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Problem

If a and b are real numbers , define$ a\prec b$ if $b-a$ is positive and rational . Show that this is a strict partial order on $\mathbb{R}$.

Attempt

It is a strict partial order because

It is easy to check that this relation is transitive and non-reflexive .

Maximal simple ordered subsets :$ \mathbb{N}, \mathbb{Q}, \mathbb{Z}$

Please comment whether this solution is right or not.

Note

This relation is not comparable .For example $1,\pi \in \mathbb{R} $ but $1\nprec \pi$ and $\pi \nprec 1$.

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  • $\begingroup$ As far as I know transitive and non-reflexive are necessary and sufficient for being a strict partial order. Being (non-)comparable plays no part in that matter. Also a linear order is a strict partial order (with an extra property). $\endgroup$
    – drhab
    Aug 18, 2018 at 13:30
  • $\begingroup$ @drhab Answer edited. $\endgroup$
    – Bluey
    Aug 18, 2018 at 14:01
  • $\begingroup$ @drhab Is maximal simply ordered subsets correct? Are there any more? $\endgroup$
    – Bluey
    Aug 18, 2018 at 14:01
  • $\begingroup$ I think that maximal simply ordered sets are here the sets: $a+\mathbb Q$ for $a\in\mathbb R$ (among them is $\mathbb Q$ but not $\mathbb N$ or $\mathbb Z$, they are not maximal). $\endgroup$
    – drhab
    Aug 18, 2018 at 14:09
  • $\begingroup$ Ok. So they are ordered but not maximal. Right? $\endgroup$
    – Bluey
    Aug 18, 2018 at 14:12

1 Answer 1

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For every $a\in\mathbb R$ the set $a+\mathbb Q:=\{a+q\mid q\in\mathbb Q\}$ is evidently simply ordered by $\prec$.

If $x\notin a+\mathbb Q$ then $a$ and $x$ are not comparable so it cannot be that a set $S\subseteq\mathbb R$ exists that is simply ordered and satisfies $a+\mathbb Q\cup\{x\}\subseteq S$.

This proves that $a+\mathbb Q$ is maximal in being simply ordered by $\prec$.

If conversely $M\subseteq\mathbb R$ is simply ordered and $a\in M$ then we find that $b-a\in\mathbb Q$ for every $b\in M$ or equivalently that $M\subseteq a+\mathbb Q$.

Then if $M$ is maximal in being simply ordered by $\prec$ it cannot be that $M$ is a proper subset of $a+\mathbb Q$ so in that case we conclude that $M=a+\mathbb Q$.

Proved is now that a subset $M$ of $\mathbb R$ is maximal simply ordered by $\prec$ if and only if $M=a+\mathbb Q$ for some $a\in\mathbb R$.

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