In the previous lectures, a number of regularity principles for sets of real numbers emerged:
- (PS) the perfect subset property,
- (LM) Lebesgue measurability,
- (BP) the Baire property.
We have seen that the Borel sets in have all these properties. In this lecture we will show how to construct counterexamples for each of these principles. The proofs make essential use of the Axiom of Choice.
One of the most famous applications of is Vitali’s construction of a non-Lebesgue measurable set.
Proof
Put
It is straightforward to check that this is an equivalence relation on . Using a choice function on the equivalence classes of intersected with the unit interval , we pick from each equivalence class a representative from , and collect them in a set .
If we let, for ,
then
Suppose is measurable. Then so is each , and .
If , then , which is impossible. On the other hand, if , then, by countable additivity,
contradiction.
Next, we use the Well-ordering Principle () to construct a set such neither nor contains a perfect subset. Such sets are called Bernstein sets.
Proof
Let be the set of perfect subsets of . We can well-order this set, say
Note that there are -many open subsets of (as every open set is a countable union of open intervals with rational endpoints) and therefore -many closed subsets of . Since every perfect set is closed, there are at most -many perfect subsets of , and it is not hard to see that there are exactly -many. Hence we can choose a well-ordering of of length .
Furthermore, we assume each is well-ordered.
Pick from . Assume for we have chosen and so that
we can choose to be the first two elements of . This is possible since a perfect subset of has cardinality , and .
Put
Neither nor has a perfect subset by construction, and since , is a Bernstein set.
Proof
Assume for a contradiction a Bernstein set has the Baire property. By an exercise in the previous chapter, we can write , where is meager and is .
At least one of , is not meager. Wlog assume is not meager. (If not, obtain the representation “meager ” above for and proceed analogously.) Then must be non-meager, too, and hence is an uncountable set. By Theorem 1, is Polish and hence must contain a perfect subset, contradiction.
As mentioned before, the existence of arbitrary choice functions appears to be a rather strong assumption. However, the weaker versions of introduced in the section The Axiom of Choice do not suffice to construct counterexamples as above. Solovay (1970) constructed (though under a large cardinal assumption) a model of in which every set of real numbers is Lebesgue measurable, has the Baire property, and has the perfect subset property.
- Solovay, R. (1970). A model of set-theory in which every set of reals is Lebesgue measurable. The Annals of Mathematics.