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) •
Question 07 (b)
Let and be non-empty subsets of the interval [0,1]. Now let be the set
(i) Prove that and exist.
(ii) Suppose that and , prove that 0 is a lower bound for and 1 is an upper bound for .
(iii) Suppose that , prove that .
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(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.
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(i) Since is bounded, we have that and exist by the completeness axiom.
(ii) Suppose that 0 was not a lower bound. Then there exists a with . Also, there exists numbers and with . As and are both positive (since and are subsets of ) we see this is a contradiction.
Suppose that 1 was not an upper bound. Then there exists a with . Also, there exists numbers and with . As and are both positive (since and are subsets of ) numbers less than or equal to 1, we see that 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 which would be the supremum existing by the completeness axiom. Now, choose a value of such that . Since 1 is the supremum of both and , there exist elements and such that . With this chosen, we see that . Thus but is larger than which was the supremum. this is a contradiction and thus the supremum must indeed be 1.
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