Fréchet ideal and finitely additive measures

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Discussion Overview

The discussion revolves around the properties of finitely additive measures on natural numbers, specifically addressing the question of why there is no finitely additive measure whose null sets form the Fréchet ideal. The conversation includes references to set theory and examples of families of subsets with specific properties.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant states that for every finitely additive measure on natural numbers, all null sets form an ideal.
  • Another participant clarifies the definition of a finitely additive measure and questions whether the Fréchet ideal consists of all finite subsets of natural numbers.
  • A third participant references a theorem from Kunen's book, presenting a contradiction that arises when attempting to define a finitely additive measure whose null sets form the Fréchet ideal.
  • One participant proposes specific families of subsets that exhibit the properties discussed, seeking confirmation of their correctness.
  • Another participant agrees with the proposed families and suggests modifications to eliminate the need for the axiom of choice in one of the examples.

Areas of Agreement / Disagreement

Participants generally agree on the definitions and properties of finitely additive measures, but the discussion remains unresolved regarding the existence of a finitely additive measure whose null sets form the Fréchet ideal. Multiple viewpoints and examples are presented without consensus.

Contextual Notes

The discussion includes references to specific mathematical properties and theorems, but does not resolve the implications of these properties on the existence of certain measures. There is also mention of the axiom of choice and its relevance to the examples provided.

liwi
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Hi!
For every finitely additive measure \eta on natural numbers, all \eta-null sets obviously form an ideal.
Why there is no finitely additive measure on natural numbers whose null sets form the Fréchet ideal?
Thanks, liwi
 
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Just to be clear, you define \eta to be a finitely additive measure iff \eta : P(\mathbb{N}) \to \mathbb{R} is defined for every subset of \mathbb{N} and satisfies:

\eta (\emptyset ) = 0, \eta (\mathbb{N}) = 1

X \subset Y \Rightarrow \eta (X) \leq \eta (Y)

\forall n \in \mathbb{N}\ \eta (\{ n\} ) = 0

X_1, \dots , X_n\ \mbox{pairwise disjoint} \Rightarrow \eta (\cup _{i=1}^n X_i) = \Sigma _{i=1} ^n \eta (X_i)

And the Fréchet ideal is the one consisting of all finite subsets of \mathbb{N}?
 
Last edited:
See Kunen's "Set Theory: An Introduction to Independence Proofs." By I Theorem 1.3 of that text, there is a family \mathcal{A} of 2^{\omega} subsets of N such that for all X, Y in \mathcal{A}, |X \cap Y| < \omega. Define \mathcal{B}_n = \{ X \in \mathcal{A} : \eta (X) > 1/n\}. By a simple counting argument, there exists n such that \mathcal{B} _n is infinite. Fix such n, and take X_1, \dots , X_n \in \mathcal{B} _n. Then it's easy to see that:

1 = \eta (\mathbb{N} ) \geq \eta (\cup X_i) = \Sigma \eta (X_i) > \Sigma 1/n = n(1/n) = 1

1 > 1, contradiction.
 
Thanks a lot!
I don't have Kunen's book, but for example \mathcal{A} = \{ \{f \restriction n; n \in \omega \}; f \in \, ^{\omega}2 \} or \mathcal{A} = \{ s(i); s(i) chosen sequence of rational numbers converging to i, i irrational \} is an uncountable family of subsets of a countable set, having the property you described. Is that correct?
thanks again,
liwi
 
Yes, both are correct. Your first A is perhaps preferable because it doesn't requiring invoking choice, however the second one can be modified to eliminate choice. Just explicitly specify what the sequence s(i) should be for given i, for instance you could specify that it be the decimal expansion approximation sequence, e.g.

s(\pi ) = \{ 3, 3.1, 3.14, 3.141, \dots \}
 

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