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Consider the rationals $\mathbb{Q}$ with the usual order $\leq$. Now let $A$ be a subset of $\mathbb{Q}$, such that foreseen with the induced order $\leq$, $(A,\leq)$ is a dense linear order.

Furthermore, suppose that the complement of $A$ in $\mathbb{Q}$ is not finite.

  • (@) Does there exist an order-automorphism $\alpha$ of
    $(\mathbb{Q},\leq)$ such that $A \subset \alpha(A) \ne A$ ?

(The condition on $\vert \mathbb{Q} \setminus A \vert$ might be too naive, but I am looking for "as mild as possible" conditions on $A$ so that (@) would be true.)

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  • $\begingroup$ Can we answer this using the ideas at mathoverflow.net/questions/9901/…? E.g. starting from $\mathbb{Q}$, "do the ordinary middle-third construction of the Cantor set, except that whenever you delete the n-th (numbered by level and then left to right, say) middle-third interval leave in exactly n points from that interval. Let's call the resulting set" $\mathbb{Q} \,\backslash\, A$. Would that work? $\endgroup$
    – user44143
    Sep 24, 2019 at 16:49
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    $\begingroup$ Answer to my question in the above comment: no, that does not provide a counterexample; even though $\mathbb{Q} \,\backslash \, A$ is rigid, with no non-trivial isomorphisms, it has structure-preserving maps into itself. Answer to the question in the post: None of the obvious counterexamples work, but I don’t see how to prove it. $\endgroup$
    – user44143
    Sep 25, 2019 at 8:26
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    $\begingroup$ I guess the answer is true, but the argument should deal carefully with the complement, which is not assumed dense and is therefore highly not homogeneous. When $A$ has dense complement it's easy: $\mathrm{Aut}(\mathbf{Q},\le)$ acts transitively on dense subsets with dense complement. $\endgroup$
    – YCor
    Sep 27, 2019 at 9:38
  • $\begingroup$ @YCor It’s so far from homogeneous that we’d need a structure theory. Dealing with the complement” is basically proving that any countably infinite linear order has a non-trivial endomorphism. $\endgroup$
    – user44143
    Sep 30, 2019 at 7:56
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    $\begingroup$ @MattF. In fact it's true that every countably infinite linear order has a non-trivial endomorphism. This is an old theorem of Dushnik and Miller (Concerning similarity transformations of linearly ordered sets, Bull. AMS 46 (1940), no. 4, pp. 322–326). See also this question. But that result does not suffice to "handle the complement", since an arbitrary endomorphism of the complement of $A$ may fail to extend to an endomorphism of $\mathbb{Q}$. $\endgroup$ Oct 3, 2019 at 19:04

1 Answer 1

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Let $f(x)=2x$ for any rational $x$, and let $A$ consist of all rationals in $(0,1)$.

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  • $\begingroup$ wrong original "answer"; another try. $\endgroup$
    – Don Monk
    Sep 27, 2019 at 2:22
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    $\begingroup$ But the question is...can we make a construction like this for any dense coinfinite A? $\endgroup$
    – user44143
    Sep 27, 2019 at 2:35

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