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Let $(K,+_K,*_K,{\leq}_K)$ be an non-Archimedean totally ordered field, i.e. ${\exists}k\,{\in}\,K:k\,{\neq}\,0_K\,{\land}\,{\forall}n\,{\in}\,\mathbb{N}:n{\vert}k{\vert}\,{\leq}\,1_K$. ${\vert}k{\vert}$ is defined as $k$ for $0_K\,{\leq}\,k$ and $-k$ otherwise.

How to prove that this field can't have the least upper bound property?

I have already proved that there exists a subset $I={\{x:0<x\,{\land}\,{\forall}n\,{\in}\,\mathbb{N}:n*_Kx<1_K}\}$ of $K$ for any such field and that $1_K$ is its upper bound. Now I don't know how to prove that there is no least upper bound $y$ for it. The source of my question, see Exercise 3., advises to consider $2y$ if $y\,{\in}\,I$ and $\frac{y}{2}$ if not, but I still don't know what to do.

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asdasdf
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1 Answers1

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Suppose $y$ is an upper bound for $I$; you want to show that $y$ is not a least upper bound.

First, you need to argue that $y$ can't be in $I$. Suppose $y\in I$; then what can you say about $2y$? Why does that contradict the assumption that $y$ is an upper bound for $I$?

So $y\not\in I$. OK, now we want to find a smaller upper bound for $I$. So let's consider something smaller but not too much smaller - $y\over 2$ is a good place to start. Certainly ${y\over 2}<y$ (why?), so it will be enough to show that $y\over 2$ is in fact an upper bound for $I$. Can you show that ${y\over 2}$ is not in $I$? (HINT: use the fact that $y\not\in I$.) Now, how does that imply that ${y\over 2}$ is an upper bound for $I$?

Noah Schweber
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  • How to show that $\frac{y}{2}$ isn't in $I$? – asdasdf Jan 10 '17 at 21:47
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    @asdasdfsss Based on the fact that $y\in I$, do you see how to find an $n\in\mathbb{N}$ such that $n{y\over 2}\ge1$? HINT: Think about the corresponding $m$ for $y$ . . . – Noah Schweber Jan 10 '17 at 22:17