By Sterling K. Berberian

ISBN-10: 0387942173

ISBN-13: 9780387942179

**Uploader's Note:** Ripped from SpringerLink.

The publication deals an initiation into mathematical reasoning, and into the mathematician's state of mind and reflexes. in particular, the basic operations of calculus--differentiation and integration of features and the summation of countless series--are equipped, with logical continuity (i.e., "rigor"), ranging from the genuine quantity approach. the 1st bankruptcy units down special axioms for the true quantity procedure, 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 e-book, specially thorough. The concluding bankruptcy establishes an important beachhead within the conception of the Lebesgue quintessential by way of easy skill.

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**Additional info for A First Course in Real Analysis (Undergraduate Texts in Mathematics)**

**Sample text**

3. 3. 1. Theorem. For each real number x, there exists a unique integer n such that n::; x < n + 1 . Proof Uniqueness: The claim is that a real number x can't belong to the interval [n, n + 1) for two distinct values of n. If m and n are distinct integers, say m < n, then n - m is an integer and is > 0, therefore n - m 2: 1 (cf. 4); thus m + 1 ::; n and it follows that the intervals [m, m + 1) and [n, n + 1) can have no element x in common. Existence: Let x E IR. By the Archimedean property, there exists a positive integer j such that j1 > -x, that is, j +x > O.

Ii) Infer from (i) that if a E lR , a ~ 0, then there exists a unique b E lR , b ~ 0 , such that b2 = a . {Hints: (i) Writing y = 1 - r , x = 1 - s , we have O:S y :S 1 and the problem is to find a real number x, O:S x :S 1, such that (1-x)2 = 1-y, that is, x = ~(Y+X2) . The formulas Xl = 0 , Xn+l = ~[Y+(Xn)2J define recursively an increasing sequence (xn ) such that O:S Xn :S 1. 6. Theorem on Nested Intervals A sequence of intervals (In) of lR is said to be nested if II ::J 12 ::J J .. 2, Exercise 2).

6) . 5 can thus be reformulated as follows: A sequence (an) in IR is convergent if and only if it is bounded and bn - Cn ! O. This is the promised reformulation of convergence in terms of monotone convergence. 7. Theorem. Let (an) be a bounded sequence in IR and let s= let c {x E = lim inf an IR : a nk ....... x for some subsequence (a nk ) }; and b = lim sup an. Then {c,b} eSc [c,b], thus c is the smallest element of Sand b is the largest. Proof The first inclusion asserts that each of c and b is the limit of a suitable subsequence of (an); for example, let's prove the assertion for b .

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

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