Science:Math Exam Resources/Courses/MATH220/December 2010/Question 07 (b)

MATH220 December 2010

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q10 (a)  •  Q10 (b)  •  Q10 (c)  •  

   Other MATH220 Exams

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

Question 07 (b)
Let $A$ and $B$ be non-empty subsets of the interval [0,1]. Now let $C$ be the set $C=\{ab\quad {\textrm {s.t.}}\quad {}a\in {A},b\in {B}\}.$ (i) Prove that $\sup(A)$ and $\sup(B)$ exist. (ii) Suppose that $\inf(A)=\inf(B)=0$ and $\sup(A)=\sup(B)=1$, prove that 0 is a lower bound for $C$ and 1 is an upper bound for $C$. (iii) Suppose that $\sup(A)=\sup(B)=1$, prove that $\sup(C)=1$.

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
(i) Completness Axiom (ii) This is a straight definition check. (iii) We know form part (ii) that 1 is an upper bound. Proceed by a contradiction to show that it must be the least upper bound.

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) Since $\displaystyle [0,1]$ is bounded, we have that $\displaystyle \sup(A)$ and $\displaystyle \sup(B)$ exist by the completeness axiom. (ii) Suppose that 0 was not a lower bound. Then there exists a $\displaystyle c\in C$ with $\displaystyle c<0$. Also, there exists numbers $\displaystyle a\in A$ and $\displaystyle b\in B$ with $\displaystyle c=ab$. As $a$ and $b$ are both positive (since $A$ and $B$ are subsets of $\displaystyle [0,1]$) we see this is a contradiction. Suppose that 1 was not an upper bound. Then there exists a $\displaystyle c\in C$ with $\displaystyle c>1$. Also, there exists numbers $\displaystyle a\in A$ and $\displaystyle b\in B$ with $\displaystyle c=ab$. As $a$ and $b$ are both positive (since $A$ and $B$ are subsets of $\displaystyle [0,1]$) numbers less than or equal to 1, we see that $\displaystyle c=ab\leq 1$ and this is a contradiction. (iii) Notice that 1 is the least upper bound hence the supremum. That 1 is a least upper bound was part (ii). If 1 was not the supremum, then there would be some $\displaystyle c<1$ which would be the supremum existing by the completeness axiom. Now, choose a value of $x$ such that $\displaystyle c<x^{2}<x<1$. Since 1 is the supremum of both $A$ and $B$, there exist elements $\displaystyle a\in A$ and $\displaystyle b\in B$ such that $\displaystyle x<a,b$. With this chosen, we see that $\displaystyle c<x^{2}=x\cdot x<ab<1$. Thus $\displaystyle ab\in C$ but is larger than $c$ which was the supremum. this is a contradiction and thus the supremum must indeed be 1.

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