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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 $\Bbb Q$? $\endgroup$
    – Angina Seng
    Aug 22, 2017 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, 2017 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, 2017 at 15:05

1 Answer 1

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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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