By John Srdjan Petrovic

ISBN-10: 1351381431

ISBN-13: 9781351381437

ISBN-10: 1466565640

ISBN-13: 9781466565647

ISBN-10: 2592642692

ISBN-13: 9782592642691

Sequences and Their Limits Computing the LimitsDefinition of the restrict houses of Limits Monotone Sequences The quantity e Cauchy Sequences restrict more advantageous and restrict Inferior Computing the Limits-Part II genuine Numbers The Axioms of the Set R results of the Completeness Axiom Bolzano-Weierstrass Theorem a few innovations approximately RContinuity Computing Limits of features A assessment of capabilities non-stop services: ARead more...

summary: Sequences and Their Limits Computing the LimitsDefinition of the restrict houses of Limits Monotone Sequences The quantity e Cauchy Sequences restrict more desirable and restrict Inferior Computing the Limits-Part II actual Numbers The Axioms of the Set R outcomes of the Completeness Axiom Bolzano-Weierstrass Theorem a few strategies approximately RContinuity Computing Limits of services A evaluate of services non-stop capabilities: a geometrical standpoint Limits of capabilities different Limits homes of continuing features The Continuity of trouble-free services Uniform Continuity houses of continuing capabilities

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**Example text**

If b − a ≤ 1, we can stretch the interval (a, b), by multiplying both a and b by a positive integer. First we will show that there is a rational number r ∈ (a, b). By the Archimedean Property, there exists a positive integer q such that q(b − a) > 1. This implies that the distance between numbers qa and qb is bigger than 1, so there must be an integer p between them. Let r = p/q. Since qa < p < qb, we obtain that√ a <√ r < b. In order to construct ρ, we consider the interval ( b2 , a2 ). Once again, the Archimedean Property yields a positive integer p such that √ √ 2 2 > 1.

This raises the question: How do we know that such an integer exists? 2. Let a ∈ R and let A = {n ∈ Z : n ≤ a}. Then the set A is bounded above and sup A is an integer. Proof. It is obvious that A is bounded above by a, so it has a least upper bound s. Suppose, to the contrary, that s is not an integer. Since s is a least upper bound, the number s − 1 cannot be an upper bound of A. Therefore, there exists N ∈ A such that N > s − 1. It follows that N + 1 > s so N + 1 ∈ / A. Therefore, N is the largest integer in A, and we conclude that N is an upper bound of A smaller than s.

7 is also known under the name Cantor’s Intersection Theorem. Georg Cantor (1845–1918) was a German mathematician, best known as the inventor of set theory. He used the nested intervals in his original proof of the uncountability of real numbers in his 1874 paper [13]. This established that there exist infinite sets of different sizes. Cantor introduced the concept of a cardinal number and the power set of A (the set of all possible subsets of A).

### Advanced Calculus : Theory and Practice by John Srdjan Petrovic

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