1
vote
1answer
62 views

Extending a Filter in a Well-Ordered Boolean Algebra to an Ultrafilter WITHOUT the Axiom of Choice

Hypothesis: Let $B$ be a well-ordered boolean algebra and let $F \subseteq B$ be a filter on $B$. Goal: Show that $F$ can be extended to an ultrafilter without the axiom of choice (or any equivalent ...
10
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1answer
204 views

Stone's Representation Theorem and The Compactness Theorem

If you're working on $\mathsf {ZF}$ and you assume the compactness theorem for propositional logic, then you have the prime ideal theorem, and thus you can show that the dual of the category of ...
1
vote
1answer
166 views

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 ...
6
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1answer
263 views

Simple functions and axiom of choice

The question I have is more of a curiosity, and that is why I decided to post here instead of Mathoverflow. Before posing the question, let me set up some background. Background: Let $\Omega$ be a ...