Figure 1.15. Projecting from two dimensions to one
successive points were projections of (m, n) and (m + 1, n). We can then
identify L with the real line, with p corresponding to the origin.
The perceptive reader will have noticed that I did not say whether S
was constructed from the open unit square or the closed unit square. If L
goes through an integer point (m, n), then both (m + 1, n) and (m, n + 1)
will be on the boundary of S. Do we include one point, the other, or both in
Λ1? The set of points p for which this problem arises has measure zero, but
is still infinite. For each irrational α, let Ωα
be the set of tilings constructed
this way in which L does not contain any integer points. Then let Ωα be
the completion of Ωα
in the tiling metric.
If m and n are integers, then the tiling obtained from p = (x0 +m, y0 +n)
is exactly the same as that obtained from p = (x0, y0). Conversely, if (x0, y0)
and (x1, y1) do not differ by an integer vector, then the tilings built from
these two choices of p are not the same. This implies that Ωα
is isomorphic
to the 2-torus, minus the points corresponding to lines L that hit integers.
This set of singular lines is itself a line in the 2-torus, albeit one that winds
To understand the points of Ωα that are not in Ωα,
we consider tilings
based on the point p = (m + , n) in the limit of 0. As
tilings converge in the tiling metric and contain the projection of the point
(m + 1, n), but not the projection of (m, n + 1). As
the tilings
contain the projection of (m, n + 1), but not that of (m + 1, n). The two
limits are otherwise identical. The upshot is that there are two tilings in Ωα
that correspond to p = (m, n), and these points are not close in the tiling
metric! Topologically, the space Ωα consists of the 2-torus with an irrational
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