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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?

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  • $\begingroup$ See this question and this question. $\endgroup$ – J.-E. Pin Oct 25 '19 at 4:44
  • $\begingroup$ See also this detailed answer. $\endgroup$ – J.-E. Pin Oct 25 '19 at 4:56
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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Oct 25 '19 at 5:41
  • $\begingroup$ 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. $\endgroup$ – William Elliot Oct 25 '19 at 5:51
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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

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