Why These Balls Have Sharp Thresholds
In the usual metric on the real line, changing a radius gradually changes the interval that forms an open ball. In a discrete metric, there are only two possible distances between points, so most radius changes do nothing. Ultrametrics also have sharp thresholds: their strong triangle inequality forces intersecting balls to fit inside one another rather than overlap in an irregular way. We will calculate these balls directly and prove the nesting property.
Throughout, an open ball in a metric space \((X,d)\) is \(B_d(x,r)=\{y\in X:d(x,y)<r\}\), where \(r>0\). The topology on each metric space below is the topology induced by its metric. The inequality in the definition is strict; in particular, a point at distance exactly \(r\) is not in the ball.
Balls in the Discrete Metric
For a set \(X\), the discrete metric is \(d(x,y)=0\) when \(x=y\) and \(d(x,y)=1\) when \(x\ne y\). For any fixed center \(x\), every point of \(X\) is therefore at one of just two distances from \(x\): distance zero if it is \(x\), and distance one otherwise.
Proof. If \(0<r\le1\), then \(d(x,x)=0<r\), so \(x\in B_d(x,r)\). For any \(y\ne x\), \(d(x,y)=1\ge r\), so \(y\notin B_d(x,r)\). Thus the ball is \(\{x\}\), including at the endpoint \(r=1\). If \(r>1\), then \(d(x,y)\le1<r\) for every \(y\in X\), so the ball is all of \(X\). These conclusions also hold when \(X=\{x\}\), since then \(\{x\}=X\). \(\square\)
Worked Example: A Four-Point Discrete Space
Let \(X=\{a,b,c,d\}\) with the discrete metric. Find \(B_d(b,0.8)\), \(B_d(b,1)\), and \(B_d(b,1.2)\). In each case, membership means the distance from \(b\) is strictly less than the stated radius.
The distances from \(b\) are \(d(b,b)=0\), \(d(b,a)=1\), \(d(b,c)=1\), and \(d(b,d)=1\). For radius \(0.8\), only the distance zero is less than the radius, so \(B_d(b,0.8)=\{b\}\). For radius \(1\), the other distances equal the radius and fail the strict inequality, so \(B_d(b,1)=\{b\}\). For radius \(1.2\), all four distances are less than the radius, so \(B_d(b,1.2)=\{a,b,c,d\}\).
Check: At radius \(1\), \(d(b,a)=1\) is not less than \(1\), so \(a\notin B_d(b,1)\). At radius \(1.2\), \(d(b,a)=1<1.2\), and the same calculation holds for \(c\) and \(d\); hence all four points are included.
The discrete metric is itself an ultrametric. If \(x=z\), then \(d(x,z)=0\le\max\{d(x,y),d(y,z)\}\). If \(x\ne z\), it is not possible for both \(x=y\) and \(y=z\); at least one of \(d(x,y)\) and \(d(y,z)\) is \(1\). Thus their maximum is \(1=d(x,z)\). This gives a first example of the strong triangle inequality.
Intersecting Balls in an Ultrametric
The defining inequality has a useful consequence that does not hold for arbitrary metric balls. If two ultrametric balls intersect, the one with the smaller radius is contained in the one with the larger radius. The centers need not be the same.
Proof. Choose a point \(w\in B_d(x,r)\cap B_d(y,s)\). To prove the inclusion, take any \(z\in B_d(x,r)\). The strong triangle inequality, applied with intermediate point \(w\), gives \(d(z,y)\le\max\{d(z,w),d(w,y)\}\). A second application gives \(d(z,w)\le\max\{d(z,x),d(x,w)\}\). Since \(z,w\in B_d(x,r)\), both \(d(z,x)<r\) and \(d(x,w)<r\), so \(d(z,w)<r\le s\). Also \(d(w,y)<s\), because \(w\in B_d(y,s)\). The maximum of these two numbers is strictly less than \(s\), so \(d(z,y)<s\), which means \(z\in B_d(y,s)\). This proves the inclusion. If the radii are equal, apply the inclusion in both directions to conclude that intersecting balls contain one another and hence are equal. \(\square\)
The strict inequalities matter in this proof: even when \(r=s\), every point in either ball is at distance strictly less than that common radius from its center. The conclusion is about open balls as defined here; it does not replace the membership test with a non-strict inequality.
Worked Example: Balls in a Four-Point Ultrametric
Let \(X=\{a,b,c,d\}\). Define \(d(x,x)=0\), \(d(a,b)=d(b,a)=1/4\), \(d(c,d)=d(d,c)=1/4\), and let every distance between a point in \(\{a,b\}\) and a point in \(\{c,d\}\) be \(1\). Find the balls centered at \(a\) for radii \(1/4\), \(0.7\), and \(1\).
First check the ultrametric inequality. In a triple with points from both pairs, at least two of the three distances are \(1\): if the triple has two points from one pair, their mutual distance is \(1/4\) and their two distances to the third point are \(1\); if all three distances cross between pairs, those distances are \(1\). A triple cannot contain three distinct points from just one of the two-point pairs. Thus, whenever a distance in a triple is \(1\), the maximum of the other two distances is also \(1\); a distance of \(1/4\) is at most the maximum of the other two distances. Triples with repeated points satisfy the inequality because a zero distance is at most every distance and the two nonzero distances, if present, are equal. Hence \(d\) is an ultrametric.
The distances from \(a\) are \(d(a,a)=0\), \(d(a,b)=1/4\), \(d(a,c)=1\), and \(d(a,d)=1\). At radius \(1/4\), the point \(b\) is excluded because \(1/4\) is not strictly less than \(1/4\); therefore \(B_d(a,1/4)=\{a\}\). At radius \(0.7\), \(0<0.7\) and \(1/4<0.7\), while \(1\not<0.7\), so \(B_d(a,0.7)=\{a,b\}\). At radius \(1\), the two points at distance \(1\) are still excluded, and \(B_d(a,1)=\{a,b\}\).
