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Hence s, E Z, n A, = Z and s, = 0 for k 2 k,. 2). Since A, is principal, we can find two relatively prime elements P(X) and Q(X) of A,[X] such that Q(0) = 1 andz,,, rkXk= P(X)/Q(X). Since Irklm+ 0 as k + KI, the rational fraction P(X)/Q(X) does not have pole in the closed disc lzlm < 1. Since 1tk1,< 1, the series tkXkisconvergent if XEQ,, 1x1, < 1; therefore l/O is the unique root of Q in 52, whose absolute value is less than one. Hence 6 is a Pisot number. (Q, (AOk))2 < KI. We put ilOk = (, r, and rk= r, - Or,-, = OE,-, - E, (k 2 1).

Rn (a depends only on Y , , .. , Yn and 8 ) . Therefore D is relatively dense. To get the cor,verse result, the following lemma is needed: + SO + which gives Lemma 24. Returning to Proposition 10, let D,,for each a in Q,, be the set of all Since D is relatively dense, d = ( x , Y ) in Q , x Rn, such that 1x1, < D,is relatively dense in aZ, x Rnprovided lalpissufficientlylarge. The projection L of Daon Rn is a discrete and relatively dense subgroup of Rn; therefore L = Z Y , + + ZY,. The only element common to D and Q , x (0) is 0 since their intersection would be a discrete subgroup of Q,.

For each harmonious and relatively dense subset M of Q,, there exists a model A and afinite subset F of Q , such that M is contained in A F. Proof: Theorem IV and Proposition 10 give Theorem VII. + Since xZ is dense in the ball of radius Jx(,,we can find a rational integer m such that ly - mxl, < I ;this implies1f ( y ) - mf(x)lp < 1. a. group G with the following properties: (a) G is not compact; (b) G is torsion-free; (c) each harmonious subset A of G is finite. Proof: Let p be a prime, p Z 2, and let Z, be the compact group of p-adic integers.

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