Science:Math Exam Resources/Courses/MATH220/December 2009/Question 09 (b)

MATH220 December 2009

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  

   Other MATH220 Exams

  •  April 2011  •  April 2005  •  December 2011  •  December 2010  •  December 2009  •  

Question 09 (b)
Let A be a non-empty proper subset of (0,3). Note $A\neq (0,3)$. Define $B=\{-a\quad {\textrm {s.t.}}\quad {}a\in {}A\}$ (i) Prove that $\inf(A)$ and $\sup(A)$ exist using the Completeness Axiom. (ii) Prove that $(-\inf(A))$ is an upper bound for B. (iii) Prove that $\sup(B)=(-\inf(A))$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Keep in mind that the infimum is the greatest lower bound and the supremum is the least upper bound. The rest is just symbol pushing.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. (i) The Completeness Axiom says that any nonempty set of real numbers must contain an infimum and a supremum which is exactly what we want to show. (ii) Assume towards a contradiction that there exists an element b of the set B that is larger than the claimed supremum, i.e. $\displaystyle -\inf(A)<b$. Now, by definition of B, there exists an element a inside A such that $\displaystyle b=-a$. Thus, $\displaystyle -\inf(A)<b=-a$ and clearing negatives gives $\displaystyle a<\inf(A)$ which contradicts the definition of the infimum. Hence $\displaystyle -\inf(A)$ is an upper bound of B. (iii) Now, to show that $\displaystyle -\inf(A)$ is the least upper bound, assume towards a contradiction that there is a real number c such that c is a smaller upper bound for B. Thus $\displaystyle -\inf(A)>c$. Multiplying by negative one gives $\displaystyle \inf(A)<-c$ and so by definition of the infimum of A, there is an element a of A such that $\displaystyle \inf(A)<a<-c$ Taking the negatives again gives $\displaystyle -\inf(A)>-a>c.$ However this element $\displaystyle -a$ lives inside B which contradicts the definition of c as an upper bound for B! Thus $\displaystyle \inf(A)$ bust be the least upper bound (the supremum).

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Infimum and supremum, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag