Class Notes on Lipschitz Extension from Finite Sets

Proof. Suppose that π ∈ Sn is such that the minimal j ∈ {1, . . . , n} for which aπ(j) ∈ BX(x, r+dX(x, y)) actually satisfies aπ(j) ∈ BX(x, r−dX(x, y)). Hence j r (x) = j and therefore aπ(j) = ar (x). Also, dX(aπ(j), y) 6 dX(aπ(j), x) + dX(x, y) 6 r, so j r (y) 6 j. But dX(x, ajπ r (y)) 6 dX(y, ajπ r (y)) + dX(x, y) 6 r+ dX(x, y), so by the definition of j we must have j r (y) > j. Thus j π r (y) = j, so that a π r (y) = aπ(j) = a π r (x). We have shown that if in the random order that π induces on A the first element that falls in the ball BX(x, r + dX(x, y)) actually falls in the smaller ball BX(x, r−dX(x, y)), then ar (y) = ar (x). If π is chosen uniformly at random from Sn then the probability of this event equals |A∩BX(x, r−dX(x, y))|/|A∩BX(x, r+dX(x, y))|. Hence, |{π ∈ Sn : ar (x) = ar (y)}| n! > |A ∩BX(x, r − dX(x, y))| |A ∩BX(x, r + dX(x, y))| .