Rudin: Supremum of Finite Decimals

I am reading Rudin. Note the following construction:

Let $x>0$ be real. Let $n_0$ be the largest integer s.t. $n_0 \leq x$. Then, having chosen $n_0, \ldots, n_{k-1}$, let $n_k$ be the largest integer s.t. $n_0 + \dfrac{n_1}{10} + \ldots + \dfrac{n_k}{10^k} \leq x$.

Let $E$ be the set of numbers $n_0 + \dfrac{n_1}{10} + \ldots + \dfrac{n_k}{10^k} \leq x$, where $k$ is a non-negative integer. Then $x = \sup E$.

Conversely, for any infinite decimal, the set $E$ is bounded above, and the infinite decimal expansion is the decimal expansion of $\sup E$.

How does one prove both directions?

I was thinking that for the first part, we must use precisely the definition of the supremum. That is, if $\epsilon > 0$, we must find a number $h \in E$ s.t. $h> x - \epsilon$. However, it isn't clear to me that such a decimal number can be constructed.

And how is the second direction the converse? I think it is a reiteration of the first direction. Set $x$ to be an infinite decimal expansion, and the result follows immediately from the proof of the first direction.

• Ad 1) Can you, given $a<x$, create such finite decimal expansion such that it's less than $a$, but the "next" element (i.e. with last digit greater by one) is greater than $a$, but still less than $x$? Ad 2) Yes, I believe you're correct as long as we agree that such infinite decimal expansion determines a real number. Feb 23, 2014 at 17:18
• It isn't clear to me, say that, if $\epsilon$ has an infinite decimal form, that the "next" element is clearly defined Feb 23, 2014 at 17:22
• No, instead of dealing with $a=x-\epsilon$ itself I asked about finite decimal $b$ (with, say, $b_n 10^{-n}$ as the last term), such that $b<a$ and $b+10^{-n}>a$, but still $b+10^{-n}<x$. No infinity for now. Feb 23, 2014 at 17:25
• Yeah, I see how your construction works where $\epsilon$ is a finite decimal. What about for infinite decimals though? Also, can it be easily shown that any real $\epsilon$ can be represented by the union of the set of finite and infinite decimals? Feb 23, 2014 at 17:28
• We don't need to deal with infinite decimals. Our $a<x$ is arbitrary (finite decimal or not, does not matter), only the $b$ is finite decimal. Essentially the argument relies on the fact that $a$ lies between two finite decimals. As for the second part of the comment: I'm afraid this is precisely the fact this exercise is supposed to teach us :) Feb 23, 2014 at 17:30