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Fréchet ideal and finitely additive measures

  1. Mar 8, 2009 #1
    For every finitely additive measure [tex]\eta[/tex] on natural numbers, all [tex]\eta[/tex]-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
  2. jcsd
  3. Mar 9, 2009 #2


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    Just to be clear, you define [itex]\eta[/itex] to be a finitely additive measure iff [itex]\eta : P(\mathbb{N}) \to \mathbb{R}[/itex] is defined for every subset of [itex]\mathbb{N}[/itex] and satisfies:

    [tex]\eta (\emptyset ) = 0, \eta (\mathbb{N}) = 1[/tex]

    [tex]X \subset Y \Rightarrow \eta (X) \leq \eta (Y)[/tex]

    [tex]\forall n \in \mathbb{N}\ \eta (\{ n\} ) = 0[/tex]

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

    And the Fréchet ideal is the one consisting of all finite subsets of [itex]\mathbb{N}[/itex]?
    Last edited: Mar 9, 2009
  4. Mar 9, 2009 #3


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

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

    1 > 1, contradiction.
  5. Mar 10, 2009 #4
    Thanks a lot!!
    I don't have Kunen's book, but for example [itex]\mathcal{A} = \{ \{f \restriction n; n \in \omega \}; f \in \, ^{\omega}2 \} [/itex] or [itex]\mathcal{A} = \{ s(i); s(i) [/itex] chosen sequence of rational numbers converging to [itex] i, i [/itex] irrational [itex] \} [/itex] is an uncountable family of subsets of a countable set, having the property you described. Is that correct?
    thanks again,
  6. Mar 10, 2009 #5


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    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.

    [tex]s(\pi ) = \{ 3, 3.1, 3.14, 3.141, \dots \}[/tex]
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