Math 450: Real Analysis
 News forum Course Information: Aims of this course IPCA Theorem Sequence (Attitudes towards thinking and learning) Elementary Logic Sets Definitions for Real Analysis (Got the book?) Assignment 1 Assignment 2 Proofs and Solutions FCRA Ch 1, IPCA Ch 2 The rules refined Assignment 3 Proofs and Solutions FCRA Ch 2, IPCA Ch3 Assignment 4 Assignment 5 Proofs for IPCA Chapter 4 Assignment 6 (Quiz 1) Proofs and solutions from IPCA Chapter 5 and FCRA ... Assignment 7 Questions and observations on real analysis Take-home test 1 Assignment 8 Assignment 9 Jump to... Take-home test 2 Proofs and solutions from IPCA Chapter 9 Proofs and solutions from IPCA Chapter 10 Assignment 10 Proofs and solutions from IPCA Chapter 11 Final Exam

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 Theorem 3 from Vancouver :)by ... - Thursday, October 28 2004, 02:25 PM Theorem 3: Let `U` be the set of all upper sums of `f`, over all nets `N`. Then `U` is bounded below. Proof: Let `U = { U(N) : N` is a net `}`. By definition: `U(N)=sum_(i=1)^n(M_i\Deltax_i)`. We know that `f ` is bounded, moreover, we know that `f` is bounded below. This means that `EE m in RR` such that: `f(x) >= m` for all `x in [a, b]` If we can show that `U(N) >= m(b - a)` then our proof will be done. ( we need to show this for all nets `N` ) Take the inequality: `M_i\Deltax_i>=\Deltax_im`. It obviously holds since m is a lower bound, then `m <= M_i` for all `i`. Let's take the finite sum of both sides, we get: `sum_(i=1)^nM_i\Deltax_i>=sum_(i=1)^n(m\Deltax_i)` The LHS of the inequality is `U(N)` by definition, and on the right hand side, we can take the constant `m` outisde to get: `U(N)>=msum_(i=1)^n(\Deltax_i)` Since the sum of all `\Deltax_i` is the interval `[a,b]` itself, we finally conclude: `U(N)>=m(b-a)` And the result follows. This proof was obviously inspired by the one E... did for exercice 4. Thanks E... Good luck all! ... Edit | Delete | Reply   Rate... 5 / 5 4 / 5 3 / 5 2 / 5 1 / 5 0 / 5

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