# Why is a topology easily described by its subbasis, but a $\sigma$-algebra cannot be expressed by its generating set?

Let $$X$$ be a set and $$\mathscr{A} \subseteq 2^X$$ a collection of subsets. Describing the topology $$\tau(\mathscr{A})$$ generated by $$\mathscr{A}$$ is easy: any open set in $$\tau(\mathscr{A})$$ is just an arbitrary union of finite intersections of elements in $$\mathscr{A}$$.

However, the $$\sigma$$-algebra $$\sigma(\mathscr{A})$$ generated by $$\mathscr{A}$$ is a completely different story. One can never hope to express $$\sigma(\mathscr{A})$$ explicitly in terms of $$\mathscr{A}$$, even if $$\mathscr{A}$$ is an algebra of sets. In fact, let's say $$\mathscr{A}$$ is an algebra of sets. Let $$\kappa(\mathscr{A})$$ be the collection of sets obtained from taking complements, countable unions and countable intersections of members of $$\mathscr{A}$$ countably many times. Then $$\kappa(\mathscr{A})$$ is still a very small subcollection compared to $$\sigma(\mathscr{A})$$, nowhere approaching all of $$\sigma(\mathscr{A})$$. The collection $$\sigma(\mathscr{A})$$ is simply too huge to have any explicit expression.

Why is it so much harder to describe a $$\sigma$$-algebra than a topology, even though the definitions of both look simple and similar?

• See this question and this question. – J.-E. Pin Oct 25 '19 at 4:44
• Definition of a topology does not involve cardinalities. Arbitrary union of open sets are open. Defintion of a sigma algebra involves countability in a crucial way. – Kavi Rama Murthy Oct 25 '19 at 5:41
• The sigma algebra (or topology) generated by a collection K of subsets, is the intersection of all sigma algebra (topologies) that contain the subset K. – William Elliot Oct 25 '19 at 5:51

There is an "explicit" description (depending on what you call explicit), by something resembling the Borel hierarchy :

Rename your generating set $$\Sigma_0$$, and the set of complements of those $$\Pi_0$$.

Then define by transfinite induction $$\Sigma_{\alpha+1} =$$ the set of countable unions of elements of $$\Pi_\alpha$$ and $$\Pi_{\alpha+1}=$$ the set of countable intersections of elements of $$\Sigma_\alpha$$ (equivalently : the set of complements of elements of $$\Sigma_{\alpha+1}$$), and at limit stages define both $$\Sigma_\alpha = \bigcup_{\beta < \alpha}\Sigma_\beta$$ and $$\Pi_\beta = \bigcup_{\beta < \alpha}\Pi_\beta$$

Then one may check that for each element in the generated $$\sigma$$-algebra, there is an ordinal $$\alpha$$ such that it belongs to $$\Sigma_\alpha$$. This sounds awful because ordinals go very far up, but in fact one may do better than that : it actually stops at $$\omega_1$$, that is : every element of the generated $$\sigma$$-algebra belongs to $$\Sigma_\alpha$$ for some countable ordinal $$\alpha$$; or in other words, the generated $$\sigma$$-algebra is precisely $$\Sigma_{\omega_1}$$ (this follows from the fact that $$\omega_1$$ is regular, i.e. -here- it is not a countable union of countable subsets)

In general, one cannot hope to do better than that : for instance if you start with the open subsets of $$\mathbb R$$ as $$\Sigma_0$$, there is no lower stage where you have all the generated $$\sigma$$-algebra (the Borelian $$\sigma$$-algebra)

Now this is "explicit", but of course more complex than the situation for topology. So what's happening ?

Well you're adding various new complex phenomena : first you're adding complements (that's not too big a deal, because they behave nicely with respect to intersections and unions : just swap them), but most importantly you're allowing for infinite intersections, with some constraints. To my mind, that's what it all comes down to.

Indeed although $$(\bigcup_{i\in I}U_i)\cap (\bigcup_{j\in J}V_j)$$ is easily described : $$\bigcup_{(i,j)\in I\times J}U_i\cap V_j$$ (which is a countable union, if both $$I,J$$ are), so that finite intersections of families of unions are easy to understand; $$\bigcap_{\alpha\in A} \bigcup_{i\in I_\alpha}U_i$$ is not so easily described : it's (assuming the axiom of choice) $$\bigcup_{f: A\to \bigcup_{\alpha} I_\alpha \mid \forall \alpha, f(\alpha) \in I_\alpha} \bigcap_{\alpha \in A} U_{f(\alpha)}$$, where the first union is not necessarily countable anymore, even if $$A$$ and all $$I_\alpha$$'s were : you're getting away from the constraint you had imposed.

So in the case of topology, all problems of non-distributivity (you can't write a union of intersections as an intersection of unions) were taken care of in one step : "oh just take finite intersections, then just all unions of that", here you have a second issue which comes from the mixing of infinite intersections with the requirement for countability. So you can't reduce to something simpler in the same way : the operations you would want to do get you far from countability