Let $S^1$ denote point whose radius is 1 from the center. Metric is given by distance between two point is the shortest distance, that is the length metric. Prove that $S^1$ with this metric is not isometric to any subset of plane having the restriction of Euclidean metric?
In the length metric the maximum distance between points is $\pi$. So it cannot be isometric to any interval of length $\pi$ as there are two point in the circle which are at a distance more than $\pi/2$. Can anyone please how to do in general?
Take three points on the circle that form an equilateral triangle. The corresponding points in the plane would also have to form an equilateral triangle. Now take a point on the circle halfway between two of the first three points. Its corresponding point in the plane would have to be halfway between the two corresponding points in the plane. Indeed, if you take the three "halfway points" on the circle, the corresponding points in the plane would have to be the midpoints of the equilateral triangle in the plane. But on the circle, the three new points are as far from each other as the three original points are from each other, whereas in the plane the new points are closer to each other than the old points are to each other. So there's no isometry.