Show that an open ball of radius 0 < r ≤ 1 in a discrete metric space contains only its centre. 

Solution:

In a discrete metric space, the distance between any two distinct points is 1. Therefore, if we consider an open ball of radius 0 < r ≤ 1, it will contain only its center, which we can denote as x.

To see why this is true, let y be any point in the ball, i.e., d(x, y) < r. Since the distance between any two distinct points is 1 in a discrete metric space, it follows that either y = x or d(x, y) = 1.

If y = x, then y is the center of the ball and the ball contains only one point.

If d(x, y) = 1, then y is not the center of the ball, and we can consider the open ball of radius 1/2 centered at y. Any point in this ball will have distance at most 1/2 from y, which implies that it has distance at least 1/2 from x (since d(x, y) = 1). Therefore, this ball does not contain x, which contradicts the assumption that y is in the original ball.

Hence, the only point in the open ball of radius 0 < r ≤ 1 is its center, and the statement is proven.