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I'm using the following definition of a Dedekind cut on an ordered set: a proper subset with no maximum and downward closed.

Let $(X,\le)$ be a totally ordered set. What is an $X$ having a Dedekind cut which is not an initial segment? I know that $X$ must be not complete, since completeness characterizes Dedekind cuts.

If I'm not mistaken, every Dedekind cut of $\Bbb Q $ is an initial segment, even if $\Bbb Q$ is not complete. I can't think of a good candidate.

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  • $\begingroup$ What's an initial segment? Is $\{a\in\Bbb Q: a<0\text{ or } a^2<2\}$ an initial segment in $\Bbb Q$? $\endgroup$ – Lord Shark the Unknown Aug 22 '17 at 10:08
  • $\begingroup$ @LordShark No, it's not. It is not the set of all and only the rationals smaller than any given rational, thank you. Please add that as an answer and I'll accept it. $\endgroup$ – Richard Aug 22 '17 at 10:26
  • $\begingroup$ @LordShark On the other hand, am I right in thinking that every Dedekind cut in $\Bbb Q^+$ is an initial segment? $\endgroup$ – Richard Aug 22 '17 at 15:05
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$\{a\in\Bbb Q: a<0\text{ or } a^2<2\}$ is a Dedekind cut of rationals that is not an initial segment.

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