/sci/ - Science & Math

>>9895263
It's easily provable that every point on ON can be uniquely mapped to a corresponding segment on or above it, for some power x^n, and vice versa, so the length of ON is fully exhausted by the sequence of lengths 1 + x + x^2 + ... and this sequence of lengths is fully exhausted by the length of ON, so they are equal.

This pic illustrates (an initial finite segment of) my mapping. You can't point to a single point of ON that isn't covered by the full mapping.