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(All the sets in this discussion are sets of positive integers.) I say a set $V$ is “rational” if it satisfies one of the following conditions:

I. $V$ is finite (possibly empty);

II. $V$ can be written as the union of a finite (possibly empty) set and finitely many arithmetic progressions of infinite length.

Letting $\Delta$ denote the symmetric difference operation, I say that a set $W$ is “almost rational” if there exists a rational set $V$ such that $V\Delta W$ has zero natural density. Clearly, every $W$ of density $0$ or $1$ is almost rational (with $V$ being any finite set in the former case, and $V$ being any set whose complement is finite, in the latter case). My question is: are there any sets of positive natural density which are not almost rational?

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    $\begingroup$ How about the complement of powers of two? Or diagonalize against an enumeration of arithmetic progressions with the next element removed at least twice as large as the previous removed element. Gerhard "Can Think Of Thinner Sets" Paseman, 2019.10.08. $\endgroup$
    – Gerhard Paseman
    Commented Oct 8, 2019 at 20:16
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    $\begingroup$ OK then. How about Thue Morse? The set of all positive numbers with an even number of bits in their binary expansion? Gerhard "Is Grabbing At Bits Now" Paseman, 2019.10.08. $\endgroup$
    – Gerhard Paseman
    Commented Oct 8, 2019 at 20:25
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    $\begingroup$ $\{n:\sqrt{2}n$ mod $2\;\in [0,1]\}$ $\endgroup$
    – YCor
    Commented Oct 8, 2019 at 21:16
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    $\begingroup$ Any set with irrational natural density works. $\endgroup$
    – LeechLattice
    Commented Oct 9, 2019 at 3:31
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    $\begingroup$ Indeed, write (mod $\pi$) instead of (mod 2) in my previous comment and this yields irrational density. $\endgroup$
    – YCor
    Commented Oct 9, 2019 at 4:45

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