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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?

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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.

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    $\begingroup$ So in fact, it's a finite topological sum of indiscrete topologies. $\endgroup$ – Nate Eldredge Jun 2 '10 at 14:36
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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 $\mathbb{N}$.

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  • $\begingroup$ 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. $\endgroup$ – Pietro Majer Jun 2 '10 at 20:25
  • $\begingroup$ Don’t you want your definition of $s{\mathcal R}t$ to have a biconditional, as in $s \in A \iff t \in A$? $\endgroup$ – LSpice Dec 17 '18 at 3:28
  • $\begingroup$ 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 $\endgroup$ – Pietro Majer Dec 17 '18 at 10:35

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