Course in Mathematical Analysis by Nikolsky S.M. PDF

By Nikolsky S.M.

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Sample text

We see that in this example the derived set E' consists of only one. pomt Onot bclonging Lo E. f the form 1fn (n = 1, 2, .. er beloogs to E 1 . Thus, a limit point of a set may or may not belong to that set. bourhood, that is any interval (e, d) such that e < a < d, con ta ins al Icast onc po int x belonging toE and distinct from a. . Jferent points X¡, X2, x 3 , •.. (x, ~ xkfor n ~ k) con vergen/toa. Indecd, according to lhc definition, (e, d) conta ins a point x 1 E E such th~t ;\':1 Jé a. not c~n­ taining x 1• Again, according to the dcfinition of lhe pomt a.

O. exit4a~ .. = O. ex}alaA (3) We shall show that this conclusion leads toa contradiction. Let us take, for every natural n, a decimal digit ex, such that the inequalities ex, > O and o:n # ex: hold, which is obviously possible. Now we construct the number x3 = X 2 x1 Here at each stage of the numbering process we should delete those elements which have already been numbered at the foregoing stages; the matter is that Ek and E1 may have sorne common elements. This procedure results in an infinite sequence of elements {y¡, y2, Ya, •..

The rightmost of these parts containing an infinite number of elements X 11 (let it be denoted as Lh). This means that if both parts contain an infinitude of elements then L1 1 is the rightmost of them and if only one of these parts contains an infinite number of elements Xn then Lit denotes that very part. Let Xn be one of the elements belonging to the interval L1 1• Dividing the interval ~ 1 into two equal parts we denote by Ll2 the rightmost of the parts containing an infinite number of elements Xn- Among these elements there is obviously an element Xn3 with n2 > n¡.

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Course in Mathematical Analysis by Nikolsky S.M.


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