# Finite Topology vs sigma Field

Suppose we have a finite $$\sigma$$ -field $$S$$, of which $$A$$ and $$B$$ are member sets. Since $$S$$ is closed under union and complementation [by definition], it follows that $$(A' \cup B')' = (A \cap B)' \in S$$. From closure under complementation, we have that $$A \cap B \in S$$, implying that $$S$$ is closed under intersections.

Does it follow that every finite $$\sigma$$ -field is a topology?

Yes, it is also a topology on its union, the largest member of $S$. Since $S$ is finite, the arbitrary union rquirement amounts to finite union, which you have.

In fact,$S$ is a Boolean algebra, and since it is finite, it is isomorphic to a powerset algebra---the power set of the atoms of $S$ (the minimal non-empty elements of $S$). Every set in $S$ is a union of a finite number of these atoms.

If these atoms are singletons, then $S$ will be the discrete topology on the underlying set. If not, however, then $S$ will clearly not separate those points, and so will not be Hausdorff and so on.

• So in fact, it's a finite topological sum of indiscrete topologies. Jun 2 '10 at 14:36

It is worth remarking that the analogous characterization of σ-algebras also holds in the case of countable underlying sets:

Any σ-algebra $$\mathcal{A}$$ on a countable set $$S$$ is atomic.

That is, it is generated by a partition (the classes being the "atoms"). The corresponding equivalence relation is

$$s\mathcal{R}t\ \Longleftrightarrow\ (\ \forall A\in\mathcal{A}\ (s\in A \Longleftrightarrow t\in A)\ ).$$

(In other words, $$s$$ and $$t$$ are equivalent precisely if they are not separated by sets $$A \in \mathcal{A}$$.) As a consequence, any element of $$\mathcal{A}$$ writes uniquely as union of atoms, making $$\mathcal{A}$$ isomorphic to the power set $$\mathcal{P}(S/\mathcal{R})$$ (in particular, of course, $$\mathcal{A}$$ is also a topology on $$S$$).

It may not be obvious that the class (or atom) $$[s]$$ of an element $$s\in S$$ in the equivalence relation $$\mathcal{R}$$ actually belongs to $$\mathcal{A}$$, for it writes as an a priori non countable intersection: $$[s]:=\bigcap_{s\in A\in \mathcal{A}} A$$ But one can also write it as a countable intersection $$[s]:=\bigcap_{t\in S} A_{s,t} ,$$ where { $$A_{s,t}$$ }$$_{(s,t)\in S\times S}$$ is a collection of elements of $$\mathcal{A}$$ chosen so that for any $$(s,t)$$ one has

$$A_{s,t}= S\$$ if $$\ s\mathcal{R}t,$$

$$s\in A_{s,t}\$$ and $$\ t\notin A_{s,t}$$ otherwise.

The above characterization has some foundational relevance in Probability: dealing with a discrete random variable $$X:\Omega\to E$$ (or a finite family of them) if we please we may assume with no loss of generality that the base probability space $$\Omega$$ is $$\mathbb{N}$$.

• Note that to define the family of sets A<sub>s,t</sub> I needed a countable choice; I suppose it's unavoidable in order to prove the statement. Jun 2 '10 at 20:25
• Don’t you want your definition of $s{\mathcal R}t$ to have a biconditional, as in $s \in A \iff t \in A$? Dec 17 '18 at 3:28
• Yes, of course, but $\mathcal{R}$ was already assumed to be an equivalence relation, in particular symmetric. But maybe the double implication is more clear. Done. Thank you Dec 17 '18 at 10:35