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Ronnie Pavlov
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I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $a$ and positive integer $b$, your set $A(a,b)$ is dense. So $C(a)$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.

EDIT: adding a few more details at request of the OP. The easiest self-contained proof uses the so-called van der Corput lemma, which states that if $(x_n)$ is a sequence in $[0,1)$ and for all $h \in \mathbb{N}$, the sequence $y^{(h)}_n = (x_{n+h} - x_n) \pmod 1$ is equidistributed, then $(x_n)$ itself is equidistributed.

Now you can easily prove the following by induction on the degree: for any non-trivial polynomial $p(n)$ with integer coefficients and any irrational $a$, the sequence $p(n) a \pmod 1$ is equidistributed. This obviously implies that $\{p(n) a + k \ : \ k,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$.

I'm sure there are many resources, but here is a nice expository paper on van der Corput and more: https://arxiv.org/pdf/1510.07332.pdf.

I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $a$ and positive integer $b$, your set $A(a,b)$ is dense. So $C(a)$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.

I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $a$ and positive integer $b$, your set $A(a,b)$ is dense. So $C(a)$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.

EDIT: adding a few more details at request of the OP. The easiest self-contained proof uses the so-called van der Corput lemma, which states that if $(x_n)$ is a sequence in $[0,1)$ and for all $h \in \mathbb{N}$, the sequence $y^{(h)}_n = (x_{n+h} - x_n) \pmod 1$ is equidistributed, then $(x_n)$ itself is equidistributed.

Now you can easily prove the following by induction on the degree: for any non-trivial polynomial $p(n)$ with integer coefficients and any irrational $a$, the sequence $p(n) a \pmod 1$ is equidistributed. This obviously implies that $\{p(n) a + k \ : \ k,n \in \mathbb{Z}\}$ is dense in $\mathbb{R}$.

I'm sure there are many resources, but here is a nice expository paper on van der Corput and more: https://arxiv.org/pdf/1510.07332.pdf.

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Ronnie Pavlov
  • 2.6k
  • 10
  • 15

I'm not sure I'm understanding your question (as for others, I'm confused about 'universe'), but for every irrational $a$ and positive integer $b$, your set $A(a,b)$ is dense. So $C(a)$ is always the empty set.

This seems to already be stated in your post, but maybe it's related to not knowing what 'universe' means.