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In Baby Rudin 1.22 (Decimals):

We conclude this section by pointng out the relation between real numbers and decimals. Let $x>0$ and $n_0$ be the largest integer such that $n_0\leq x$. (Note that the existence of $n_0$ depends on the Archimedean property of $\mathbb R$.) Having chosen $n_0,n_1,\dots,n_{k-1},$ let $n_k$ be the largest integer such that $$n_0+\frac{n_1}{10}+\dots+\frac{n_k}{10^k}\leq x$$ Let E be the set of these numbers $$n_0+\frac{n_1}{10}+\dots+\frac{n_k}{10^k}$$

Here is my question: I want to prove that $x=\sup E$.

$x \geq \sup E$ is explained easily, because x is upper bound of E from the definition, I wonder how we can get contradiction, if we suppose $x>\sup E$.

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  • $\begingroup$ Suppose that set of numbers has a smaller upper bound y < x. Derive a contradiction. $\endgroup$ Commented Jun 5, 2018 at 7:05
  • $\begingroup$ Because I want to prove that $x=\sup E$, I think it is more proper to search a element of E is bigger than $\sup E$ here. $\endgroup$
    – 백주상
    Commented Jun 5, 2018 at 7:34

1 Answer 1

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Clearly, $$ x<n_0+1 $$ Inductively, one can similarly prove that $$ x<n_0+\frac{n_1}{10}+\cdots+\frac{n_k+1}{10^k} $$ thus $$ x-\frac{1}{10^k}<n_0+\frac{n_1}{10}+\cdots+\frac{n_k}{10^k}\le \sup E, $$ for all $k$, which implies that $$ x\le \sup E. $$

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