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On the strength of the Sikorski extension theorem for Boolean algebras

Published online by Cambridge University Press:  12 March 2014

J.L. Bell
J.L. Bell*
Affiliation:
The London School of Economics and Political Science, London WC2, England
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Extract

The Sikorski Extension Theorem [6] states that, for any Boolean algebra A and any complete Boolean algebra B, any homomorphism of a subalgebra of A into B can be extended to the whole of A. That is,

Inj: Any complete Boolean algebra is injective (in the category of Boolean algebras).

The proof of Inj uses the axiom of choice (AC); thus the implication AC → Inj can be proved in Zermelo-Fraenkel set theory (ZF). On the other hand, the Boolean prime ideal theorem

BPI: Every Boolean algebra contains a prime ideal (or, equivalently, an ultrafilter)

may be equivalently stated as:

The two element Boolean algebra 2 is injective,

and so the implication Inj → BPI can be proved in ZF.

In [3], Luxemburg surmises that this last implication cannot be reversed in ZF. It is the main purpose of this paper to show that this surmise is correct. We shall do this by showing that Inj implies that BPI holds in every Boolean extension of the universe of sets, and then invoking a recent result of Monro [5] to the effect that BPI does not yield this conclusion.

Type
Research Article
Information
The Journal of Symbolic Logic , Volume 48 , Issue 3 , September 1983 , pp. 841 - 846
Copyright
Copyright © Association for Symbolic Logic 1983

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References

REFERENCES

[1]Bell, J.L., Boolean-valued models and Independence proofs in set theory, Oxford Logic Guides, Oxford University Press, Oxford, 1977.Google Scholar
[2]Felgner, Ulrich, Models of ZF-set theory, Lecture Notes in Mathematics, vol. 223, Springer-Verlag, Berlin, 1971.CrossRefGoogle Scholar
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