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.