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+ 2022-04-19 How the Axiom of Choice Gives Sizeless Sets

Apr 24, 202212 min read

And additionally, that a single point has a measure of zero: Therefore and by Sigma Additivity, a set like the integers also has a measure of 0

Then, although it is difficult, we can construct a non-measurable set.

Proof

Let and be equivalent if their difference is a rational number with

This spans the Equivalence Relation That is, if is a rational number , then they go into the same “bin” (i.e. the fall within the same Equivalance Class). If their difference is irrational they go into different bins.

Properties of the rational numbers

let always rational always rational always irrational always irrational rational if contains the irrational part of (for example, , irrational else.

Therefore

  1. one Equivalance Class contains all rational numbers, because a rational number minus another rational number is a rational number and by definition of equivalence classes, equivalence classes are unique and can be represented by any representative. In this case, every rational number is a valid representative.
  2. for every irrational number there is a single separate bin. That is, because the difference of two irrational numbers is only rational, if their decimal-part is equal, e.g., . This would still be the bin for . Since , the number is in a different Equivalance Class. Although this seems lile it limits the amount pf irrational numbers, this is not the case. The amount of irrationals is still uncountably infinite. Actuallyit is the very same irrational number, just shofzed by a rational therefore the set of irrationals is not reduced

By we see that every such Equivalance Class is a shifted copy of all rational numbers in the interval:

Let be the set of all such equivalence classes

By the Axiom of Choice, we construct a new set , where contains a single random element from each of the equivalence classes in .

Let be elements of this set: , therefore, they are also representatives of the respective equivalence class they were chosen from

CLAIM: The new set is non-measurable. PROOF:

First, we list the rational numbers in the interval : This is possible, because is countably infinite.

Based on these rational numbers we construct copies of the set .

Each such copy is constructed by the rule with

For example

  1. Since the ‘s are shifted versions of and consists of a single rational and an irrational (and themselves disjoint) equivalence classes, every the ‘s are disjoint: .
  2. Since and and there is a single equivalence class representing the rational numbers and one equivalence class for every irrational number between , the interval from is a subset of the union of ‘s:

Because the ‘s are disjoint, the rule of Sigma Additivity gets applied and its measure We chose and . Therefore, and the union of ‘s and the measure must be less than 3 since and greater than 1 because .

If has a measure, than each of the ‘s must be of the same measure, because they are just shifted copies and by translational invariance, shifting a set preserves its size.

Since there are countably infinite rational numbers in the interval

if if

Since the measure of the union must be between 1 and 3 we conclude that is non-measurable.

The Axiom of Choice states that

Definition

Given a (possibly) infinite collection of nonempty sets, we can form a new set that contains one (random) element from each set.

We do not know, which element was chosen.

Link to original

In this proof we made use of it when we constructed ‘s. Without the axiom of choice, it is impossible to create a non-measurable set

Todo:https://youtu.be/SJ8YoV6YZFA ab 40min

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