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Motivation. Recently a group of researchers found an aperiodic monotile in $\mathbb{R}^2$, answering a long-standing question. There are many results in higher dimensions, so let's explore the lower dimension.

Formalisation. For any set $S\subseteq \mathbb{R}$ and $a\in \mathbb{R}$, we set $a+S = \{a+s: s\in S\}$, and for any collection of subsets ${\frak S}\subseteq {\cal P}(\mathbb{R})$ we let $a + {\frak S} = \{a+S: S\in{\frak S}\}$.

We say that a partition ${\frak T}$ of $\mathbb{R}$ is a mono-tiling if for any $T_0\neq T_1\in {\frak T}$ there is $a\in\mathbb{R}$ such that $T_0 = a + T_1$. We call the mono-tiling ${\frak T}$ aperiodic if for all $y\in \mathbb{R}$ we have ${\frak T}\neq y + {\frak T}$.

Question. Does $\mathbb{R}$ have an aperiodic mono-tiling?

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    $\begingroup$ If the 'tile' is bounded and measurable then no - every mono-tiling is periodic. This (and much more) is in a paper of Lagarias and Wang (math.hkust.edu.hk/~yangwang/Reprints/tiling-1d-inv.pdf) $\endgroup$
    – Thomas Bloom
    Commented Mar 24, 2023 at 10:08
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    $\begingroup$ Note that the notion of tiling discussed here (translational tiling) differs from the notion of tiling used in the R^2 result, where rotations and reflections of the tile are permitted. Its not immediately obvious to me that the Lagarias-Wang results continue to hold if one is permitted to reflect the one-dimensional tile. $\endgroup$
    – Terry Tao
    Commented Mar 25, 2023 at 9:47
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    $\begingroup$ @dominic Thank you for asking this nice question. $\endgroup$
    – mathworker21
    Commented Mar 27, 2023 at 19:49
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    $\begingroup$ An "aperiodic monotile" in the plane means a single tile that admits a tiling of the plane, but no periodic tiling. It seems you're asking for something much weaker in $\mathbb R$: a single tile that admits an aperiodic tiling (but it may also admit periodic tilings). In my opinion, this weaker version is still a good question, but the title is misleading. $\endgroup$
    – Will Brian
    Commented Mar 29, 2023 at 13:33
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    $\begingroup$ @TerryTao: If we allow reflections, I think it's possible to find an example of an aperiodic mono-tiling of $\mathbb R$. The idea is to use $\{1,5,6\}$ as a tile. Three copies of this tile can be used to cover $\{1,2,\dots,9\}$ in two distinct ways: there is a "two forwards one backwards" configuration and a "two backwards one forwards" configuration. Using these two configurations, we can tile $\mathbb Z$ in a way that codes an arbitrary infinite sequence of $0$'s and $1$'s. $\endgroup$
    – Will Brian
    Commented Mar 29, 2023 at 13:40

1 Answer 1

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It all depends on what one is willing to accept as a "tile". Bounded measurable set is out of question (if we use translations only, which the OP requested), so we have to drop either boundedness, or measurability. I leave playing with Hamel Basis to somebody else and just construct a countable union of intervals of infinite measure (so periodic tiling is certainly out of question). This is, of course, a too easy way to cheat, but one has to start somewhere :-)

Just start with an interval and place it somewhere. Now suppose we have congruent (by translation) tiles $T_1,\dots, T_n$ consisting of finitely many intervals placed already and have some hole (also an interval) we want to cover. Split this hole into $N$ small intervals of equal length so that their length is much less than the shifts between the already placed tiles. Now add to each tile $N$ intervals of the same length far away on the right and very far from each other in the positions to be chosen later. The tiles still won't overlap if "far" is far enough. Now use shifts of one tile to the left to cover the hole with different added intervals: just add them inductively left to right, so that when using each one, you'll move the rest of the tile (including the previously added intervals) far to the left where it cannot interfere with anything. You'll end up with more tiles consisting of more (but still finitely many intervals) and a filled hole. Then take the next hole you want to fill, and so on.

So, let's assume that we do not want a too easy way out and request that the measure of each tile is finite for the next attempt. Anyone up to the challenge?

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