PureMath.pdf

# Suppose that we are given a set of intervals in a

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Suppose that we are given a set of intervals in a straight line, that is to say an aggregate each of whose members is an interval [ α, β ]. We make no restriction as to the nature of these intervals; they may be finite or infinite in number; they may or may not overlap; * and any number of them may be included in others. It is worth while in passing to give a few examples of sets of intervals to which we shall have occasion to return later. (i) If the interval [0 , 1] is divided into n equal parts then the n intervals thus formed define a finite set of non-overlapping intervals which just cover up the line. (ii) We take every point ξ of the interval [0 , 1], and associate with ξ the interval [ ξ - δ, ξ + δ ], where δ is a positive number less than 1, except that with 0 we associate [0 , δ ] and with 1 we associate [1 - δ, 1], and in general we reject any * The word overlap is used in its obvious sense: two intervals overlap if they have points in common which are not end points of either. Thus [0 , 2 3 ] and [ 1 3 , 1] overlap. A pair of intervals such as [0 , 1 2 ] and [ 1 2 , 1] may be said to abut .

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[V : 105] LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE 224 part of any interval which projects outside the interval [0 , 1]. We thus define an infinite set of intervals, and it is obvious that many of them overlap with one another. (iii) We take the rational points p/q of the interval [0 , 1], and associate with p/q the interval p q - δ q 3 , p q + δ q 3 , where δ is positive and less than 1. We regard 0 as 0 / 1 and 1 as 1 / 1: in these two cases we reject the part of the interval which lies outside [0 , 1]. We obtain thus an infinite set of intervals, which plainly overlap with one another, since there are an infinity of rational points, other than p/q , in the interval associated with p/q . The Heine-Borel Theorem. Suppose that we are given an inter- val [ a, b ] , and a set of intervals I each of whose members is included in [ a, b ] . Suppose further that I possesses the following properties: (i) every point of [ a, b ] , other than a and b , lies inside * at least one interval of I ; (ii) a is the left-hand end point, and b the right-hand end point, of at least one interval of I . Then it is possible to choose a finite number of intervals from the set I which form a set of intervals possessing the properties (i) and (ii) . We know that a is the left-hand end point of at least one interval of I , say [ a, a 1 ]. We know also that a 1 lies inside at least one interval of I , say [ a 0 1 , a 2 ]. Similarly a 2 lies inside an interval [ a 0 2 , a 3 ] of I . It is plain that this argument may be repeated indefinitely, unless after a finite number of steps a n coincides with b . If a n does coincide with b after a finite number of steps then there is nothing further to prove, for we have obtained a finite set of intervals, selected from the intervals of I , and possessing the properties required. If a n never coincides with b , then the points a 1 , a 2 , a 3 , . . . must (since each lies to the right of its predecessor) tend to a limiting position, but this limiting position may, so far as we can tell, lie anywhere in [ a, b ].
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