are open. Further, each open set $O\in\mathcal T$ can be written uniquely in the form $O=O_1\cup O_2$ where $O_1\subset X_1$ and $O_2\subset X_2$ are open. This means that the lattice $L$ of open sets of $X$ is isomorphic to the lattice $L_1\times L_2$ where $L_1$ is the lattice of open sets of the subspace $X_1\subset X$ and $L_2$ is the lattice of open sets of the subspace $X_2\subset X$.

Conversely, suppose that $L\cong L_1\times L_2$, and that $L$ is the lattice of open sets of $(X,\mathcal T)$. We have that $X$ corresponds to the element $(1,1)$ of the lattice $L_1\times L_2$, and we also have $(1,1)=(1,0)\lor (0,1)$, meaning that $X=X_1\cup X_2$ where $X_1$ is the open set corresponding to $(1,0)$ and $X_2$ is the open set corresponding to $(0,1)$. This means that $X$ is disconnected by definition. $\blacksquare$