Metric Spaces, The Open Ball Topology

The Open Ball Topology

If a set of points has a valid metric, as described in the previous page, then the set has an induced topology. The set, with its metric topology, is called a metric space. The topology is also called the open ball topology, and is defined as follows.

For each point p and each positive ε, let the set of points satisfying |x,p| < ε, i.e. the open ball about p, act as a base for the topology. We need to show this is a valid base. Let p be a point inside two balls. Thus p is within distance q of the center of the first ball, and within distance r of the center of the second ball, where q and r are the two radii. Let d be the difference between q and the distance from p to the center of the first ball. Think of d as the distance from p to the "edge" of the first ball. Find a second distance separating p from the edge of the second ball. Let ε be smaller than either of these distances. Now all the points within ε of p are contained within the two larger balls, thanks to the triangular inequality. Thus we have a valid base and a valid topology.

Don't forget to define the empty set as open; it isn't characterized by an open ball.

Rational Radii

We can restrict radii to rational numbers; the topology is unchanged. Consider an open ball with radius r, where r is an irrational number. Every point p in the ball is a certain distance away from the edge of the ball, and can be enclosed in a smaller ball with a rational radius. Thus our ball of radius r is covered by open sets, and is open, reproducing the original topology.


A metric space is separable if it possesses a countable dense set. We will show this is equivalent to having a countable base.

Let the metric space T have a dense set D. Let the base be the set of open balls with centers at D and rational radii, a countable set. We need to show that such a base covers every open set. Take any point x in an arbitrary open set W, build a ball of radius e about x, such that the ball is inside W, find a point y from D within e/3 of x, choose a rational radius between e/3 and e/2, and we have a base set centered at y that contains x, and is contained in W. Thus W is covered, and we have a countable base.

Conversely, given a countable base, choose any point from each base set to build the dense set D. Every open set contains base sets, and points from D.

Discrete Metric Space

The discrete topology can be given the metric 1 or 0, for points different or the same respectively. Thus every point is an open set, with ε set to ½. This is not the only discrete metric. If |x,y| is bounded above some ε for x different from y, then every point is open.


If a metric exists for a given space, with a given topology, and that metric defines the same topology, then the space is metrizable. We may as well think of it as a metric space.

Compressing the Metric

If f() is a distance metric, and g() is a monotonically increasing real-valued function, with g(0) = 0, and g′′ ≤ 0, then g(f()) also determines a metric. Log(x+1) is an example. The only tricky part is the triangular inequality.

Let a b and c be the lengths of the sides of a triangle, thus c ≤ a+b. Since g is increasing, g(c) ≤ g(a+b). We only need show that g(a)+g(b) is even larger than g(a+b).

Hold a fixed and let x be a continuous variable that starts at 0 and increases through b and beyond. Let h(x) = g(a)+g(x)-g(a+x). Note that h(0) = 0. The derivative of h is g′(x)-g′(a+x). If this is negative then g′(a+x) is larger than g′(x). By the mean value theorem, g′′ is positive somewhere between x and a+x. Yet g′′ is never positive, hence h′ is never negative. Again, the mean value theorem tells us h can never get smaller, else h′ would be negative. Thus h(b) is at least as large as h(0), which is 0. Thus g(a)+g(b) ≥ g(a+b), and the triangular inequality holds.

If g is strictly increasing, with g′ = 0 only at isolated points (if ever), Then g is invertible. The set of points that are less than ε distance from a given point p, under the original metric, can also be described as the set of points that are less than g(ε) distance from p, in the new metric. The open balls are the same, and the topology is the same.

If you want to make an unbounded space bounded, set g(x) = x/(1+x), or any other bounded, increasing function.