Check: The ball at radius \(0.7\) is contained in the ball at radius \(1\), and both are \(\{a,b\}\). The radius-\(1/4\) ball is \(\{a\}\), not \(\{a,b\}\), because the boundary point \(b\) has distance exactly \(1/4\).
A Divisibility-Based Ultrametric
A useful model on the integers records how many powers of a prime divide the difference between two integers. Fix a prime \(p\). For a nonzero integer \(n\), let \(v_p(n)\) be the largest nonnegative integer \(k\) such that \(p^k\) divides \(n\), and put \(v_p(0)=\infty\). Define \(d_p(m,n)=p^{-v_p(m-n)}\), with \(p^{-\infty}=0\). The space is \(\mathbb{Z}\) with this metric and its induced topology.
For completeness, this formula gives an ultrametric. It is nonnegative, equals zero exactly when \(m=n\), and is symmetric because \(m-n\) and \(n-m\) have the same divisibility by powers of \(p\). If \(p^k\) divides both \(m-n\) and \(n-\ell\), it divides their sum \(m-\ell\). Consequently \(v_p(m-\ell)\ge\min\{v_p(m-n),v_p(n-\ell)\}\), with the same statement valid when one of the differences is zero. Taking negative powers reverses this comparison and gives \(d_p(m,\ell)\le\max\{d_p(m,n),d_p(n,\ell)\}\). The strong triangle inequality implies the ordinary triangle inequality, since the maximum of two nonnegative numbers is at most their sum. Thus \(d_p\) is an ultrametric.
The possible nonzero distances are \(1,p^{-1},p^{-2},\ldots\). The open-ball condition \(d_p(a,n)<r\) means \(p^{-v_p(a-n)}<r\), so it becomes more restrictive when the radius passes one of these distance values. In particular, when \(0<r\le1\), there is a unique integer \(N\ge1\) such that \(p^{-N}<r\le p^{-(N-1)}\). The strict inequality then says exactly that \(v_p(a-n)\ge N\), or \(p^N\) divides \(a-n\). When \(r>1\), every distance is less than \(r\), so the ball is all of \(\mathbb{Z}\).
Worked Example: Congruence-Class Balls for \(p=2\)
Use \(d_2(m,n)=2^{-v_2(m-n)}\) on \(\mathbb{Z}\) to compute \(B_{d_2}(3,1)\), \(B_{d_2}(3,1/2)\), and \(B_{d_2}(3,1/4)\). Here \(v_2(k)\) counts the largest power of \(2\) dividing a nonzero integer \(k\).
For radius \(1\), the condition is \(2^{-v_2(3-n)}<1\). This holds exactly when \(v_2(3-n)\ge1\), which means \(2\) divides \(3-n\). Therefore \(B_{d_2}(3,1)=3+2\mathbb{Z}\), the integers congruent to \(3\) modulo \(2\).
For radius \(1/2\), the condition is \(2^{-v_2(3-n)}<1/2\). It holds exactly when \(v_2(3-n)\ge2\), so \(4\) divides \(3-n\) and \(B_{d_2}(3,1/2)=3+4\mathbb{Z}\). For radius \(1/4\), the condition \(2^{-v_2(3-n)}<1/4\) requires \(v_2(3-n)\ge3\), so \(8\) divides \(3-n\) and \(B_{d_2}(3,1/4)=3+8\mathbb{Z}\).
Check: The integer \(7\) differs from \(3\) by \(4\), so \(d_2(3,7)=2^{-2}=1/4\). Thus \(7\in B_{d_2}(3,1/2)\), since \(1/4<1/2\), but \(7\notin B_{d_2}(3,1/4)\), since \(1/4\) is not less than \(1/4\). The computed congruence classes give exactly these membership results.
A Common Threshold Mistake
A frequent error is to treat the boundary of an open ball as included. For example, one might calculate \(B_{d_2}(3,1/2)\) by allowing distance \(1/2\), and conclude that it is \(3+2\mathbb{Z}\). That is incorrect: \(n=5\) satisfies \(d_2(3,5)=2^{-1}=1/2\), but the ball requires \(d_2(3,5)<1/2\), so \(5\) is excluded. The correct answer is \(3+4\mathbb{Z}\). Likewise, in the discrete metric, the ball of radius \(1\) is a singleton, not the whole space. At each threshold, test the strict inequality against the distance value itself.
The two main patterns can be summarized as follows.
| Metric setting | Radius behavior | Ball shape |
|---|---|---|
| Discrete metric | \(0<r\le1\) or \(r>1\) | Singleton or whole space |
| Ultrametric, general | Intersecting balls with \(r\le s\) | The radius-\(r\) ball lies inside the radius-\(s\) ball |
| Divisibility metric on \(\mathbb{Z}\) | \(p^{-N}<r\le p^{-(N-1)}\) | A congruence class modulo \(p^N\) |
Check Your Understanding
For each question, use the strict inequality in the definition of an open ball.
- In the discrete metric on \(X=\{u,v,w\}\), compute \(B_d(u,1)\) and \(B_d(u,1.01)\).
- Suppose two ultrametric balls of radii \(0.2\) and \(0.6\) intersect. Which ball is contained in the other?
- In the four-point ultrametric above, compute \(B_d(c,0.25)\) and \(B_d(c,0.5)\).
- For \(d_2\) on \(\mathbb{Z}\), describe \(B_{d_2}(5,1/8)\) as a congruence class.
- In \(d_2\), decide whether \(9\) belongs to \(B_{d_2}(1,1/4)\), and show the distance calculation that determines your answer.