I think the answer is a no. If you take the topologists sine curve, together with adjoining the two end points of the sine curve via a different route, then the resulting set is rectifiably connected but has infinite diameter (with respect to the rectifiable curve metric you introduced).
My guess is to look at a closed, bounded subset of R^2 with an inward cusp. Like the closure of the bounded component of a standard cardioid. There can be no bilipschitz map of this set to the same set but with the length metric.
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I think the answer is a no. If you take the topologists sine curve, together with adjoining the two end points of the sine curve via a different route, then the resulting set is rectifiably connected but has infinite diameter (with respect to the rectifiable curve metric you introduced).
Wonderful! That will do.
My guess is to look at a closed, bounded subset of R^2 with an inward cusp. Like the closure of the bounded component of a standard cardioid. There can be no bilipschitz map of this set to the same set but with the length metric.
Yes. But is the length distance not compact? For example, is it not true that every sequence has a convergent subsequence?
Good work from you!
We came across your course on Udemy and it's such an outstanding course with so much opportunity of gaining a stronger feet amidst other courses in your niche.
Are you open to for us to explore the path to the success of your course together?