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Consider some gap (i,j) with i > 0 and j > 0. We already know {iα} < {jα}. The two points p1 = {(i − 1)α} and p2 = {(j − 1)α} are simply a left-shift of {iα} and {jα} by amount α, wrapping around 0 if necessary. If both or none of the two points wrap around 0, we see that they will form a region from p1 to p2 having the same length as gap (i,j). This region is (i − 1,j − 1).
We can prove that it is impossible to have only {iα} wrap around 0 with p2 > 0. Because then, 0 will be a point within this “wrapped region” from p1 to p2. That means, right-shift of p1, 0 and p2 by α will give us gap (i,j) but with point {α} enclosed by the gap. That is impossible since a gap cannot enclose a point.
Though, we still can have only {iα} wrap around 0 with p2 = 0. This is possible only with j = 1. In this case, we will define the region formed by p1 and p2 to be from endpoint p1 to 1, which is denoted by (i − 1,0). Since j = 1, it is (i − 1,j − 1).
Thus, we can always associate a well-defined region with points p1 and p2, which is precisely the region (i − 1,j − 1) and is of the same length as gap (i,j).
So, given any gap (i,j) with i > 0 and j > 0, we can obtain the region (i − 1,j − 1) from it. We will say, the gap (i,j) was “reduced” to the region (i − 1,j − 1).
The obtained region (i − 1,j − 1) may enclose a point and so need not be a gap. Suppose it encloses a point {kα} with k < N. But that means, {(k + 1)α}, which is a right-shift of {kα} by α, must be enclosed by gap (i,j). That is impossible since a gap cannot enclose a point. So we must have k = N. We conclude that, the region (i − 1,j − 1) encloses either no point (so, is a gap) or exactly one point {Nα}.
Further, since there is only one final point {Nα}, there must be exactly one region (f1,f2) which encloses this point and no other point.
Thus, the region (i − 1,j − 1) must be either a gap or the unique region (f1,f2).
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