A First Course in Real Analysis (Undergraduate Texts in by Sterling K. Berberian

By Sterling K. Berberian

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The e-book bargains an initiation into mathematical reasoning, and into the mathematician's way of thinking and reflexes. in particular, the elemental operations of calculus--differentiation and integration of services and the summation of countless series--are equipped, with logical continuity (i.e., "rigor"), ranging from the genuine quantity procedure. the 1st bankruptcy units down unique axioms for the true quantity method, 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 booklet, specially thorough. The concluding bankruptcy establishes an important beachhead within the concept of the Lebesgue fundamental by way of straight forward ability.

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Exercises 1. Let (an) be a sequence in lR. Prove that (an) is unbounded if and only if there exists a subsequence (a nk ) such that lank I ~ k for all k. 2. Let (an) be a sequence in lR . Prove that the following conditions are equivalent: (a) (an) is divergent; (b) for every a E lR there exist an E > 0 and a subsequence (a nk ) such that lank - al ~ E for all k. 3. *True or false (explain): The sequence an subsequence. * = sin n has a convergent 4. True or false (explain): If (an) is any sequence in lR, then the sequence has a convergent subsequence.

9, Exercise 5), the pair (X, d) is called a metric space. A metric space is said to be compact if every sequence in the space has a convergent subsequence (cf. 4, Exercise 12). 2), where a < b, and let d(x, y) = Ix - yl be the usual metric on X. Prove: X is compact if and only if X = [a , b]. 10. } (ii) If (X, d) is a compact metric space, show that the subset {d(x , y) : x, y EX} of JR is bounded. {Hint: For every pair of sequences (x n ) and (Yn) in X , the sequence of real numbers d(xn, Yn) has a convergent subsequence.

Sequences continuing in this way, we construct a subsequence (a nk ) such that lank - bl < 1/ k for all k; then a nk -+ b, so b E S. The proof that c E S is similar. To prove the second inclusion, assuming a nk -+ x we have to show that c:::; x :::; b; let's show, for example, that x:::; b, that is, x - b :::; o. 2). 8. The proof that c:::; x is similar. <> Exercises 1. Find lim sup an and lim inf an for each of the following sequences: (i) an = (-l)n + l/n (ii) an = (-1)n(2+3/n) (iii) a n =1/n+(-1)n/n 2 (iv) an = [n + (-1)n(2n + l)]/n 2.

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