Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.