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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
You haven't defined "The Maximum Principle in V^{(B)}$" in your question. – Asaf Karagila Dec 8 '12 at 21:30
is called "set theory, boolean-valued models and indipendence proofs". – Tonno Dec 8 '12 at 21:30
(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 $V^{(B)} \models ∃x.φ(x)$, then $V^{(B)} \models φ(u)$ for some $u ∈ V^{(B)}$. sorry I forgot to post before, – Tonno Dec 8 '12 at 21:38
You can (and should) edit this into the question. – Asaf Karagila Dec 8 '12 at 21:43
up vote 0 down vote accepted

First show that the Maximality Principle (MP) is implied by $\Pi_B$. So $\Sigma_B$ implies MP and the axiom of choice implies $\Sigma_B$. This is not a difficult exercise.

It is left to show that MP implies the axiom of choice. Assume that MP holds, then. And let $\{A_i\mid i\in I\}$ be a family of disjoint non-empty sets. We will show that MP implies that there is a choice function.

Consider $B=\mathcal P(I)$ the complete Boolean algebra. Let $\check x\in V^{(B)}$ denote the canonical name for $x\in V$. Let $\dot G$ denote the canonical name for the generic ultrafilter, namely $[[b\in\dot G]]=b$ for all $b\in B$. I will confuse between $i$ and $\{i\}$ which is an element in our algebra. Note that $\{i\in I\}$ is a maximal disjoint antichain in $B$.

$$\left[\left[\exists x.\sum_{i\in I}[[x\in\check A_i]]\cdot[[i\in\dot G]]\right]\right]=1$$

Therefore there is some $u$ such that $\sum_{i\in I} [[u\in A_i]] = 1$. Therefore there is $a_i\in A_i$ such that $[[u=a_i]]=i$. This is a choice function as wanted.

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