Is It Possible to Construct Non-Measurable Sets Without the Axiom of Choice?

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Discussion Overview

The discussion revolves around the possibility of constructing non-measurable sets without relying on the Axiom of Choice (AoC). Participants explore the implications of the Axiom of Choice on the existence of non-measurable sets within the context of set theory and measure theory.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant recalls Vitali's proof, noting its dependence on the Axiom of Choice for establishing non-measurable sets.
  • Another participant asserts that without the Axiom of Choice, it is possible that all sets could be measurable, referencing ongoing investigations into such theories.
  • A participant questions whether the use of AoC limits the construction of non-measurable sets to merely existential claims, suggesting the possibility of special cases where a choice function could be explicitly constructed.
  • A later reply challenges this notion, stating that constructing a choice function for non-measurable sets is not possible, as all operations involving non-measurable sets require the Axiom of Choice.
  • One participant reflects on the implications of AoC guaranteeing the existence of a choice function without explicitly defining it, leading to confusion about the necessity of AoC for constructing non-measurable sets.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of the Axiom of Choice for constructing non-measurable sets. While some argue that AoC is essential, others explore the possibility of special cases where it might not be required, indicating an unresolved debate.

Contextual Notes

The discussion highlights the complexity of the relationship between the Axiom of Choice and the construction of non-measurable sets, with participants acknowledging the potential for differing interpretations and implications based on foundational assumptions in set theory.

Kreizhn
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"Concrete" non-measurable sets

I've had Vitali's proof of the existence of non-(Lebesgue) measurable sets branded into the side of my brain over the years. However, the proof always critically relies on evoking the axiom of choice. Has anybody every demonstrated a non-AoC construction of a non-measurable set? Or do the intuitionist logicians just avoid measure theory all together?
 
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Hi Kreizhn! :smile:

The axiom of choice is necessary for the existence of non-measurable sets. If you don't adopt the axiom of choice (or a similar principle), then it can happen that all sets are measurable! There is a current investigation of some theories in which all sets are measurable, so it certainly can happen!
 


Good to know. Then does using AoC necessarily preclude the ability to construct non-measurable sets that are anything but "existential?"

For example, (unless it's just been so long since I've done set theory) I imagine there are times when we must use the AoC in general, but for special cases we could explicitly construct a choice function. Are there special cases wherein this can be done for non-measurable sets? Maybe that's just a silly question.
 


Kreizhn said:
Good to know. Then does using AoC necessarily preclude the ability to construct non-measurable sets that are anything but "existential?"

For example, (unless it's just been so long since I've done set theory) I imagine there are times when we must use the AoC in general, but for special cases we could explicitly construct a choice function. Are there special cases wherein this can be done for non-measurable sets? Maybe that's just a silly question.

I don't quite understand what you're asking. Do you want to construct a choice function for a non-measurable set?? This is not possible, everything you do with non-measurable sets involves the axiom of choice. It is not possible to construct a choice function for them.
 


I guess what I'm thinking is that AoC always guarantees the existence of a choice function, without explicitly defining it. However, there are instances in which AoC is not required to give the choice function.

But then I guess the answer is that if we must always use AoC, then it is impossible to explicitly construct a choice function, since then AoC would not be necessary to construct nonmeasurable sets, and you have just told me that it is.
 

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