Generalized free products of boolean algebras with an amalgamated subalgebra

Dwinger, Ph.; Yaqub, Fawzi M.

doi:10.1016/s1385-7258(63)50024-9

Cited by 14 publications

(6 citation statements)

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“…Now suppose X has been assigned some xed wellordering W. Then de ned by x y , x = y or the W-least member of x y lies in y (where is`symmetric di erence') is a linear extension of . 2 We remark that by contrast, verifying OE in our model, where in particular we cannot appeal even to BPIT, is quite involved.…”

Section: Preliminariesmentioning

confidence: 76%

“…o-transitively on X, so if R 6 = ;, we either have (x; y) 2 R where x < y, in which case R = < jX, or (x; y) 2 R where x > y, giving R = > jX. 2 Now we shall consider increasing n-tuples of members of U. Let X f(u 1 ; u 2 ; : : :; u n ) 2 U n : u 1 < u 2 < : : : < u n g. A relation R on X is called a lexicographic ordering if for some ordering fi 1 ; i 2 ; : : :; i n g of f1; 2; : : :; ng and < j 2 f<; >g:…”

Section: Veri Cation Of Oe In Mostowski's Modelmentioning

confidence: 99%

“…To say that B is the free product of subalgebras Bi and B2, written B = B] * B2, means that B = (Bi U B2) and for any homomorphisms 6 t : B, -> C of Bi, B2 to a boolean algebra C there is a unique extension of 6\ U 8 2 to a homomorphism of B to C. This is equivalent to sayingthatB = (Bi UB2) and for any b\ e Bj -{ 0 } and b 2 € B 2 -{0}, b\ n b 2 ^ 0. It is shown in [2] that for any Bi and B 2 , Bi * B 2 exists and is uniquely determined up to isomorphism. If Bi = ^(Xi) and B2 = &>…”

mentioning

confidence: 99%

“…We also say that B is the relatively free product of its subalgebras B] and B2, and that B = (B!,B 2 ) is relatively free. Again a construction of Bi * A B2 where A is a common subalgebra of Bi and B2 is given in [2], LEMMA 2.2. Let B be a finite boolean algebra, B], B2 subalgebras ofB such that B = (Bi U B2), and A = Bi n B2.…”

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confidence: 99%

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The independence of the Prime Ideal Theorem from the Order-Extension Principle

Felgner

Truss

1999

J. symb. log.

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It is shown that the boolean prime ideal theorem BPIT: every boolean algebra has a prime ideal, does not follow from the order-extension principle OE: every partial ordering can be extended to a linear ordering. The proof uses a Fraenkel–Mostowski model, where the family of atoms is indexed by a countable universal-homogeneous boolean algebra whose boolean partial ordering has a ‘generic’ extension to a linear ordering. To illustrate the technique for proving that the order-extension principle holds in the model we also study Mostowski's ordered model, and give a direct verification of OE there. The key technical point needed to verify OE in each case is the existence of a support structure.

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Section: Preliminariesmentioning

confidence: 76%

Section: Veri Cation Of Oe In Mostowski's Modelmentioning

confidence: 99%

mentioning

confidence: 99%

mentioning

confidence: 99%

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The independence of the Prime Ideal Theorem from the Order-Extension Principle

Felgner

Truss

1999

J. symb. log.

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“…For a -0 the algebras to be considered are just Boolean algebras so the conclusion follows from Dwinger and Yaqub [3], For a -1 notice first that we can clearly amalgamate the simple algebras of each of the classes. The amalgamation property for the classes now follows by the same argument as used in Theorem 2.7 of Daigneault [2] for the class of all locally-finite polyadic algebras of infinite dimension.…”

Section: Let X = C (Am{m]) (A R Bj ΐ[ Imentioning

confidence: 94%