By Choudhary P.
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2aX = 4 Therefore 11(V) cannot cover the segment [(1 +a/3)X, (1 + 3a/4)X] of length (5/12)aX, so there is a to in that segment not belonging to 11(V). The line 2' of slope 2E through (to, 0) must, from the figure, cut G, say at a point (Y, n(Y) - n(X)), with (1 + a/3)X S Y < (1 + a)X. Since 2', passing through that point, does not touch any part of V (to 18 IX B Converse to Polya's gap theorem; general case being 0 11(V) ), we have n(Y) - n(t) 5 2e(Y - t) for X < t S Y. Calling Y/X = 1 + /3, we have the lemma.
By the same token, there is an x2, x1 5 x2 n(x2) - n(x1) > e (x2 - xl) (so in particular n(x2) >, n(x1) + 1 ). The process continues, yielding x3, x2 S x3 5 (1 + a)x2, x4, and so forth, with n(xk+1)-n(xk) > 2(xk+1-xk), as long as the number Xk already obtained is < (1 + a)M-'R. Since n(xk+1) n(xk) >, 1, Xk cannot remain 5 (1 + a)M'R indefinitely (we must eventually have n(xk) > n((1 +)M-'R)) ). Let x, be the last xk - which is ,< (1 + a)M -' R; then we can still get an x, + 1 between (1 + a)M -' R and (1 + a)MR, such that n(x,+ 1) - n(xi) > 2 (xi+ 1 - xi)- Adding to this the corresponding inequalities already obtained, we get n(x,+ 1) - n(R) > 2-(x,+,-R).
Since (1 + a)M = 1 + c, (1 + c)R, so x, + 1 n((1 + c)R) 3 n(x,+1). And x,+1-R >, (l+a)M-'R-R = +C)(M-1)IM (1 -1 cR. ), we have (1 +c)(M-1)/M - 1 c > 1 2 This would make the left-hand side of the previous relation > e/4, in contradiction with our choice of the number R. For such large M, then, a number X with the properties specified above must exist. This establishes our claim, and proves the lemma. 2 E not measurable; start of Fuchs' construction 17 Lemma. Given E > 0, let a > 0 and X = Xi be as in the statement of the previous lemma.
Abstract algebra by Choudhary P.