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

Let A be a nonempty proper subset of (0,3). Note . Define (i) Prove that and exist using the Completeness Axiom. (ii) Prove that is an upper bound for B. (iii) Prove that . 
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.

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. . Now, by definition of B, there exists an element a inside A such that . Thus, and clearing negatives gives
which contradicts the definition of the infimum. Hence is an upper bound of B. (iii) Now, to show that 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 . Multiplying by negative one gives and so by definition of the infimum of A, there is an element a of A such that Taking the negatives again gives However this element lives inside B which contradicts the definition of c as an upper bound for B! Thus bust be the least upper bound (the supremum). 