Comparing Metric Balls
A metric ball is specified by a center and a positive radius. Changing the radius while keeping the center fixed has a direct effect: a point that is closer than the smaller radius is also closer than any larger radius. Changing the center requires an estimate, because a point’s distance to the new center may increase. The triangle inequality provides that estimate and gives a useful rule for proving one ball lies inside another.
Throughout, \((X,d)\) is a metric space, with its metric topology understood. For \(x\in X\) and \(r>0\), the open ball is \(B(x,r)=\{z\in X:d(x,z)<r\}\). The strict inequality is part of the definition: a point at distance exactly \(r\) is not in the ball. The triangle inequality says that \(d(u,w)\le d(u,v)+d(v,w)\) for all \(u,v,w\in X\).
Increasing the Radius
Proof. Let \(z\in B(x,r)\). By the definition of an open ball, \(d(x,z)<r\). Since \(r\le s\), we have \(d(x,z)<r\le s\), and therefore \(d(x,z)<s\). Thus \(z\in B(x,s)\). Every point of \(B(x,r)\) is in \(B(x,s)\), proving the inclusion. \(\square\)
This theorem asserts inclusion, not necessarily strict inclusion. The balls can be equal even when \(r<s\), because a metric space need not have points at every distance from the center. Whether the inclusion is proper depends on the particular space, center, and radii.
Worked Example: Increasing a Radius in the Real Line
Use the usual metric \(d(x,y)=|x-y|\) and the standard topology on \(\mathbb{R}\). Compare \(B(1,2)\) and \(B(1,4)\).
For the first ball, \(|z-1|<2\) is equivalent to \(-2<z-1<2\), so \(-1<z<3\). Hence \(B(1,2)=(-1,3)\). For the second, \(|z-1|<4\) is equivalent to \(-4<z-1<4\), so \(-3<z<5\). Hence \(B(1,4)=(-3,5)\).
Conclusion and check: \((-1,3)\subset(-3,5)\). For instance, every \(z\) in the first interval satisfies \(-1<z<3\), which also implies \(-3<z<5\). The endpoints are excluded: \(z=-1\) and \(z=3\) have distance exactly \(2\) from \(1\), so neither belongs to \(B(1,2)\).
Moving the Center
Suppose a point \(z\) is in a ball centered at \(x\), but we want to measure its distance from another center \(y\). Applying the triangle inequality along the route from \(z\) to \(x\) to \(y\) gives \(d(y,z)\le d(y,x)+d(x,z)\). Since \(d(x,z)\) is strictly less than the original radius, this estimate leads to a ball inclusion.
Proof. Assume \(d(x,y)+r\le s\), and take any \(z\in B(x,r)\). Then \(d(x,z)<r\). By the triangle inequality and symmetry of the metric,
Thus \(d(y,z)<s\), so \(z\in B(y,s)\). This holds for every \(z\in B(x,r)\), proving the inclusion. Notice that the hypothesis allows equality \(d(x,y)+r=s\): the strict inequality \(d(x,z)<r\) still makes the final distance strictly less than \(s\). \(\square\)
The condition \(d(x,y)+r\le s\) is sufficient: whenever it holds, the inclusion follows. In an arbitrary metric space it need not be necessary. A space may have too few points near the edge of a ball for the triangle-inequality bound to be sharp. The examples below illustrate both the usefulness and the limits of the rule.
Worked Example: Two Shifted Balls in the Real Line
In \(\mathbb{R}\) with the usual metric and standard topology, show that \(B(0,2)\subseteq B(3,5)\), including the case where the radius bound is an equality.
The center distance is \(d(0,3)=|0-3|=3\). The sum of this distance and the first radius is \(3+2=5\), which equals the target radius. Theorem 2 therefore gives \(B(0,2)\subseteq B(3,5)\).
We can also check the sets directly. The first ball is \((-2,2)\), and the second is \((-2,8)\), since \(|z-3|<5\) is equivalent to \(-5<z-3<5\), then to \(-2<z<8\). Every \(z\in(-2,2)\) satisfies \(-2<z<2<8\), so it lies in \((-2,8)\).
Check by the triangle inequality: for \(z\in B(0,2)\), \(|z|<2\), and hence \(|z-3|\le |z|+3<2+3=5\). Thus \(z\in B(3,5)\), as required. Equality in the radius bound causes no problem because the estimate for \(|z|\) is strict.
Worked Example: A Euclidean Ball in the Plane
Give a guaranteed ball centered at \(y=(3,4)\) that contains the Euclidean ball \(B(x,3)\), where \(x=(0,0)\) in \(\mathbb{R}^2\). Use the Euclidean metric and standard topology.
