The proof of existence of supremum for a non-empty set which has an upper bound

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.

$\textbf{Theorem}:$ Every non-empty subset $X \subset \mathbb{R}$ that has an upper bound has a least upper bound $(sup X)$.

$\textbf{Proof}:$ Suppose we have $x \in X$ and $x \geq0$. Also, suppose $\alpha$ is a given upper bound.

If $x \neq \alpha$, then there is a first $j$ such that $j^{th}$ digit of $x$ is smaller than the $j^{th}$ of $\alpha$. 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 $X=\{0.5, 2, 2.5, \cdots, 3.23\}$ where $3.23$ is the largest value in the set and $\alpha = 6.2$. I select $2$ as my $x$. Then, the value of $j$ is one. Hence, the set is defined as $[2.1,2.2., \cdots, 2.9,6.2]$. In this case, $b_j$ is 2.9.

If create a new set based on $b_j$, doesn't that become $[2.91,2.92., \cdots, 2.99,6.2]$. 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."

Answer 1

Using the proof in Hubbard and your example.

$x = 2$ and $a = 6.2$. $2 < 6$ 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 $b_4 = 3.299$ and so on $b_5 = 3.2999$ ....And $\sup X$ is $3.22999999999....... = 3.23$

Answer 2

You're approaching this the wrong way.

In fact, the first digit of $x$ which is smaller than a digit of $\alpha$ might occur before the decimal place.

Consider the numbers $21.\overline{1}$ and $12.\overline{2}$. The only place where $12.\overline{2}$ has a smaller value than $21.\overline{1}$ is the tens place, where $j = -1$.

In the case you presented, it's the ones place digit. So $j = 0$.

Thus, we start out with $b_j = 3$.

We then get $b_{j + 1} = 3.2$.

We then get $b_{j + 2} = 3.22$.

We hten get $b_{j + 3} = 3.229$.

In general, we have $b_{j + 2 + n}$ is the decimal beginning with $3.22$ and followed by $n$ copies of $9$.

Taking the infinite digit string gives us $3.22\overline{9} = 3.23$ as required.

Answer 3

I think it's worth pointing out that the proof is incorrect because there is no guarantee that $b_j$ exists; there might not be any number in $$S = \{ u: u \in [x,a] \text{ and } u \text{ can be written using only } j \text{ digits after the decimal, then all zeros} \},$$ that is not an upper bound. Consider the case when $X=\{2\}$, $x=2$ and $a=3$, then $S = \{2,3\}$ and only contains upper bounds to $X$. Or when $X$ is the set of all real numbers less than $1$ and $a=10$, then $S = \{10\}$ and again only contains upper bounds.

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Edit: The proof and some terminology is changed in the fifth edition of the book which corrects the proof:

• $[x]_k$ denotes the number formed by keeping all digits to the left of and including the $10^k$th position, setting all others to $0$.
• $j$ is the largest integer such that $[x]_j < [a]_j$
• $S$ is defined as all numbers that have the same $k$th digit as $x$ for $k > j$ and that have $0$ as the $k$th digit for $k < j$, that are in $[[x]_j, a]$
• $b_j$ is defined as the largest element in $S$ such that $X \cap [b_j, a] \neq \emptyset$

This ensures $S$ is non-empty and that $b_j$ exists.