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The Boolean prime ideal theorem is strictly stronger than ZF, and strictly weaker than ZFC. I'm looking for nice examples (like the existence of non-measurable set) that request at least that theorem (ZF+BPI), but weaker than AC (ZFC).

The context is that I'm trying to gain some intuition, but most literature I could get access to is very technical and being lost in the details I can't see the big picture. Any help (like examples or your intuition behind it) or references that provide examples would be greatly appreciated.

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4 Answers 4

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The Boolean Prime Ideal theorem has a lot of useful equivalents. Two important ones are:

  1. The completeness theorem for first-order logic.
  2. The compactness theorem for first-order logic.

What you wrote in your question, however, is not fully accurate. The existence of a non-measurable subset does not require "at least" the Boolean Prime Ideal theorem. It is in fact much much weaker than that; and is implied by weaker principles (e.g. Hahn-Banach theorem) as well very different principles (e.g. $\aleph_1\leq2^{\aleph_0}+\sf DC$ implies the existence of a non-measurable set).

If you are looking for consequences of $\sf BPI$ which are unprovable from $\sf ZF$ itself then there are plenty. Here are a few:

  1. Every set can be linearly ordered.
  2. Every infinite set has a non-trivial ultrafilter.
  3. If $V$ is a vector space, and $V$ has a basis $B$ then every basis of $V$ has the same cardinality as $B$.
  4. Hall's marriage theorem.
  5. Every partial order can be extended to a linear order.
  6. Hahn-Banach theorem.
  7. Every field has an algebraic closure, which is unique up to isomorphism.
  8. Every family of finite non-empty sets admits a choice function.

And many many more.

Some of these examples you can find in the surprisingly not-very-technical book by Herrlich, The Axiom of Choice.

If you are looking for principles which are unprovable from $\sf BPI$, but true in $\sf ZFC$, then there are plenty of these as well:

  1. The axiom of countable choice.
  2. Every infinite set is Dedekind-infinite.
  3. More generally, $\sf DC_\kappa$, for any $\kappa$.
  4. The statement "For every infinite cardinal $\frak a$, $\frak a+a=a$".

And many many others.

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In the second paragraph I think you meant $\aleph_1<2^{\aleph_0}$. –  Camilo Arosemena Oct 8 '13 at 23:21
    
@Camilo: No. I didn't. I meant $\leq$. –  Asaf Karagila Oct 8 '13 at 23:22
1  
@Camilo: If $2^{\aleph_0}$ is an $\aleph$-number then it's extremely easy to prove the existence of non-measurable sets (e.g. Vitali sets, Bernstein sets, Hamel bases and their consequences, ultrafilters on $\omega$, and so on). –  Asaf Karagila Oct 8 '13 at 23:29
    
Even more examples I was looking for ;-) Indeed, I was searching for consequences of $\mathsf{BPI}$ which are unprovable in $\mathsf{ZF}$. Could you possibly add an example that is weaker than $ZFC$, but unprovable in $\mathsf{BPI}$ (if such is known)? –  dtldarek Oct 9 '13 at 7:01
    
@dtldarek: Of course. I've added a few. –  Asaf Karagila Oct 9 '13 at 7:22

An example from topology: $\mathsf{BPI}$ is equivalent to the Tikhonov product theorem for compact Hausdorff spaces, while $\mathsf{AC}$ is equivalent to the full Tikhonov product theorem.

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Well, that's exactly a kind of example I was looking for, thank you! –  dtldarek Oct 8 '13 at 21:37
    
@dtldarek: You’re welcome! –  Brian M. Scott Oct 8 '13 at 21:39

There is a book, Equivalents of the Axiom of Choice (two volumes) by Rubin and Rubin that has an exhaustive catalog of such things. Also there is a book The Axiom of Choice by Jech that discuses the prime ideal theorem.

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You mean this (and the second volume here) and this? –  dtldarek Oct 8 '13 at 21:48
    
@dtldarek: You may also be interested in Herrlich's The Axiom of Choice book which contains a lot of nice equivalence of $\sf BPI$. –  Asaf Karagila Oct 8 '13 at 22:43

A great reference for Consequences of the Axiom of Choice and their relations to each other is the book by Paul Howard and Jean Rubin, Consequences of the Axiom of Choice.

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While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. –  Lost1 Jan 14 at 22:57

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