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Let $B=\{b_n: n\geq 1\}$ be a set of positive integer numbers with positive upper asymptotic density and let $\alpha$ be a real irrational number.

Is it true that $\{b_n \alpha\}$ is equidistributed mod 1?

Of course this is true for $B=\mathbb{N}$ (equiditribution theorem) and even for zero density sets as the set of prime numbers (Vinogradov) or the perfect squares.

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  • $\begingroup$ I'm voting to close this question as off-topic because it has an obvious negative answer (given already). $\endgroup$
    – fedja
    Commented Mar 31, 2018 at 1:58

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No, take $B=\{k:\{k\alpha\}>\frac12 \}.$

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  • $\begingroup$ Thanks, I didn't see that. But for example, if I have two asymptotic equivalent sequences $x_n$ and $y_n$. If $x_n$ is equidistributed then so $y_n$? $\endgroup$
    – Jean
    Commented Mar 29, 2018 at 22:22
  • $\begingroup$ If by "asymptotic equivalent" you mean that $x_n-y_n$ tends to $0$ as $n\to\infty$, then yes, it is an easy consequence of the definition of equidistribution that $\{x_n\}$ is equidistributed if and only if $\{y_n\}$ is. $\endgroup$
    – Greg Martin
    Commented Mar 30, 2018 at 1:45
  • $\begingroup$ @GregMartin I mean $x_n/y_n\to 1$ as $n\to \infty$. $\endgroup$
    – Jean
    Commented Mar 30, 2018 at 9:03
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    $\begingroup$ @Jean you mean equidistributed modulo 1? Of course no, take $x_n=n\sqrt{2}, y_n=[x_n]$. $\endgroup$
    – Fedor Petrov
    Commented Mar 30, 2018 at 9:50

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