First compute the distance between the centers:
Theorem 2 applies with \(r=3\) and \(s=5+3=8\), because \(d(x,y)+r=5+3=8\). Thus \(B((0,0),3)\subseteq B((3,4),8)\). In pointwise form, if \(z\in B((0,0),3)\), then \(\|z-(0,0)\|<3\), and
Check: the computed center distance is \(5\), and the chosen target radius is \(8\), so the required inequality is \(5+3\le8\), which holds with equality. The strict inequality in the displayed estimate confirms that each point of the source open ball is inside the target open ball, not merely at its boundary.
What the Bound Does Not Say
It is important not to reverse Theorem 2 or interpret its hypothesis as an exact test for every metric space. The proof gives an upper estimate for the distance to the new center. It does not claim that points in the original ball attain distances arbitrarily close to \(d(x,y)+r\). In spaces with gaps between possible distances, a smaller target ball may already contain the entire original ball.
Worked Example: A Finite Metric with Distance Gaps
Let \(X=\{a,b,c\}\), with the topology induced by the metric given by \(d(a,b)=1\), \(d(a,c)=2\), and \(d(b,c)=2\), together with \(d(u,u)=0\) and symmetry. This is a metric: distances between distinct points are positive, symmetry holds, and each triangle inequality follows from \(2\le1+2\) or \(1\le2+2\). Its metric topology is discrete, since for each point \(u\), the ball \(B(u,1/2)\) contains only \(u\).
Take \(x=a\), \(y=b\), \(r=1/2\), and \(s=11/10\). Every point other than \(a\) is at distance at least \(1\) from \(a\), so \(B(a,1/2)=\{a\}\). From \(b\), the distances to \(a,b,c\) are respectively \(1,0,2\). Since \(1<11/10<2\), we have \(B(b,11/10)=\{a,b\}\). Therefore \(B(a,1/2)\subseteq B(b,11/10)\).
However, \(d(a,b)+r=1+1/2=3/2\), and \(3/2>11/10\). The hypothesis of Theorem 2 fails even though the inclusion holds. This verifies concretely that the theorem gives a sufficient condition, not a necessary one.
The same space also shows why increasing a radius need not strictly enlarge a ball. The distances from \(a\) to the other points are \(1\) and \(2\). Since \(1<11/10<19/10<2\), both \(B(a,11/10)\) and \(B(a,19/10)\) equal \(\{a,b\}\). The radius increased, but the ball did not change.
A Common Mistake and a Reliable Procedure
Common mistake. One may think the center-and-radius condition must be strict, and reject a valid inclusion whenever \(d(x,y)+r=s\). For the real-line balls in the previous example, that would require \(3+2<5\), which is false, and would incorrectly suggest that the inclusion had not been established. The correction is to use the definition of the source ball: \(d(x,z)<r\), not \(d(x,z)\le r\). Consequently the triangle inequality gives \(d(y,z)<d(x,y)+r=s\), so equality in the condition is sufficient. The direct calculation \(|z-3|\le |z|+3<2+3=5\) confirms the result.
For a concrete containment problem, use the following sequence of checks:
Compute \(d(x,y)\) using the metric in the problem; do not substitute a different notion of distance.
For a common center, compare \(r\) and \(s\). For different centers, test whether \(d(x,y)+r\le s\).
A point in \(B(x,r)\) satisfies \(d(x,z)<r\). Use that strict inequality to conclude \(d(y,z)<s\), even when \(d(x,y)+r=s\).
When the space is explicit, compute the balls or test their possible points. If the sufficient bound fails, that alone does not disprove containment.
These comparisons are useful whenever a geometric or algebraic argument needs to place one collection of points inside a controlled neighborhood of another center. The radius bound packages the triangle inequality into a form that can be applied without knowing the shape of the ball. The finite example also gives a practical warning: the bound is reliable as a sufficient test, but if it fails, a direct examination of the metric may still settle the inclusion.
Check Your Understanding
For each question, keep the strict inequality in the definition of an open ball explicit.
- In any metric space, prove that \(B(x,3)\subseteq B(x,7/2)\).
- In \(\mathbb{R}\) with the usual metric, does \(B(2,1)\subseteq B(5,4)\)? Verify by computing the center distance and applying Theorem 2.
- In a metric space, what conclusion follows from \(d(x,y)+r=s\) when \(r,s>0\)? Explain why the equality case works.
- Can \(B(x,r)=B(x,s)\) hold when \(r<s\)? Give a metric-space example and verify the two balls.
- If \(d(x,y)+r>s\), does this prove that \(B(x,r)\) is not contained in \(B(y,s)\)? Explain using the finite metric example or another explicit metric space.