I was going through a simple proof written for the existence of supremum. When I tried to write a small example for the argument used in the proof, I got stuck. The proof is presented in Vector Calculus, Linear Algebra, And Differential Forms written by Hubbard. Here is the theorem.Theorem: Every non-empty subset X ⊂ R that has an upper bound has a least upper bound ( s u p X ).Proof: Suppose we have x ∈ X and x ≥ 0. Also, suppose α is a given upper bound.If x ≠ α, then there is a first j such that j t h digit of x is smaller than the j t h of α. Consider all the numbers in [ x , a ] that can be written using only j digits after the decimal, then all zeros. Let b j bt the largest which is not an upper bound. Now, consider the set of numbers [ b j , a ] that have only j + 1 digits after the decimal points, then all zeros.The proof continues until getting b which is not an upper bound.Now, let's assume that X = { 0.5 , 2 , 2.5 , ⋯ , 3.23 } where 3.23 is the largest value in the set and α = 6.2. I select 2 as my x. Then, the value of j is one. Hence, the set is defined as [2.1,2.2.,⋯,2.9,6.2]. In this case, b j is 2.9.If create a new set based on b j , doesn't that become [2.91,2.92.,⋯,2.99,6.2]. If so, b j + 1 = 2.99 and I'd just add extra digits behind this number. In other words, I'd never go to the next level. I am probably misinterpreting the proof statement.Two important details after receiving some comments are as follows.1. "By definition, the set of real numbers is the set of infinite decimals: expressions like 2.95765392045756..., preceded by a plus or a minus sign (in practice the plus sign is usually omitted). The number that you usually think of as 3 is the infinite decimal 3.0000... , ending in all zeroes."2. "The least upper bound property of the reals is often taken as an axiom; indeed, it characterizes the real numbers, and it sits at the foundation of every theorem in calculus. However, at least with the description above of the reals, it is a theorem, not an axiom."

Question

I was going through a simple proof written for the existence of supremum. When I tried to write a small example for the argument used in the proof, I got stuck. The proof is presented in Vector Calculus, Linear Algebra, And Differential Forms written by Hubbard. Here is the theorem.Theorem: Every non-empty subset   X  ⊂      R   that has an upper bound has a least upper bound   (  s  u  p  X  ).Proof: Suppose we have   x  ∈  X and   x  ≥  0. Also, suppose   α is a given upper bound.If   x  ≠  α, then there is a first   j such that       j          t      h       digit of   x is smaller than the       j          t      h       of   α. Consider all the numbers in   [  x  ,  a  ] that can be written using only   j digits after the decimal, then all zeros. Let       b    j   bt the largest which is not an upper bound. Now, consider the set of numbers   [      b    j    ,  a  ] that have only   j  +  1 digits after the decimal points, then all zeros.The proof continues until getting   b which is not an upper bound.Now, let&#039;s assume that   X  =  {  0.5  ,  2  ,  2.5  ,  ⋯  ,  3.23  } where 3.23 is the largest value in the set and   α  =  6.2. I select 2 as my   x. Then, the value of   j is one. Hence, the set is defined as [2.1,2.2.,⋯,2.9,6.2]. In this case,       b    j   is 2.9.If create a new set based on       b    j  , doesn&#039;t that become [2.91,2.92.,⋯,2.99,6.2]. If so,       b          j      +      1        =  2.99 and I&#039;d just add extra digits behind this number. In other words, I&#039;d never go to the next level. I am probably misinterpreting the proof statement.Two important details after receiving some comments are as follows.1. &quot;By definition, the set of real numbers is the set of infinite decimals: expressions like 2.95765392045756..., preceded by a plus or a minus sign (in practice the plus sign is usually omitted). The number that you usually think of as 3 is the infinite decimal 3.0000... , ending in all zeroes.&quot;2. &quot;The least upper bound property of the reals is often taken as an axiom; indeed, it characterizes the real numbers, and it sits at the foundation of every theorem in calculus. However, at least with the description above of the reals, it is a theorem, not an axiom.&quot;

barbesdestyle2k · Accepted Answer

Using the proof in Hubbard and your example.x=2 and a=6.2.   2  &amp;lt;  6 so the first set to consider is {2,3,4,5,6}.4 is an upper bound of x but 2,3 are not. So so       b    0    =  3 which is the largest that is not an upper bound.j=1 and we consider all the numbers between       b    0    =  3 and a=6.2 with one decimal where we consider j=1 decimal point. That set is {2.1,2.2,2.3,.........,6.0,6.1,6.2}.3.3 is an upper bound of X but 3.2 is not. So       b    1    =  3.2.Now for j=2 we consider all the numbers between 3.2 and 6.2 with two decimals. That set is {3.20,3.21,3.22,...........6.19,6.20}. Now 3.23 is an upper bound but 3.22 is not. So       b    2    =  3.22.So consider 3 decimals 3.220,.......,6.199,6.200 and get       b    3    =  3.229.Then for we consider 3.290,3.291,.....,6.1999,6.20 and b4=3.299 and so on       b    5    =  3.2999....And supX is 3.22999999999.......=3.23

I was going through a simple proof written for the existence of supremum. When I tried to write a sm

Answered question

Answer & Explanation

New Questions in College algebra