It is shown that if there exists a supercompact cardinal then every set of reals, which is an element of L(R), is the projection of a weakly homogeneous tree. As a consequence of this theorem and recent work of Martin and Steel [Martin, D. A. & Steel, J. R. (1988) Proc. Nall. Acad. Sci. USA 85, 6582-65861, it follows that (if there is a supercompact cardinal) every set of reals in L(R) is determined.The subtle relationships between the existence of certain large cardinals and regularity properties of various simple sets of reals is one of the striking developments in modem set theory.One of the first results in this direction is that of R. I show here that the existence of a supercompact cardinal implies that every set of reals that belongs to L(R) has a certain structural representation from which the regularity results, such as measurability, easily follow. This is made more precise through a sequence of definitions.For our purposes the set of reals, R, is the set 0d of all functions f:o -t) , wherew = {0, 1, . . ., k, . . .}is the set of nonnegative integers. We let I'm denote the set of all finite sequences of elements of w and for s E i< 'let N, be the set, Ns = {f 8 a': f [ 1(s) = s}, where 1(s) = length(s). The set {N, : s E A<"} generates a topology on of it is the product topology derived from the discrete topology on 0. Endowed with this topology a'xis homeomorphic to the Euclidean space of irrationals. Suppose X is a set. We denote by X 'the set of all functions f: t-s Xand we denote byX<' the set of all finite sequences of elements ofX. We adopt the usual convention that X"' is the set of all functions f: dom f xsuch that dom f Ecowand ifs E X" then dom s = i(s) = length(s). Suppose A is an ordinal, A > 0. A tree on o x A is a subset T C a' x AX"' ' such that for all pairs (s, t) E T, Suppose X is a nonempty set. We denote by m(X) the set of countably complete ultrafilters on the Boolean -algebra P(X). , is a measure on X if ,u E m(X-). For A E m(X) and A C X we write ,u(A) = 1 to indicate A E Au. Suppose that X = Y<' and that ,t E m(Y< '). Since A is countably additive, there is a unique k E w such that IL(yk) = 1. Suppose .ul, y, E m(Y< '), ,ut(Y') = 1, and ,2(Ykl) = 1. Then A < 2 (u2 projects to ,ul) if k1 < k2 and, for all A C yki, p 1(A) = 1 if and only if 1,2(A*) = 1 where A* = {s E Yk: s t ki E A}. such that for all k E w, (x rk, yrk) E domirand (ir(xrk, y [ k) : k E a) is a countably complete tower.
I(s) = I(t) and (s t i, t [ i) E Tfor all i < I(s), i
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