How do I show that $x$ is the supremum of set $S$? (decimal representation of reals)

Question

Let $x$ be a fixed positive real number.

Let $l_0 = a_0$ be the largest integer less than x (that is, $a_0\in Z$ such that $a_0 \le x$), $a_1$ be the largest integer such that $l_1 = a_0+\frac{a_1}{10^1}\le x$, $a_2$ be the largest integer such that $l_2 = a_0+\frac{a_1}{10^1}+\frac{a_2}{10^{2}}\le x$ and so on until $a_n$ be defined similarly as to let us have $l_n = a_0+\frac{a_1}{10^1}+\frac{a_2}{10^2}+\ldots +\frac{a_n}{10^n} \le x$.

We define the set $S$ as the set that contains $l_n$ for all $n\ge0$ ($n$ is a nonnegative integer).

We know that S is non-empty since we know that there's an unique integer $a_0$ such that $a_0\le x \lt a_0+1$ (I managed to prove that) and it is bounded above since $x$ is a upper bound, then by the supremum axiom, we know that S has a supremum $b = sup S$ in which $b \in \Re$.

The question is, how do I show that $b = x$?

I tried to use the fact that $l_n \le b$ for all $n \ge 0$, $l_n \le x \lt l_n + \frac{1}{10^n}$ for all $n \ge 0$ and the tricotomy to show that $b \gt x$ and $b \lt x$ both lead to a contradiction, thus $b = x$, but I haven't had the ideas to deal with this information to lead me to the contradiction in each case... So any help is very much appreciated!

Answer 1

As $S$ is bounded above by $x$ you know $\sup S$ exists and $\sup S \le x$.

If we assume $\sup S < x$ the $x - \sup S > 0$. Let's call $x - \sup S = d$.

Now make, and prove, the claim that there is an $m\in \mathbb N$ so that $0 < \frac 1{10^m} < d$. (Note: This has nothing to do with $d=x-\sup S$.... this has only to do with $d > 0$. This claim is true for all positive real numbers.)

Consider $l_m = a_0 + ......$.

Now make, and prove, the claim that $x - l_m < \frac 1{10^m}$. (That should be simply a matter of how $l_m$ was created.)

That means $\sup S = x- d < x-\frac 1{10^m} < l_m \le x$.

So we have $l_m > \sup S$ but $l_m \in S$.

That's a contradiction.

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So the job I leave to you is to prove that for any $d > 0$ there is a $m\in \mathbb N$ so that $0 < \frac 1{10^m} < d$.

(Hint: $0< \frac 1{10^m} < d \iff 10^m > \frac 1d> 0\iff m \ge \log_{10} \frac 1d$)

ANd to prove that for any $m$ that $x - l_m < \frac 1{10^{m}}$. .... But that was how $l_m$ was constructed and that is the definition of $l_m$ so that is already proven!

Answer 2

As you statet we have $x\geq l_n$ for all $n\in\mathbb{N}$.

Now what could go wrong?

Suppose $x>sup(S)$, is that possible? If we decipher the $l_n$ it is just the decimal representation up to the n-th place of $x$! So we can there see, (and proof) that $\lim_{n\rightarrow \infty}l_n=x$. So if $x$ would be bigger than our suppremum we could find an bigger $l_n$ and have a contradiction.

Answer 3

A supremum is a least upper bound. So can you show $x$ is an upper bound? Then can you show there is no upper bound that is less than $x$?

You should be able to see $l_0 \leq l_1 \leq l_2 \leq \cdots \leq l_n \leq \cdots \leq x$. (Notice that $l_n \leq l_{n+1}$. Induct.)

You should be able to see $0 = \frac{0}{10^n} \leq \frac{a_n}{10^n} < \frac{10}{10^n} = \frac{1}{10^{n-1}}$. So for any $\varepsilon > 0$, there is an $N$ such that for all $n > N$, $x - \varepsilon < l_n \leq x$, so any proposed upper bound that is less than $x$, $x- \varepsilon$, isn't actually an upper bound.

Answer 4

Since $x$ is an upper bound for $S$, it follows that $b\le x$.

It remains to show $x\le b$.

For each positive integer $n$ we have \begin{align*} x & < I_{n-1}+\frac{a_n+1}{10^n} \\[4pt] &= I_n+\frac{1}{10^n} \\[4pt] &\le b+\frac{1}{10^n} \\[4pt] \end{align*} Thus $x < b+{\large{\frac{1}{10^n}}}$ for all positive integers $n$.

But if $x > b$, then for some positive integer $n$ we would have $$ x-b > \frac{1}{10^n} \qquad\qquad\;\;\;\; $$ contradiction.