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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