Let me begin by giving the relevant definitions. A set $A \subset \mathbb{N}$ is said to be central if and only if there exists a topological system $(X,T)$ (with $X$ a compact metric space, $T$ a continuous map on $X$) and a pair of points $x,y \in X$ with $y$ uniformly recurrent and proximal to $x$, such that for some neighbourhood $U$ of $y$ one has $A = \{n \in \mathbb{N} \ : \ T^n(x) \in U \}$. That $x$ and $y$ are supposed to be proximal means that there is a sequence $n\_i \to \infty $ such that $d(T^{n\_i}x,T^{n\_i}y) \to 0$. That $y$ is uniformly recurrent means that for any neighbourhood $U$ of $y$, the set $\{n \in \mathbb{N} \ : \ T^n(y) \in U \}$ is syndetic (has bounded gaps).


Quite amazingly the above definition of centrality turns out to be equivalent to $A$ being a member of a minimal idempotent ultrafilter in $\beta \mathbb{N}$. A good reference for more details is [this article](http://www.dm.unipi.it/~dinasso/ULTRABIBLIO/Bergelson%2520-%2520Minimal%2520idempotents%2520and%2520ergodic%2520Ramsey%2520theory%2520%25282003%2529.pdf) by Vitaly Bergelson. Let me give some details. The ultrafilters $\beta \mathbb{N}$ are a compact left-topological semigroup (meaning, addition is continuous in the left argument; and of course some authors have the reverse convention). Any such group has a unique minimal two-sided ideal, call it $K$, and an ultrafilter is miniamal if and only if it belongs to $K$. An ultrafilter $p$ is idempotent if and only if $p+p = p$. One of course needs some work to get the whole theory up and running, but it can be done.


It is not hard to verify that if $A$ is central then so is $dA$ for a positive integer $d$. For this, replace $X$ by $X' := X\_1 \cup \dots \cup X\_2 \cup X\_d$, with $X\_i = X\times\{i\}$ a copy of $X$, and define $T'$ by $T'(x,i) = (x,i+1)$ for $i < d$, and $T'(x,d) = (T(x),1)$, and finally let $x',\ y' = (x,1),\ (y,1)$. Likewise, it can be shown that $A/d = \{n \ : \ nd \in A\}$ is central by considering the system $(X,T^d)$, and the same points. Of course, there are a few things to check along the way that I skipped. Because centrality is a notion of largeness, in the sense that if $A$ is central and $A \subset B$ then $B$ is central, it is clear that also the set $\{ \lfloor \frac{n}{d} \rceil \ : \ n \in A \}$ is again central, because it contains $A/d$ ($\lfloor x\rceil$ stands for the closest integer).


Given the above, it appears to be reasonable that if $A$ is a central set, then central is also the set$\{ \lfloor \alpha n \rceil \ : \ n \in A \}$, where $\alpha \in \mathbb{R}^+$ is a real number. I think a possible way to attack the problem is by considering the "suspended" system $X' = X \times \mathbb{T}$ ($\mathbb{T} = \mathbb{R}/\mathbb{Z}$) with $T'(x,t) = (T^{\lfloor \alpha + t \rfloor}(x), \{ \alpha + t \} )$, $A' = A \times \mathbb{T}$, $x',\ y' = (x,0),\ (y,0)$. However, I can't quite seem to work out the details in a way that I would find convincing (and they say that oneself is the easiest to fool...).


**Question:** If $A$ is a central, then does $\{ \lfloor \alpha n \rceil \ : \ n \in A \}$ need to be central? 



Are irrational multiples of central sets again central?
The answer is yes, it was pointed out to me by Vitaly Bergelson that it follows from Theorem 6.1 (d) of a paper by Bergelson, Hindman and Kra (Trans. Amer. Math. Soc. 348 (1996), no. 3, 893–912).
The notation $g\_{\alpha,\gamma}[A]$ is defined in the previous section and means
$$g\_{\alpha,\gamma}[A]:=\{\lfloor \alpha n+\gamma\rfloor:n\in A\}$$
where $\lfloor\cdot\rfloor$ represents the floor function.
Note that $\lfloor x+1/2\rfloor=\lfloor x\rceil$


