The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$, you can define a new distance $D'$ as $$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}.$$ It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets, what you end up with is the Jaccard distance (which actually is known by many other names as well).

For more information and references, check out Ken Clarkson's survey Nearest-neighbor searching and metric space dimensions (Section 2.3).

The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$, you can define a new distance $D'$ as $$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}.$$ It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets and empty set as $a$, what you end up with is the Jaccard distance (which actually is known by many other names as well).

For more information and references, check out Ken Clarkson's survey Nearest-neighbor searching and metric space dimensions (Section 2.3).

The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$, you can define a new distance $D'$ as

$$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}$$

It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets, what you end up with is the Jaccard distance (which actually is known by many other names as well).

For more information and references, check out Ken Clarkson's survey (Section 2.3)

The trick is to use a transform called the Steinhaus Transform. Given a metric $(X, d)$ and a fixed point $a \in X$, you can define a new distance $D'$ as $$D'(x,y) = \frac{2D(x,y)}{D(x,a) + D(y,a) + D(x,y)}.$$ It's known that this transformation produces a metric from a metric. Now if you take as the base metric $D$ the symmetric difference between two sets, what you end up with is the Jaccard distance (which actually is known by many other names as well).

For more information and references, check out Ken Clarkson's survey Nearest-neighbor searching and metric space dimensions (Section 2.3).