I only need the answer for final home work question at the end of the document.
ID: 3184472 • Letter: I
Question
I only need the answer for final home work question at the end of the document. We need to prove only one definition from the top using "definition of real numbers". Thanks so much!
8 Supremum and Infimum A maximum of a subset A of the real numbers is an element in A that is greater than or equal to every element in A. A minimum of a subset A of real numbers is an element in A that is less than or equal to every element in A. An upper bound of a subset A of the real numbers is an element that is greater than or equal to every element of A. A supremum of A is an element that is a minimum of the set of all upper bounds for A. A lower bound of a subset A of the real numbers is an element that is less than or equal to every element of A. An infimum of A is an element that is a maximum of the set of all lower bounds for A Exercise 1. Prove that 0 is the unique infimum of the set {1/n:n EN) Exercise 2. Prove that if the maximum or mininum exists for a set, then it is unique. Exercise 2. Prove that if the infimum or supremum exists for a set, then it is unique. Exercise 3. Give an example of a set that does not have a supremum. Prove your assertion You will prove some correct form of the following theorem: Theorem 3. Any subset of the real numbers that mum. Any subset of the real numbers that has a supre has an infimum. Exercise 4. Fill in the blanks above. You do not have to prove your assertions here, but be prepared to give some reasoning for your guess. Homework Using our definition of the real numbers, prove one of the statements in the correct form of the theorem above.Explanation / Answer
Theorem 3. Any subset of real numbers that is bounded from above has a supremum. Any subset of real numbers that is bounded from below has an infimum.
Proof:-
I will prove first statement i.e. Any subset of real numbers that is bounded from above has a supremum.
We need a good notation for a real number given by its decimal representation. A real number has the form
a = a0.a1a2a3a4... where a0 is an integer and a1, a2, a3, ... {0, 1, 2, ...9}
To eliminate ambiguity in defining real numbers by their decimal representation, let us decide that if the sequence of decimals ends up with nines: a = a0.a1a2...an9999... (where an < 9) then we choose this number’s decimal representation as a = a0.a1a2...(an + 1)0000.... (For example, instead of 0.4999999.. we write 0.5.)
Let S be a nonempty set of real numbers, bounded above.Let us construct the least upper bound of S.
Consider first all the approximations by integers of the numbers a of S: if a = a0.a1a2... collect the a0’s. This is a collection of integer numbers. It is bounded above (by assumption). Then there is a largest one among them, call it B0.
Next collect only the numbers in S which begin with B0. (There are some!) Call their collection S0.
Any number in S S0 (number of S not in S0) is smaller than any number in S0
Look at the first decimal a1 of the numbers in S0. Let B1 be the largest among them. Let S1 be the set of all numbers in S0 whose first decimal is B1.
Note that the numbers in S1 begin with B0.B
Also note that any number in S S1 is smaller than any number in S1.
Next look at the second decimal of the numbers in S1. Find the largest, B2 etc.
Repeating the procedure we construct a sequence of smaller and smaller sets S0, S1, S2, ...Sn, ... such that
S S0 S1 S2 ... Sn ...
Note that every set Sn contains al least one element (it is not empty).
At each step n we have constructed the set Sn of numbers of S which start with B0.B1B2...Bn; the rest of the decimals can be anything. Also all numbers in S Sn are smaller than all numbers of Sn. (The construction is by induction!)
We end up with the number B = B0.B1B2...BnBn+1....
We need to show that B is the least upper bound.
To show it is an upper bound, let a S. If a0 < B0 then a < B. Otherwise a0 = B0 and we go on to compare the first decimals. Either a1 < B1 therefore a < B or, otherwise, a1 = B1. Etc. So either a < B or a = B. Therefore B is an upper bound.
To show it is the least (upper bound), take any smaller number t < B. Then t differs from B at some first decimal, say at the nth decimal: t = B0.B1B2...Bn1 tn tn+1... and tn < Bn. But then t is not in Sn and Sn contains numbers bigger than t. Hence B is least upper bound.