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Suppose a gap (a,b) maps to (0,s). Then, we must have b > a and b − a = s. Similarly, if the gap maps to (e,0), we must have a > b and a − b = e.
Notice that the region (f1,f2) encloses a single point {Nα}. So, it consists of two adjacent gaps (f1,N) and (N,f2). But each of these two gaps must map to either (0,s) or (e,0) (they can’t map to region (f1,f2)). Since gap (f1,N) has f1 < N, it must map to (0,s). Similarly, gap (N,f2) must map to (e,0). Thus we must have: N − f1 = s and N − f2 = e. Hence, f2 − f1 = s − e.
If a gap (a,b) maps to region (f1,f2), we must have b−a = f2 −f1 (both a and b were decremented by same amount to give f1 and f2). So due to above, b − a = s − e.
In summary, each gap (a,b) must have the difference b−a one among: s, −e, s − e.
Also note that the gap-length corresponding to region (f1,f2) is simply the sum of lengths of the gaps (0,s) and (e,0).
Theorem 1 and 2 in article [1] include the same results we just proved above; s and e are referred as u1 and uN there, and the three gap lengths as L1, L2 and L1 + L2.
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