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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).)

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  • $\begingroup$ Do you mean the usual disjoint union or simply that F and l + S are always disjoint disjoint? $\endgroup$ – William Elliot May 15 '18 at 20:05
  • $\begingroup$ @WilliamElliot that l+S are always disjoint from $F$ and each other. $\endgroup$ – PyRulez May 15 '18 at 20:09
  • $\begingroup$ FYI - You are misusing "fractal" here. Self-similarity is a technique for building fractals, but it is not what fractals are. $\endgroup$ – Paul Sinclair May 16 '18 at 16:20
  • $\begingroup$ Incidentally, this question becomes a little more interesting (though also with several trivial answers) if you allow S to contain not just exact copies of itself, but 'scaled' copies of itself; that actually maps more directly onto the traditional way of looking at fractals like the Sierpinski gasket. Since 'shrinking' a set in $\mathbb{Z}$ corresponds to making it less dense, it corresponds most directly to the set $d\cdot S$ for some $d$. (cont.) $\endgroup$ – Steven Stadnicki May 16 '18 at 19:38
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    $\begingroup$ This means that you should probably look at $S$ which are the disjoint union of $F$ and the sets $S_i=\ell_i+d_i\cdot S = \{\ell_i+d_i\cdot s : s\in S\}$, for some set of ordered pairs $\langle d_i, \ell_i\rangle$ representing the 'scales' and 'offsets' of the copies of $S$ within $S$ itself. $\endgroup$ – Steven Stadnicki May 16 '18 at 19:41
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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.

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    $\begingroup$ I had just worked up a similar solution, then realized a simpler variation works: By the union requirement $S + \ell_1 \subset S \implies (S + \ell_1) + \ell_2 \subset S + \ell_2$, and similarly $(S + \ell_2) + \ell_1 \subset S + \ell_1$. Since these two sets are in fact the same, it cannot be that $S+\ell_1 \cap S + \ell_2 = \emptyset$. $\endgroup$ – Paul Sinclair May 17 '18 at 10:11
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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.


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.

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  • $\begingroup$ I think OP's intention is that $S=\cup\{F, \{s+\ell: s\in S, \ell\in L\}\}$, rather than that $\forall \ell\in L, S=F\cup\{s+\ell: s\in S\}$, but that's not entirely clear from the question. $\endgroup$ – Steven Stadnicki May 16 '18 at 4:15
  • $\begingroup$ @StevenStadnicki that's correct. Sorry if the question was vague. $\endgroup$ – PyRulez May 16 '18 at 16:21

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