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This question pertains to BBFSK Vol I, page 143. The topic is the definition of the greatest lower bound of a non-empty set $M\subset{\mathbb{R}}$ which is bounded from below.

For $1<g\in\mathbb{N},$ let $r_n$ be the greatest integral multiple of $g^{-n}$ which is a lower bound of the set of real numbers $M.$ The claim is that the existence of $r_n$ and the Archimedean ordering of $\mathbb{R}$ proves that every number which is less than some lower bound of $M$ is also a lower bound of $M$.

I get why $r_n$ is a lower bound of $M$, and why every number less than $r_n$ is also a lower bound of $M$. But, for every $r_n$ there may exist some lower bound of $M$ which is greater than $r_n$. So what, if anything, is special about $r_n$ over any other rational lower bound of $M?$

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    $\begingroup$ Surely the fact that every number which is less than some lower bound of $M$ is also a lower bound of $M$ follows directly from the definition of lower bound? $\endgroup$
    – TonyK
    Commented May 9, 2019 at 14:35
  • $\begingroup$ That's why I'm wondering if there is anything special about $r_n$ or if we just need some rational number from which to establish relative ordering. I'm guessing that they are appealing to the well-ordering theorem which they proved for natural numbers a while back. $\endgroup$
    – Steven Thomas Hatton
    Commented May 9, 2019 at 14:44
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    $\begingroup$ What's special about $r_n$, besides being a lower bound of $M$, is that it's pretty close to $M$. Specifically, $r_n+g^{-n}$ is (by definition of $r_n$) not a lower bound of $M$. So there is an element of $M$ in the little interval $[r_n,r_n+g^{-n})$. (A consequence of this will be that $\lim_{n\to\infty}r_n$ exists and is the greatest lower bound of $M$.) $\endgroup$
    – Andreas Blass
    Commented May 9, 2019 at 16:53
  • $\begingroup$ @AndreasBlass If that is the intended meaning, then perhaps some nuance was lost in translation. I will argue that your interpretation is not a statement about the existence of $r_n.$ It is, instead, a statement about the definition of $r_n.$ That is, the claim is about the pair $\left<r_n,r_n+g^{-n}\right>.$ $\endgroup$
    – Steven Thomas Hatton
    Commented May 10, 2019 at 5:24

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