So I'm looking to prove $x = \sup E$ where $E= \{ a_0 + 10^{-1}a_1 + \cdots + 10^{-n}a_n : n \in \mathbb{N} \}$.
The definition of $a_i$ is as follows:
$a_0: [x]$
$a_1=[10^1(x-a_0)]$
$\vdots$
$a_n=[10^n(x-(a_0+10^{-1}a_1+ \cdots + 10^{-n}a_{n-1}))]$
I was able to prove that $x$ is an upper bound from an earlier proof. I'm trying to prove x is the least upper bound with the fact that $10^{-n} < x < 10^{n}$, which I have already proved. Here's what I have:
By definition of upper bound, we know that upper bound $\beta$ is an upper bound of E if $B \geq e$ $ \forall e \in E$. Rewriting E as $ \Sigma^{n}_{i=0} 10^{-n} d_n $, we have that $ 10^n > \Sigma^{n}_{i=0} 10^{-n} a_n $, and so $ \beta \geq 10^{n} > \Sigma^{n}_{i=0} 10^{-n} a_n$. Since we proved earlier that $10^n > x$, we have that $ \beta \geq 10^{n} > x$ $\rightarrow \beta > x$, and so x is the least upper bound.
I'm primarily worried that my proof seems very ... circular at best. Another alternative approach I was thinking of trying was doing a proof by contradiction. Here's what I had in mind:
Assume that upper bound $\beta > x$. Then $0 > x - \beta$.
And I'm thinking of somehow using the fact that we know $10^{-n} < x < 10^{n}$ to contradict the statement.
Again, any help or feedback would be much appreciated.
Obviously $x$ is an upper bound of $E$ so $y=\sup E\leqslant x. $ Assume for the sake of contradiction that $y<x. $ Then there is some positive integer $n$ such that $10^{-n}<1/n<x-y.$ Thus $a_0+(a_1/10)+\cdots+(a_n+1)/10^n\leqslant y+(1/10^n)<x,$ which contradicts the definition of $a_n. $
Now, regarding your "proof". It is wrong. The first little but important detail is that you first write $a_n$'s and then you re-label them as $d_n$'s. But let's ignore this for a moment. Now, the part which is really wrong is when you say "$\beta>10^n$" because there is no justification for this. For instance, take $x=2,\;n=1.$ then, in this particular case, $\sup E=2$ but clearly $\sup E$ is not greater than $10^n.$