By H. G. Dales, F.K. Dashiell Jr., A.T.-M. Lau, D. Strauss

This e-book offers a coherent account of the speculation of Banach areas and Banach lattices, utilizing the areas C_0(K) of continuing services on a in the neighborhood compact house okay because the major instance. The learn of C_0(K) has been a huge zone of useful research for a few years. It provides a number of new buildings, a few concerning Boolean earrings, of this area in addition to many effects at the Stonean house of Boolean earrings. The ebook additionally discusses while Banach areas of constant services are twin areas and after they are bidual spaces.

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Each Borel subset B of a Polish space is the image of a zerodimensional Polish space; in the case where B is uncountable, B contains a subset that is homeomorphic to Δ , and so |B| = c. An analytic space is a continuous image of a Polish space. Every Borel subset of an analytic space is analytic, and it follows from Souslin’s separation theorem that a subspace Y of an analytic space X is Borel if and only if both Y and X \ Y are analytic. Let X be an uncountable Polish space. Then there are analytic subsets of X that are not Borel.

First, ϕ ◦ ι = θ , and so θ ◦ β ◦ ι = ϕ ◦ ι = θ . Now assume that β ◦ ι is not the identity map on F. 4, there is a proper closed subspace C of F such that F = C ∪ (β ◦ ι )−1 (C). But then θ (F) = θ (C) ∪ (θ ◦ (β ◦ ι )−1 )(C) = θ (C) ∪ (θ ◦ (β ◦ ι )(β ◦ ι )−1 )(C) = θ (C) , a contradiction of the irreducibility of θ . Thus β ◦ ι is the identity map on F, and so β : β Kd → F is a retraction. 3, (b) ⇒ (c), F is projective. Set GK = F and πK = θ . Then the pair (GK , πK ) has the required properties.

Proof. Denote by A the set of atoms of F, and set A0 = {a ∈ A : 0 < a ∧ b < a}. For each a ∈ A0 , choose ca ∈ C with 0 < ca < f (a); this is possible because f (a) > 0 and C is atomless. The set of atoms of the subalgebra, say Fb , of B generated by F ∪ {b} is {a ∧ b : a ∈ A0 } ∪ {a − b : a ∈ A0 } ∪ (A \ A0) . We define f : Fb → C by first setting fb (a ∧ b) = ca and fb (a − b) = f (a) − ca for a ∈ A0 . Note that a1 ∧ a2 = 0 and f (a1 ) ∧ f (a2 ) = 0 whenever a1 , a2 ∈ A0 with a1 = a2 ; this implies that fb is well defined and injective on the atoms of Fb .