7 the Completeness Axiom Definition 7.1. Let S⊆R. If There Exists a Real

7 the Completeness Axiom Definition 7.1. Let S⊆R. If There Exists a Real

Math 320 { September 20, 2020 7 The completeness axiom Definition 7.1. Let S ⊆R. If there exists a real number M, such that M ≥s for all s2S, then M is called an upper bound for S, and we say that S is bounded above. If there exists a real number m, such that m≤s for all s2S, then m is called a lower bound for S, and we say that S is bounded below. A set is called bounded, if it is bounded above and below. If M 2S is an upper bound for S, then M is called the maximum of S, M =maxS, and if m2S is a lower bound for S, then m is called the minimum of S, m=minS. Definition 7.2. Let S ⊆R be nonempty. If S is bounded above, then the least upper bound is called the supremum of S, denoted by supS. If S is bounded below, then the greatest lower bound is called its infimum, denoted by infS. Thus, M = supS is equivalent to the conditions (similar conditions hold for the infimum with reversed inequalities) (a) M ≥s for all s2S (b) If M0 ≥s for all s2S, then M0 ≥M. The contrapositive of the last condition is: (b0) If M0 <M, then there exists s2S, such that s≥M0. Axiom 7.3 (The completeness axiom). Every nonempty subset of R that is bounded above has a least upper bound. That is, supS exists, and supS 2R. The existence of the infimum for sets bounded below follows from the completeness axiom by considering the set −S =f−s:s2Sg, and observing that infS =−sup(−S). A consequence of the completeness axiom is the following property. Theorem 7.4 (Archimedean property). The set N of natural numbers is unbounded above in R. We have the following equivalent statements to the Archimedean property. Theorem 7.5. The Archimedean property is equivalent to: (a) 8z 2R;9n2N, such that n>z (b) 8x>0 and y2R;9n2N such that nx>y 1 (c) 8x>0;9n2N such that 0< n <x. Another consequence of the completeness axiom is the following. Theorem 7.6 (Nested interval property). Every sequence of nested closed intervals has a nonempty intersection. T1 That is, if In =[an;bn]=fx2R:an ≤x≤bng for every n2N, and I1 ⊇I2 ⊇I3 ⊇I4 ⊇:::, then n=1In =6 ;. Using the completeness one can also prove the following theorem. Theorem 7.7. If p2N is a prime number, then there exists a real number x, such that x2 =p. Moreover, x2 = Q. Using the equivalent statements to the Archimedean property, one can show that both the rational and irrational numbers are dense in R. Theorem 7.8. If x;y 2 R, with x < y, then there exists a rational number r 2 Q, and an irrational number w2RnQ, such that x<r<y, and x<w<y..

View Full Text

Details

  • File Type
    pdf
  • Upload Time
    -
  • Content Languages
    English
  • Upload User
    Anonymous/Not logged-in
  • File Pages
    1 Page
  • File Size
    -

Download

Channel Download Status
Express Download Enable

Copyright

We respect the copyrights and intellectual property rights of all users. All uploaded documents are either original works of the uploader or authorized works of the rightful owners.

  • Not to be reproduced or distributed without explicit permission.
  • Not used for commercial purposes outside of approved use cases.
  • Not used to infringe on the rights of the original creators.
  • If you believe any content infringes your copyright, please contact us immediately.

Support

For help with questions, suggestions, or problems, please contact us