By Sterling K. Berberian

ISBN-10: 0387942173

ISBN-13: 9780387942179

Uploader's Note: Ripped from SpringerLink.

The e-book bargains an initiation into mathematical reasoning, and into the mathematician's frame of mind and reflexes. in particular, the basic operations of calculus--differentiation and integration of capabilities and the summation of limitless series--are equipped, with logical continuity (i.e., "rigor"), ranging from the true quantity procedure. the 1st bankruptcy units down designated axioms for the genuine quantity process, from which all else is derived utilizing the logical instruments summarized in an Appendix. The dialogue of the "fundamental theorem of calculus," the focus of the publication, specially thorough. The concluding bankruptcy establishes an important beachhead within the conception of the Lebesgue necessary by means of user-friendly potential.

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Extra resources for A First Course in Real Analysis (Undergraduate Texts in Mathematics)

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One calls (Yk) a subsequence of (xn). This is also expressed by saying that x n1 ' x n2 , xn3 , . is a subsequence of Xl,X2,X3, ... , or that (x nk ) is a subsequence of (xn). Forming a subsequence amounts to choosing a sequence of indices; what is essential is that the chosen indices must march steadily to the right-no stuttering, no dropping back. 2. Remark. Let (nk) be a strictly increasing sequence of positive integers: nl < n2 < n3 < .... For every positive integer N, there exists a positive integer k such that nk > N (therefore nj > N for all j::::: k).

Choose n3 E A so that n3 > n2, and so on. The subsequence Xnl , Xn2 , xn3 ' . constructed in this way has the desired property. (b) =} (a): By assumption, nk E A for all k. Given any index N, the claim is that A contains an integer n 2: N. 2). 7. Example. Suppose we're trying to show that Ian - al < € ultimately. 6) lank - al 2: € for some subsequence (a nk ) of (an); it may be more convenient to prove that the alternative is false. 8. Theorem. Every sequence in IR has a monotone subsequence. 2).

2, Exercise 2). 1. Theorem. If (In) is a nested sequence of closed intervals, then the intersection of the In is nonempty. More precisely, if In = [an' bnJ , where an:S bn and II ::J 12 J h J . . , and if a = sup{a n : n E lP'}, b = inf{bn : n E lP'}, n::'=l [an , bnJ = [a, bJ . {The notation n::'=l [an' bnJ means the intersection n S then a:S band Proof set S of all the intervals [an , bnJ (cf. } From [an+l' bn+lJ c [an, bnJ we see that of the 24 2. First Properties of lR it follows that the sequence (an) is increasing and bounded above (for example by bl ), whereas (b n ) is decreasing and bounded below (for example by al) .

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