Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be the set of all partitions of $[0,1]$ such that each element of the partition is Lebesgue-measurable.

Let $Y$ be the set of all partitions of $[0,1]$ such that each element of the partition is a Borel set.

Is there a standard topology for a set like $X$ or $Y$? If so, is $X$ (or $Y$) compact in this topology?

I thought that the set of all partitions of $[0,1]$ (including those with Vitali sets as elements) is a more complicated object, but as @JDH points out in the proposed answer, this may not be the case.

The broader context for this question is my interest in convergence of measures on such a set (hence the focus on Borel or measurable partitions). If $Y$ is compact and metrizable, then the set of measures on $Y$ is itself a compact metric space in the weak topology, which opens the door to standard results on convergence of such measures.

share|improve this question
4  
Just a remark on the edit, not all non-measurable sets are Vitali sets. There are ultrafilters; coded graphs for discontinuous functionals; Sierpinski sets; etc. those are different animals than Vitali sets (and their existence follows from different choice principles too). –  Asaf Karagila Aug 8 '12 at 21:38
2  
are your partitions required to be finite? countable? arbitrary? –  J. Loreaux Aug 8 '12 at 22:12
    
@JDH - Thanks, that's a very clear and helpful answer. Any thoughts on whether the set of all partitions (or sets X and Y above) is compact in the lower-cone topology? –  exk Aug 9 '12 at 3:06
    
@AsafKaragila: The comments about Vitali sets was simply meant to illustrate one of the ways that the set of all partitions differs from the set of partitions whose elements are measurable; there was no claim that the Vitali sets are the only difference. Re: J.Loreaux's comment: Both $X$ and $Y$ are sets of all (thus arbitrary) partitions satisfying a given condition on the elements. –  exk Aug 9 '12 at 3:50
1  
Unfortunately, $X$ and $Y$ are not actually lattices, but rather lower semi-lattices (we only have meet $\wedge$ and not join $\vee$) and they are only $\sigma$-complete as lower semi-lattices. If you use the uniform formulation as I describe in the last paragraph of my answer, however, they those will result in lattices, but again only $\sigma$-complete and not complete. But the space of all partitions is a complete lattice and hence will be compact in this topology. –  JDH Aug 9 '12 at 10:55

1 Answer 1

up vote 8 down vote accepted

There are several natural topologies to put on the space of partitions, which can also be thought of as the space of equivalence relations.

First, the collection of partitions of a set is a partial order under the refinement relation, where one partition $\mathcal{A}$ refines another $\mathcal{B}$ when every element of $\mathcal{A}$ is a subset of a set in $\mathcal{B}$. What's more, the space of partitions for each of your spaces is a lower semi-lattice, since any two partitions $\mathcal{A}$ and $\mathcal{B}$ have a coarsest common refinement, the collection of nonempty $A\cap B$, where $A\in\mathcal{A}$ and $B\in\mathcal{B}$. Indeed, the cases of $X$ and $Y$ lead to a $\sigma$-complete lower semi-lattice, since one may similarly intersect countably many sets and remain Borel or Lebesgue measurable as required. Since every partial order has a natural topology, the lower-cone topology, where the open sets are simply those that are closed downwards with respect to the order, we may place this topology on the space of partitions. In your case, the open sets of partitions would be those that include all refinements of any partition in the set.

This topology is compact---in the cases of $X$ and $Y$ and also in the case of the set of all partitions---simply because there is a coarsest partition, the partition with only one component consisting of the whole set. The only open neighborhood of this partition in the lower-cone topology is the entire space of partitions, since every partition refines it.

Secondly, a different dual topology would arise from turning the order upside down, and for that, the open sets would be closed under encoarsening rather than refinement. This topology also is compact, since there is a finest topology, consisting of the collection of singletons, and every partition is coarser than it.

Note that the space of all partitions, despite your remark on its complexity, actually exhibits much nicer lattice-theoretic properties than the Borel partitions or the measurable partitions. This is because the space of all partitions is actually a complete lattice. This is because any family of partitions has a largest common refinement, obtained simply by intersecting the equivalence relations as sets of ordered pairs. Similarly, one may take the equivalence relation generated by any family of partitions to find the smallest common encoarsening.

Another natural topology arises on the class of equivalence relations by saying that a basic open set is determined by finitely many equivalences and non-equivalences. Thus, one specifies finitely many equivalencies and non-equivalencies on the points $x_i\sim x_j$, $y_s\not\sim y_t$, and then the basic open set is the collection of all partitions that respect those finitely many requirements. This is a common kind of topology to place on the set of models of a first-order theory, where the basic open sets specify finitely much information about the predicates of the model; the collection of partitions amounts to the collection of equivalence relations, which is first-order.

One can imagine more elaborate topologies in this line, which take more into account your context of Lebesgue measurable or Borel partitions, by moving beyond finitely many point requirements to the case of finitely many requirements on Borel sets, or on measurable sets.

It would of course depend on your purpose with the topology to know which is best for that purpose.

Finally, let me mention that the well-developed theory of Borel equivalence relations is deeply concerned with the space of Borel partitions of the reals, but not exactly in your sense. What is usually required in that theory is not merely that every element of the partition is Borel, that is, that every equivalence class is Borel, but rather one insists that the binary relation itself, thought of as a subset of the plane, should be Borel. This is a more uniform kind of Borel, and the theory is extremely robust and active.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.