Uniqueness of Topology on Finite Set

Question

Prove that if $X$ is a finite set with preorder $R$, then exactly one topology $\mathcal{T}$ on $X$ satisfies $\trianglelefteq_{\mathcal{T}} =R$.

> Here is the definition the mentioned preorder: > $\trianglelefteq_{\mathcal{T}} = \{(a,b) \in X^2 : a \in \overline{\{b\}}\}$.

At first, I wanted to show that there is a base for this topology as follows:

For each $x \in X$, let $U_x$ be the intersection of all open sets of $X$ which contain $x$. Since $X$ is finite, this is a finite intersection; thus $U_x$ is open. Let $\mathcal{U}$ be the collection of all $U_x$. Then $\mathcal{U}$ is a base for $\mathcal{T}$.

Now that I have a base for the topology, this how I am thinking of the problem now:

Since $\mathcal{U}$ is a base for $\mathcal{T}$, then $\mathcal{U} \subset \mathcal{T}$ and $\forall p \in U \subset \mathcal{T}, \exists V \in \mathcal{U}$ such that $p \in V \subset U$. I believe this to be the definition of a base for a topology.

From here, is it possible to show that two topologies are equal? For example, suppose that there were at least two topologies $\mathcal{T}$ and $\mathcal{S}$ on $X$ that satisfy $\trianglelefteq_{\mathcal{T}}=R$ and $\trianglelefteq_{\mathcal{S}}=R$, repectively. If we define $\trianglelefteq_{\mathcal{S}}=\{(a',b')\in X^2 : a' \in\overline{\{b'\}}\}$, then how do we accomplish a proof of $a = a'$ since this should imply uniqueness?

Thank you very much in advance for your time in reading this post, any assistance will be much appreciated.

Answer 1

Your basic approach of looking at the base $\{U_x:x\in X\}$ is fine. The next step is to prove that for any $x,y\in X$, $x\trianglelefteq_{\tau}y$ if and only if $y\in U_x$, which is pretty straightforward.

Given a pre-order $\trianglelefteq$ on $X$, for each $x\in X$ define $U_x=\{y\in X:x\trianglelefteq y\}$. Use the properties of $\trianglelefteq$ to verify that $\{U_x:x\in X\}$ is a base for a topology $\tau$ on $X$ such that $U_x=\bigcap\{V\in\tau:x\in V\}$ for each $x\in X$; the result of the first paragraph then shows that $\trianglelefteq_{\tau}=\trianglelefteq$, and this shows that $\tau$ is uniquely determined by the pre-order.

I’ve left you quite a bit to check here; give it a try, and if you get stuck, leave a question for me, and I’ll expand it a bit.

Answer 2

The sets $U_x$ and the definition of topological base are not so useful here. All you have to show is that if $T$ and $S$ are any two topologies on $X$ such that $\trianglelefteq_T\;=\; \trianglelefteq_S\;=R$, then $T = S$. It will not even matter that both $\trianglelefteq_T$ and $\trianglelefteq_S$ are equal to $R$. All we'll need is that they are equal to each other. Finiteness of $X$ will be sufficient; have a look at the duality between Alexandroff spaces and preordered sets.

For this proof, the most helpful way to view a topology $T$ is via its closure operator, $$\text{cl}_T:2^X\to 2^X \\ A\mapsto \bigcap\{F\subseteq X: F\text{ is closed w.r.t. $T$ and } F\supseteq A\}.$$ Closure operators define topologies and vice versa; if we can show that $\text{cl}_T = \text{cl}_S$, it will follow that $T = S$. To show that $\text{cl}_T = \text{cl}_S$, we can test them against every subset $A$ of $X$. The most important step here is that in the following (almost-tautological) identity $$A = \bigcup_{a\in A}\{a\},$$ the union is finite. And closure operators distribute over finite unions! In particular, if $\text{cl}$ is any closure operator and if $A$ is finite, then $$\text{cl}(A) = \text{cl}\left(\bigcup_{a\in A}\{a\}\right) = \bigcup_{a\in A}\text{cl}(\{a\}).$$ So if we can show that $\text{cl}_T$ and $\text{cl}_S$ agree on singletons, we'll be done.

Now, your preorder $\trianglelefteq_T$, called the specialization preorder, is defined in terms of $\text{cl}_T$: $$\trianglelefteq_T\; = \{(a, b)\in X\times X : a \in \text{cl}_T(\{b\})\}.$$ This means that $a \trianglelefteq_T b \iff (a, b) \in\;\trianglelefteq_T\,\iff a \in \text{cl}_T(\{b\})$. This is true for any topology $T$ on $X$. So if $T$ and $S$ are two topologies on $X$ with identical specialization preorders $\trianglelefteq_T$ and $\trianglelefteq_S$, then, for any $b$ in $X$, $$a \trianglelefteq_T b \iff a \trianglelefteq_S b$$ which is to say $$a \in \text{cl}_T(\{b\}) \iff a \in \text{cl}_S(\{b\})$$ i.e. $$\text{cl}_T(\{b\}) = \text{cl}_S(\{b\}).$$ Since $b$ was arbitrary, we're done!