Maximal packings of the integers

Once you have it for some packing (and the standard mumbo-jumbo about achieving (as true density) the supremum of upper densities over all packings, periodic or not), the (upper) density of certain not too long intervals $A---A$ ($A$ here is some not too short packing pattern) is positive (because every sufficiently long interval contains an interval of that sort with some $A$ and there are only finitely many possible $A$ and finitely many possible pattern lengths). If $---A$ has strictly lower density than the supremum, you can improve the upper density of your packing by replacing each occurrence of $A---A$ by a single $A$, which we assumed impossible. Otherwise $---A---A---A---$ is the packing you want.

The difference between this and higher dimensions is that the evolving boundary is finite.

I think I could make this explanation tighter, but maybe this will be clear enough: I will describe a particular finite directed graph $G.$ A packing corresponds to a certain walk in $G$ and, if it makes the (or an) optimum choice at a certain node, then there is no reason not to have it do the same thing every future time it arrives that node.

Given $B$ with minimum element $0$ and maximum element $b$, a packing is some union $P=\bigcup_{i \in \mathbb{Z}}B+x_i.$ We can stipulate that $x_i \lt x_j$ when $i \lt j.$ Then an initial segment is $P_j=\bigcup_{i \lt j}B+x_i.$ For convenience, call an integer an empty space or full space of $P$ or $Pj$ according as it is not or is an element. To get to from $P_j$ to $P_{j+1}$ a choice has to be made of $x_{j} \in [x_{j-1}+1,x_{j-1}+b+1]$ with $B+x_{j}$ disjoint from $P_j.$

The state at $P_j$ is the pattern of empty and full spaces relevant to the choice of of $x_j.$ Specifically, the set $S_j \subseteq\{{1,2,\cdots,b-1\}}$ determined as follows. let $z$ be leftmost empty space greater than $x_{j-1}.$ Then $S_j=\{{t-z \mid t \in P_j ,t \gt z\}}.$

The graph $G$ consists of the possible states with a directed edge from $S$ to $S'$ if a single copy of $B$ could turn a partial packing with state $S$ to one with state $S'.$ Then a packing corresponds to an infinite (in the past and future) walk in this graph.