Set of numbers that is essentially a discrete fractal

Question

I'm looking for a set $S \subseteq \mathbb N$ with the following property: There is an infinite set of linear shifts $L \subseteq \mathbb N$ and a finite set $F \subseteq \mathbb N$, such that $S$ is equal to the disjoint union of $F$ and $\{s+l:s \in S\}$ for each $l \in L$.

If we drop the requirement that $L$ is infinite, then $S=\{0,1,2,3,\dots\}$ works (with $L=\{1\}$ and $F=\{0\}$).

(In essence, I'm looking for a discrete fractal, a subset of $\mathbb N$ that is made up of infinitely many copies of itself (minus a finite number of points).)

Answer 1

Such a set cannot exist, given the requirement that the union be disjoint; in fact, it cannot even exist with $|L|\gt 1$.

Let $d_L$ be the GCD of $L$; then by the solution to the Frobenius problem, there exists some $N_0$ such that $\forall n\geq N_0, n\cdot d_L$ can be expressed as a linear combination of elements in $L$. Now, fix the choice of some element $s$ of $S$ (for instance, any of the members of $F$ will do); then this implies that every element $s+n\cdot d_L, n\geq N_0$ must be in $S$.

Now, choose two elements $\ell_1, \ell_2\in L$. Then $(s+N_0\cdot d_L+\ell_1)\in S$ (because this is $s+n\cdot d_L$ for an $n\geq N_0$), so $(s+N_0\cdot d_L+\ell_1+\ell_2)\in S+\ell_2$. Similarly, since $(s+N_0\cdot d_L+\ell_2)\in S$, $()s+N_0\cdot d_L+\ell_1+\ell_2)\in S+\ell_1$. But since this element is in $S+\ell_1$ and $S+\ell_2$, it's impossible for the union $\bigcup\{S+\ell: \ell\in L\}$ to be disjoint.

Answer 2

Edit: based on the final sentence in the post, I think Steven Stadnicki is correct that I have the wrong interpretation, but I will leave this answer anyway in case something useful can be gleaned from it.

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If $L$ contains $2$ members $a, b$, then you are requiring that $S + a \cap F = \emptyset$ and $S + a \cup F = S$, so $S + a = S \setminus F$. But the same thing is true of $S + b$, so $S + a = S + b = S \setminus F$.

Since $S \subset \Bbb N$, it has a least element $s$, and therefore the least element of $S + a$ is $s + a$, and of $S + b$ is $s+b$. But since these two sets are the same, $s + a = s + b$, and therefore $a = b$.

I.e. for any three non-empty sets $S, L, F$ with these properties, $L$ can have only one element. An infinite $L$ is impossible.