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

By Sterling K. Berberian

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The ebook bargains an initiation into mathematical reasoning, and into the mathematician's way of thinking and reflexes. in particular, the basic operations of calculus--differentiation and integration of services and the summation of endless series--are outfitted, with logical continuity (i.e., "rigor"), ranging from the true quantity approach. the 1st bankruptcy units down specific axioms for the true 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 point of interest of the publication, in particular thorough. The concluding bankruptcy establishes an important beachhead within the concept of the Lebesgue indispensable via hassle-free ability.

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

**Example text**

Every constant sequence (xn bounded. The sequence Xn = (_1)n is bounded. 33 S. K. Berberian, A First Course in Real Analysis © Springer Science+Business Media New York 1994 x for all n) is 3. 4. Example. The sequence xn = n is unbounded. 5. Theorem. If (xn) and (Yn) are bounded sequences in IR, then the sequences (xn + Yn) and (xnYn) are also bounded. K Proof If IXnl : : : K and IYnl::::: K' then IXn + K' and IXnYnl = IXnllYnl : : : KK' . 0 + Ynl ::::: IXnl + IYnl : : : Exercises 1. Show that if (xn) and (Yn) are bounded sequences and c E IR, then the sequences (cxn) and (x n - Yn) are also bounded.

8). 3) . Note that A contains numbers > 0; for , if c ~ 1 then 1/2 E A (because 1/4 < 1 S c), whereas if 0 < c < 1 then c E A (because c2 < c). It follows that , > O. Next, we assert that , E B; by the arguments in the preceding section, we need only show that A has no largest element. Assuming a E A, let's find a larger element of A. If a S 0 then any positive element of A will do. Suppose a > O. We know that a 2 < C; it will suffice to find a positive integer n such that (a + 1/n)2 < c. 4 (with 2 replaced by c).

If m > 1 then m - 1 is a positive integer smaller than m , so it can't belong to S; this means that m - 1 ::; x , thus x E [m - 1, m) and n = m - 1 fills the bill. 2. Definition. With notations as in the theorem, the integer n is denoted [x] and the function IR --+ Z defined by x 1-+ [x] is called the bracket function (or the greatest integer function , since [x] is the largest integer that is ::; x ). Exercises 1. 3] and [-2]. 2. Sketch the graph of each of the following functions f: IR (i) f(x) = [x].