Boolean prime ideal theorem and the axiom of choice

Question

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.

score 6 · Answer 1 · answered Oct 08 '13 at 21:30

6

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.

answered Oct 08 '13 at 21:30

Brian M. Scott

616,228

Well, that's exactly a kind of example I was looking for, thank you! – dtldarek Oct 08 '13 at 21:37
@dtldarek: You’re welcome! – Brian M. Scott Oct 08 '13 at 21:39

score 6 · Accepted Answer · edited May 07 '23 at 21:36

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

The completeness theorem for first-order logic.
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:

Every set can be linearly ordered.
Every infinite set has a non-trivial ultrafilter.
If $V$ is a vector space, and $V$ has a basis $B$ then every basis of $V$ has the same cardinality as $B$.
Marshall Hall's marriage theorem.
Every partial order can be extended to a linear order.
Hahn-Banach theorem.
Every field has an algebraic closure, which is unique up to isomorphism.
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:

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

And many many others.

In the second paragraph I think you meant $\aleph_1<2^{\aleph_0}$. — Camilo Arosemena-Serrato, Oct 08 '13 at 23:21
@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 08 '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 09 '13 at 07:01

Carl Mummert · Answer 3 · 2019-04-23T00:49:16.833

2

There is a book, Equivalents of the Axiom of Choice (two volumes) by Rubin and Rubin that has an exhaustive catalog of principles equivalent to AC. The book Consequences of the Axiom of Choice by Howard and Rubin has many more principles that are weaker than AC. There is also a book The Axiom of Choice by Jech that discuses the prime ideal theorem.

edited Apr 23 '19 at 00:49

answered Oct 08 '13 at 21:40

Carl Mummert

81,604

You mean this (and the second volume here) and this? – dtldarek Oct 08 '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 08 '13 at 22:43

score -1 · Answer 4 · answered Jan 14 '14 at 22:26

-1

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.

answered Jan 14 '14 at 22:26

Robert Cowen

1

Boolean prime ideal theorem and the axiom of choice

4 Answers4

Linked