# Problem 1.30 (The Maximum Principle is equivalent to the axiom of choice) (J.Bell, Boolean-valued Models and Independence Proofs, 3rd edition)

Problem 1.30 (The Maximum Principle is equivalent to the axiom of choice)

(i) Let $\{a_i : i ∈ I\} ⊆ B$ satisfy $\bigvee_{i∈I} a_i = 1$. A partition of unity $\{b_i : i ∈ I\}$ in B is called a disjoint refinement of $\{a_i : i ∈ I\}$ if $∀i ∈ I[b_i ≤ a_i]$. Define $u ∈ V^{(B)}$ by $dom(u) = \{\hat{i} : i ∈ I \}, u(\hat{i}) = a_i$ for i ∈ I. Let R be the set of disjoint refinements of $\{a_i : i ∈ I\}$ and $U = \{v ∈ V^{(B)} : [[ v ∈ u ]]= 1\}$. Show that the map ${b_i : i ∈ I} → \Sigma_{i∈I} b_i ·\hat{i}$ from R to U is one–one and ‘onto’ U in the sense that, for any v ∈ U there is a unique $\{b_i : i ∈ I\} ∈ R$ such that $[[ \Sigma_{i∈I} b_i ·\hat{i} = v]]$= 1.

(ii) Let $Σ_B$ be the assertion $$∀u ∈ V^{(B)}( [[u \neq Ø ]]= 1 → ∃v ∈ V^{(B)}([[v ∈ u ]]= 1))$$ (every nonempty B-valued set has an element) and $Π_B$ the assertion: ‘for any set, I, every I-indexed family of elements of B with join 1 has a disjoint refinement’. Show without using the axiom of choice that $Σ_B$ and $Π_B$ are equivalent. (Use (i).) Deduce that the assertions ‘$Σ_B$ holds for every complete Boolean algebra B’, and ‘the Maximum Principle holds in $V^{(B)}$ for every complete Boolean algebra B’ are each equivalent to the axiom of choice. (Confine attention to the case in which B is of the form PX for an arbitrary set X.)

(The Maximum Principle)

If φ(x) is any B-formula, then there is $u∈V^{(B)}$ such that $$[[∃x.φ(x)]]=[[φ(u)]]$$. In particular, if $V^{(B)}⊨∃x.φ(x)$, then $V^{(B)}⊨φ(u)$ for some $u∈V^{(B)}$.

I have some problem just with the last part of the second point, could you help me?

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Isn't the third ediio simply called "Set Theory"? –  Michael Greinecker Dec 8 '12 at 21:24

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