Honors Theses

Representing Generalized Boolean Algebras in Riesz Spaces

2011

Mathematics

Gerard Buskes

Relational Format

Dissertation/Thesis

Abstract

Proof. Suppose (x^) is an increasing sequence with G B for all n G A/” which has a supremum x G B. Since A preserves order, A(x„) is increasing and A(x„) < A(x) for all n G Af. By virtue of the fact that J is cr-complete and A(x„) is bounded above, the sequence A(x,i) has a supremum K(y) for some y G: B. Claim: A(x) ^ A(^). Assume that A(x) ^ A(y). Since A(y) is the supremum of A(x„), K[y) < A(x). It also follows that x ^ y because A is injective. We claim that y is an upper bound for (xn)- If not then there exists an A G AA such that x^ Ay ^ xj^. By the injectivity of A, it follows that 29 \{xy A y) = A(xiv) A A(y) A(xjv). It follows that A(y) is not an upper bound for all A(x„) which is a contradiction of our assumption that A(y) is the supremum of A(x„). Thus y is an upper bound for (xn). It follows that x < y as x is the supremum of (xn). However, the latter implies A(x) < A(y) which contradicts our assumption that A(y) < A(x). Therefore A(x) = A(y).

Searchable text

COinS