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8. Let {an} be a sequence which is bounded above and define a new sequence {bn} by bn = max{a1, a2, ..., an}. (The maximum of a finite set A is the element of A which is greater than or equal to all elements of A.) Warning: the sequence {bn} may not be a subsequence of {an}. a.) Show that the sequence {bn} is increasing. (Hint: Notice that if if A and B are finite sets of real numbers with A ? B, then max A ? max B.) b.) Assume that U is an upper bound for {an}, show that U is an upper bound for {bn}. c.) From the Property of Completeness we know that {bn} must have a limit, call it L. Prove that an ? L for all n ? ?, i.e. L is an upper bound for {an}. Also prove that if U is any upper bound for {an} then L ? U. Thus L is the least upper bound of the sequence {an}. We denote this least upper bound by sup an, where sup is short for the Latin word supremum . Similarly, if the sequence {an} is bounded below, we can use the Property of Completeness to prove that it has a greatest lower bound . We denote this by inf an, where inf is short for the Latin word infimum .

          8. Let {an} be a sequence which is bounded above and define a new sequence {bn} by bn = max{a1, a2, ..., an}. (The maximum of a finite set A is the element of A which is greater than or equal to all elements of A.) Warning: the sequence {bn} may not be a subsequence of {an}.
a.) Show that the sequence {bn} is increasing. (Hint: Notice that if if A and B are finite sets of real numbers with A ? B, then max A ? max B.)
b.) Assume that U is an upper bound for {an}, show that U is an upper bound for {bn}.
c.) From the Property of Completeness we know that {bn} must have a limit, call it L. Prove that an ? L for all n ? ?, i.e. L is an upper bound for {an}. Also prove that if U is any upper bound for {an} then L ? U. Thus L is the least upper bound of the sequence {an}. We denote this least upper bound by sup an, where sup is short for the Latin word supremum .
Similarly, if the sequence {an} is bounded below, we can use the Property of Completeness to prove that it has a greatest lower bound . We denote this by inf an, where inf is short for the Latin word infimum .

8. Let an be a sequence which is bounded above and define a new sequence bn by bn = maxa1, a2, ..., an. (The maximum of a finite set A is the element of A which is greater than or equal to all elements of A.) Warning: the sequence bn may not be a subsequence of an.
a.) Show that the sequence bn is increasing. (Hint: Notice that if if A and B are finite sets of real numbers with A ? B, then max A ? max B.)
b.) Assume that U is an upper bound for an, show that U is an upper bound for bn.
c.) From the Property of Completeness we know that bn must have a limit, call it L. Prove that an ? L for all n ? ?, i.e. L is an upper bound for an. Also prove that if U is any upper bound for an then L ? U. Thus L is the least upper bound of the sequence an. We denote this least upper bound by sup an, where sup is short for the Latin word supremum .
Similarly, if the sequence an is bounded below, we can use the Property of Completeness to prove that it has a greatest lower bound . We denote this by inf an, where inf is short for the Latin word infimum .

Added by Sharon G.

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