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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:0Exercise 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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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}

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    How to show that $\frac{y}{2}$ isn't in $I$? – 2017-01-10
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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 $y$ . . . – 2017-01-10