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Call a number abnormal if its decimal expansion doesn't feature every digit an infinite number of times. Call a triangle in ${\Bbb R}^2$ abnormal if at least one of its angles spans an abnormal fraction of $2\pi$ radians.

Assuming the Continuum Hypothesis, by the obvious transfinite induction, one gets a set $M$ so small that no three of its points form an abnormal triangle, but so large that every point of ${\Bbb R}^2$ either falls in $M$ or else occurs as the midpoint of two points in $M$.

Does such an $M$ exist in ZFC?

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    $\begingroup$ Can't you just perform your recursion in length $\frak{c}$ rather than length $\omega_1$? This will avoid the need for CH. $\endgroup$
    – Joel David Hamkins
    Commented Feb 26, 2018 at 3:34
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    $\begingroup$ Not something I feel comfortable doing in ZFC :) $\endgroup$
    – David Feldman
    Commented Feb 26, 2018 at 5:54
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    $\begingroup$ May I confess my pedagogical motivation: to devise the most elementary sounding statement I could that depends on CH for its proof. $\endgroup$
    – David Feldman
    Commented Feb 27, 2018 at 21:06
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    $\begingroup$ @David I have a candidate for that: $\mathrm{CH}$ holds if and only if the plane $\mathbb R^2$ can be covered by 3 clouds. (See here.) With a little more work you can actually squeeze out a characterization of $2^{\aleph_0} = \aleph_n$ for all $n < \omega$ -- a result we rediscovered over lunch during least year's Arctic Set Theory Conference. $\endgroup$
    – Stefan Mesken
    Commented Feb 28, 2018 at 10:43
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    $\begingroup$ There is a mean called Davies trees which could enable you to remove CH. One reason in which usually we cannot go up beyond \aleph_1 when CH fails is that we cannot get a countable elementary substructure at stage \omega_1 if of course a proof using countable elemenatry substructures exists. Now, everything here is definable and you may use elementary substructure in order to get M under CH. Now Davies tree gives you again a countable elementary substructure at level \omega_1. You can take a look at the following interesting paper. arxiv.org/abs/1705.06195 $\endgroup$
    – Rahman. M
    Commented Mar 5, 2018 at 12:03

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