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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 R\Real 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 AC\AC is Vitali’s construction of a non-Lebesgue measurable set.

Proof

Put

xy   if and only if   xyQ.x \sim y \; \text{ if and only if }\; x-y \in \Rat.

It is straightforward to check that this is an equivalence relation on R\Real. Using a choice function on the equivalence classes of \sim intersected with the unit interval [0,1][0,1], we pick from each equivalence class a representative from [0,1][0,1], and collect them in a set SS.

If we let, for rQr \in \Rat,

Sr={s+r ⁣:sS},S_r = \{s+r \colon s \in S \},

then

SrSt for rt.S_r \cap S_t \quad \text{ for $r \neq t$}.

Suppose SS is measurable. Then so is each SrS_r, and λ(Sr)=λ(S)\lambda(S_r) = \lambda(S).

If λ(S)=0\lambda(S) = 0, then λ(R)=0\lambda(\Real) = 0, which is impossible. On the other hand, if λ(S)>0\lambda(S) > 0, then, by countable additivity,

2=λ([0,2])λ(rQ[0,1]Sr)=rQ[0,1]λ(S)=,2 = \lambda([0,2]) \geq \lambda\left(\bigcup_{r\in \Rat\cap[0,1]} S_r\right) = \sum_{r\in \Rat\cap[0,1]} \lambda(S) = \infty,

contradiction.

Next, we use the Well-ordering Principle (WO\WO) to construct a set BRB\subseteq \Real such neither BB nor RB\Real\setminus B contains a perfect subset. Such sets are called Bernstein sets.

Proof

Let P\mathcal{P} be the set of perfect subsets of R\Real. We can well-order this set, say

P={Pξ ⁣:ξ<20}.\mathcal{P} = \{P_\xi \colon \xi < 2^{\aleph_0} \}.

Note that there are 202^{\aleph_0}-many open subsets of R\R (as every open set is a countable union of open intervals with rational endpoints) and therefore 202^{\aleph_0}-many closed subsets of R\R. Since every perfect set is closed, there are at most 202^{\aleph_0}-many perfect subsets of R\Real, and it is not hard to see that there are exactly 202^{\aleph_0}-many. Hence we can choose a well-ordering of P\mathcal{P} of length 202^{\aleph_0}.

Furthermore, we assume each PξP_\xi is well-ordered.

Pick a0b0a_0 \neq b_0 from P0P_0. Assume for ξ<20\xi < 2^{\aleph_0} we have chosen {aβ ⁣:β<ξ}\{a_\beta\colon \beta < \xi \} and {bβ ⁣:β<ξ}\{b_\beta\colon \beta < \xi \} so that

β<ξ    aβ,bβPβ and all aβ,bγ pairwise distinct,\forall \beta < \xi \; \; a_\beta, b_\beta \in P_\beta \quad \text{ and } \quad \text{all $a_\beta, b_\gamma$ pairwise distinct},

we can choose aξ,bξPξa_\xi, b_\xi \in P_\xi to be the first two elements of Pξγ<ξ{aγ,bγ}P_\xi \setminus \bigcup_{\gamma < \xi} \{a_\gamma, b_\gamma\}. This is possible since a perfect subset of R\Real has cardinality 202^{\aleph_0}, and ξ<20\xi< 2^{\aleph_0}.

Put

A={aξ ⁣:ξ<20}B={bξ ⁣:ξ<20}.A = \{a_\xi \colon \xi < 2^{\aleph_0} \} \qquad B = \{b_\xi \colon \xi < 2^{\aleph_0} \}.

Neither AA nor BB has a perfect subset by construction, and since ARBA \subseteq \Real\setminus B, BB is a Bernstein set.

Proof

Assume for a contradiction a Bernstein set BB has the Baire property. By an exercise in the previous chapter, we can write B=MGB = M \cup G, where MM is meager and GG is GδG_\delta.

At least one of BB, RB\Real\setminus B is not meager. Wlog assume BB is not meager. (If not, obtain the representation “meager \cup GδG_\delta” above for RB\Real\setminus B and proceed analogously.) Then GBG \subseteq B must be non-meager, too, and hence is an uncountable GδG_\delta set. By Theorem 1, GG 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 AC\AC 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 ZF+DC\ZF+\DC in which every set of real numbers is Lebesgue measurable, has the Baire property, and has the perfect subset property.

References
  1. Solovay, R. (1970). A model of set-theory in which every set of reals is Lebesgue measurable. The Annals of Mathematics.