cute
one quote "the tile admits uncountably many tilings": is it even possible to have a set of tiles that aperiodically tile the plane without having uncountably many tilings?
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RT @cs_kaplan
In a new paper, David Smith, Joseph Myers, Chaim Goodman-Strauss and I prove that a polykite that we call "the hat" is an aperiodic monotile, AKA an einstein. We finally got down to 1! https://arxiv.org/abs/2303.10798 4/6
https://twitter.com/cs_kaplan/status/1637996332475359232
ok i guess i'd stipulate "only exists an aperiodic tiling" (i think necessary) and "snaps on to a grid" (i suspect unnecessary) here
like the space S of aperiodic tilings comes with a compact topology where the basis of open sets is "we are given where a finite number of where the tiles are is"
and Z^2 acts freely on this set by translations because the tiling is aperiodic
hrmm. construct for any ordinal α S_α
S_0=S
S_{α+1} is S_α minus all its isolated points
S_λ is the intersection of S_α α<λ
S_α is always closed, nonempty, and infinite
S_α is also eventually constant if for no other reason than that eventually you run out of points. call the resulting set T
pretty sure because T is perfect and this is a nice enough topological space T has cardinality continuum
i think this reasoning still works if we're talking tiles where there's only a finite number of ways to glue 2 tiles together sensibly
you don't quite get the group action but you do still get a compact topology on all the ways to build a tesselation around some base tile
you feel like there has to be a one-line proof citing some theorem in model theory or something
i feel like you can repeat the argument in full generality with some bespoke metric on the space of all tilings where the tiles are allowed to slide around but that seems like effort
and one point in this space always gives you an infinite number because aperiodic