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 that has an upper bound has a least upper bound .
Proof: Suppose we have and . Also, suppose is a given upper bound.
If , then there is a first such that digit of is smaller than the of . Consider all the numbers in that can be written using only digits after the decimal, then all zeros. Let bt the largest which is not an upper bound. Now, consider the set of numbers that have only digits after the decimal points, then all zeros.
The proof continues until getting which is not an upper bound.
Now, let's assume that where 3.23 is the largest value in the set and . I select 2 as my . Then, the value of is one. Hence, the set is defined as [2.1,2.2.,⋯,2.9,6.2]. In this case, is 2.9.
If create a new set based on , doesn't that become [2.91,2.92.,⋯,2.99,6.2]. If so, 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."