By R.F. Dickman, P. Fletcher
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We select such a half and call it Zk. This completes the inductive proof of the existence of the contracting sequence. By the completeness of R (see Section 5), there is a point x such that 2 E z k for all k. Since x E Z and C covers I , there is an open set U of the open covering C such that x E U . Hence there is a - 0 . 0 , 40 number r II EXISTENCE THEOREMS IN DIMENSION 1 > 0 such that N ( x , r, I ) C U. Now the intervals 10, 11,* * * , Ik, * * * contain x and have decreasing lengths (b - a ) , (b - a ) / 2 , **=, (b - a)/2k, **a.
PI NEIGHBORHOODS, CONTINUITY 21 can nevertheless be obtained with a bit of algebra by considering various similar triangles (see Exercise 9 below). Geometrically, we can obtain a 6 thus: Let t be a point of intersection of S and the sphere with center f x a n d radius E. T h e cross-section on the plane through the three points z, f x , and t is shown in Fig. 4. L e t 6 be the perpendicular distance from x to the line zt. Then each point of the interior of the sphere N ( x , 6) projects into N ( f x , E, S).
Let c be the 0 . - 44 EXISTENCE THEOREMS IN DIMENSION 1 I;r midpoint of Ik-1 = [ak-l, bk-11. If t is an upper bound of X, we take Ik = Cak-1, c], and if t is not an upper bound, we set Ik = [c, bk-11. In either case, Ik has the required properties. By the completeness of R (see Section S), there is a number M such that M E Ik for every k = 0, 1, 2, We shall show fist that M belongs to X. Suppose the contrary were true. Since X is a closed set, X is open; then there would be an r > 0 such the complement R that N ( M , I ) C R X.