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Given any gap (a,b) (a or b may be 0), we can repetitively reduce it to obtain regions (a,b), (a− 1,b− 1), (a− 2,b− 2) etc. until either of these two happen: (1) the obtained region (a′,b′) encloses a point, (2) the obtained region (a′,b′) has a′ or b′ equaling 0.
Upon termination, we will map the initial gap (a,b) to the final obtained region (a′,b′).
If (1) holds upon termination, region (a′,b′) must be the region (f1,f2).
Otherwise ((2) holds, (1) doesn’t hold), the region (a′,b′) must be a gap, either (0,b − a) (if b > a), or (a − b,0) (if a > b). Note that gap (a − b,0) represents the gap ending at 1. We know that there is exactly one gap (0,s) starting at 0, and exactly one gap (e,0) ending at 1. So, in this case, the region (a′,b′) must be either (0,s) or (e,0).
Since reducing a gap doesn’t change its length, the above process will end up mapping any gap to a gap/region of the same length. But we found that, any gap can map to only one of these three: the region (f1,f2), the gap (0,s), the gap (e,0). Thus, there can be only upto three distinct gap lengths.
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