Proving that $0.999\ldots=1$ using the supremum axiom

Question

Use the supremum axiom to show that $0.999\ldots=1$.

Hints: Consider the set $C=\{0.9, 0.99, 0.999, \ldots\}$.

Discuss if $C$ is upper bound, and find the supremum(s) of $C$.

We know that every number with decimal expansion $x=0.a_1a_2a_3\ldots$ is such that $0\leq x \leq 1$.

What would happen if $0.999\ldots < 1$?

My attempt:

Clearly, $0.999\ldots=\operatorname{sup}{(C)}$ since $0.999\ldots=\sum_{i=1}^{\infty}9\cdot10^{-i}$ has a $n+1$-th positive term while $\sum_{i=1}^{n}9\cdot10^{-i}, \forall n \in \mathbb{N}$ doesn't, so $c \leq 0.999\ldots, \forall c \in C$. We know that $0.999\ldots \leq 1$, so all we need to do is show that $1=\operatorname{sup}{(C)}$. I thought about showing there is no $k\in\mathbb{R}$ such that $0.999\ldots < k < 1$, but I got stuck here.

My question:

How do I prove this with just elementary set theory? I don't want an answer based on arithmetic or analysis, since that's not how this question was meant to be solved. I would also like to know if showing $1=\operatorname{sup}{(C)}$ is the right path and if so how to proceed.

My research:

First, I asked my colleagues, but nobody I talked to got this question in the exam right. I tried to contact the professor but he did not respond. Online, I've found this thread but it did not get anywhere. I've wondered about this question for months so help would be really appreciated.

Answer 1

What would happen if 0.999…<1?

Then that would mean $1-0.9999.... > 0$.

The real question is why is that a problem?

Well let $1-0.9999999 =h > 0$.

Well, as $0.9999....$ is an upper bound then every $0.\underbrace{9999.....9}_{n\ nines}\in C$ si so that $0.99999....9 \le 0.9999..... < 1$

So $1 - 0.9999.....9= 0.00000.....1=\frac 1{10^n} \ge 1-0.99999...... = h > 0$.

So $\frac 1{10^n} \ge h > 0$ no matter how big $n$ is.

Is that problem?

That means that $\frac 1h \le 10^n$ no matter how large $n$ is.

Is that a problem?

Let $M = \frac 1h$ is a valid number as $h$ is a valid positive number. And $M$ is larger than any possible $10^n$. Every possible $10^n$. Is that possible? Is there a number that is bigger than any possible $10^n$.

.... Now's the time for you to pipe in "No, that'd be ridiculous".

So we can not have $0.99999...... = \sup C < 1$.

So $0.99999.... = \sup C \ge 1$.

Answer 2

I'm not much of a fan of using decimals to establish fundamental real analysis facts like this. I prefer the existence of properties of decimal expansions be developed over a real analysis course. So, this answer will assume some things about decimal expansions that you may or may not have developed in your course.

A decimal expansion of a number $0 \le x < 1$ (which we assume exists, even if it may not be unique) must have a $0$ in the units digit, i.e. its most significant digit is in the $10^{-1}$ digit. Otherwise, if the units digit is any other digit $1 \le d \le 9$, then $x \ge d \ge 1$, as $x \ge 0$. Therefore, it must take the form $$x = 0.d_1d_2d_3\ldots$$ where $d_1,d_2,d_3, \ldots$ are digits.

Now, increasing these digits (individually) increases the number. With this in mind, we have $$x = 0.d_1d_2d_3\ldots \le 0.999\ldots,$$ by maximising every digit. This shows that, for all $0 \le x < 1$, we have $$x \le 0.999\ldots \le 1,$$ where the final inequality is from the result you explicitly stated. But, $[0, 1)$ has the supremum $1$, and since $0.999\ldots$ is an upper bound of $[0, 1)$, we must also have $$1 \le 0.999\ldots$$ Consequently, $$1 = 0.999\ldots$$ Hope that helps.

Answer 3

Let $x_n=9\sum_{k=1}^n 10^{-k}$

Then $x_{n+1}-x_n=\frac{9}{10^{n+1}}>0$. So the sequence is increasing.

$x_n=(10-1)\sum_{k=1}^n10^{-k}=\sum_{k=1}^n 10^{-(k-1)}-\sum_{k=1}^n10^{-k}=1+\frac{1}{9}x_{n-1}-\frac{1}{9}x_n=1-\frac{1}{10^n}$

So $x_n=1-\frac{1}{10^n}$.

The set of $x_n$'s is non-empty and bounded above by $1$. So by the Least Upper Bound Property, the sequence has a least upper bound.

$\forall \epsilon>0$ and $1-\epsilon<x_n<1$, we have $1-\epsilon<1-\frac{1}{10^n}\implies \frac{1}{\epsilon}<10^n$

We want to find an $N$ so that $n>N\implies \frac{1}{\epsilon}<10^n$.

If $1\ge1/\epsilon$, then $N=0$. Otherwise $1<1/\epsilon$. Let $A=\{n \in N|10^n <\frac{1}{\epsilon} \}$. So $A$ is non-empty and bounded above, so it has a least upper bound. The least upper bound of a subset of the integers is an element of that subset. So we are guaranteed our $N$.

None of our $x_n$ exceed $1$. For any amount below one, however small, we can find an $x_n$ between that number and 1. It follows that $\sup \ \{x_n\}=1$.