Open Balls

From Distance to Open Balls

In “Distance on the Real Line,” distance was defined by \(d(x,y)=|x-y|\). That definition measures the separation of two points. An open ball turns the same distance into a set: it collects all points whose distance from a chosen centre is less than a chosen positive radius. The strict inequality is important because points exactly one radius away are not included.

The interval description follows by solving the absolute-value inequality. For \(r>0\), the inequality \(|x-a|<r\) is equivalent to \(-r<x-a<r\). Adding \(a\) throughout gives \(a-r<x<a+r\). Thus a real-line open ball is exactly an open interval centred at \(a\), with endpoints \(r\) units to the left and right. This agrees with the description of open balls given in “The Real Line as a Metric Space.”

The centre always belongs to its ball, because \(d(a,a)=0<r\). The two endpoints \(a-r\) and \(a+r\) do not belong: each has distance exactly \(r\) from \(a\), not distance less than \(r\). A radius must be positive; with radius zero the strict inequality \(d(x,a)<0\) has no solutions, and a negative number cannot serve as a distance threshold for a ball of the intended kind.

Membership and the Meaning of Radius

To check whether a point \(x\) belongs to \(B_r(a)\), compare its distance from the centre to the radius. The point is inside precisely when \(d(x,a)<r\). Equality means it lies at the boundary of the interval, so it is not in the open ball; a distance greater than \(r\) places it outside.

Changing the radius while keeping the centre fixed changes the size of the ball. If \(0<r\leq s\), then \(B_r(a)\subseteq B_s(a)\): whenever \(d(x,a)<r\), we also have \(d(x,a)<s\) if \(r<s\), and the inclusion is immediate when \(r=s\). This nesting is useful when a problem asks for a ball small enough to fit inside another one.

When Does One Ball Fit Inside Another?

For balls with different centres, comparing radii alone is not enough. A smaller ball can fail to fit inside a larger one if its centre is too far away. On the real line, both the gap between the centres and the two radii give an exact containment test.

Proof. Using the interval formula, the inclusion in question is \((a-r,a+r)\subseteq(b-s,b+s)\). Inclusion of these open intervals holds exactly when the left endpoint of the first is no farther left than the left endpoint of the second, and the right endpoint of the first is no farther right than the right endpoint of the second. In inequalities, this says

Rearranging the first inequality gives \(s-r\geq b-a\), and rearranging the second gives \(s-r\geq a-b\). Together these are equivalent to \(s-r\geq\max\{a-b,b-a\}=|a-b|=d(a,b)\), or \(d(a,b)+r\leq s\). This proves the criterion in both directions. The endpoint comparisons are valid even though the intervals omit their endpoints: if either comparison failed, points of the first interval sufficiently close to the offending endpoint would lie outside the second; if both hold, every point strictly between the first interval’s endpoints lies strictly between the second interval’s endpoints. \(\square\)

The non-strict inequality in the criterion can be easy to overlook. When \(d(a,b)+r=s\), the smaller ball’s rightmost or leftmost limiting endpoint coincides with an endpoint of the larger ball, but neither endpoint belongs to the smaller open ball. Containment still holds. If the sum is greater than \(s\), points in the smaller ball near that limiting endpoint extend beyond the larger ball.

When Do Two Open Balls Overlap?

Containment asks whether every point of one ball lies in another. A different question is whether the balls have any point in common. Two balls overlap exactly when the distance between their centres is less than the sum of their radii.

Proof. First suppose there is a point \(z\in B_r(a)\cap B_s(b)\). Then \(d(a,z)<r\) and \(d(z,b)<s\). By the triangle inequality,

This proves that overlap implies the stated strict inequality. Conversely, suppose \(d(a,b)<r+s\). If \(a=b\), the common centre \(a\) belongs to both balls because \(0<r\) and \(0<s\). If \(a\neq b\), let \(z=(a+b)/2\), the midpoint of the centres. Then

Since \(d(a,b)=|a-b|<r+s\), we have \(\frac{|a-b|}{2}<\frac{r+s}{2}\). This alone does not guarantee that the midpoint is within each radius if \(r\) and \(s\) differ, so instead choose a point on the segment between the centres as follows. Set \(\alpha=d(a,b)>0\) and choose \(t=\min\{r,\alpha\}\). Define \(z=a+\frac{t}{\alpha}(b-a)\). Then \(d(a,z)=t\) and \(d(z,b)=\alpha-t\). If \(\alpha\leq r\), then \(d(a,z)=\alpha\leq r\), with strict inequality unless \(\alpha=r\); in the equality case, the assumption \(\alpha<r+s\) does not suffice for this choice. A more direct choice is to place \(z\) at distance \(u\) from \(a\), where \(u\) satisfies \(0\leq u\leq\alpha\), \(u<r\), and \(\alpha-u<s\). Such a \(u\) exists: the interval of allowable distances includes a choice because \(\alpha<r+s\); for example, take \(u=\max\{0,\alpha-s\}\). Then \(u<r\) by \(\alpha<r+s\), and \(\alpha-u\leq s\). If \(\alpha-u=s\), replace \(u\) by a slightly larger number still less than \(r\) and no greater than \(\alpha\); this is possible because \(\alpha-s=u<r\) and \(\alpha>u\). Thus we can arrange both \(u<r\) and \(\alpha-u<s\). Taking \(z=a+\frac{u}{\alpha}(b-a)\) gives \(d(a,z)=u<r\) and \(d(z,b)=\alpha-u<s\), so \(z\) belongs to both balls. This proves the converse and the theorem. \(\square\)

The strict inequality in this result reflects the fact that both balls are open. If \(d(a,b)=r+s\), the intervals can meet only at the two touching endpoints, and those endpoints are excluded. They therefore have no common point.

A Ball Has Room Around Each of Its Points

A useful feature of open balls is that a point strictly inside one is not trapped at its boundary. There is always a smaller positive-radius ball around that point that stays within the original ball. This property is a basic way to recognize the local meaning of openness.

Proof. Since \(y\in B_r(a)\), the definition gives \(d(a,y)<r\). Therefore \(\rho=r-d(a,y)>0\). Let \(z\in B_\rho(y)\). Then \(d(y,z)<\rho\), and the triangle inequality gives

Thus \(z\in B_r(a)\). Since every \(z\in B_\rho(y)\) has this property, \(B_\rho(y)\subseteq B_r(a)\), as required. \(\square\)

The chosen radius is the full remaining distance from \(y\) to the boundary as measured from the centre. Any smaller positive radius also works. A common mistake is to choose \(\rho=r\) for every \(y\) in the ball: that need not work, because a point close to the boundary may have only a small amount of room left. The subtraction \(r-d(a,y)\) accounts for how much of the original radius has already been used.

Open balls provide a concrete language for describing closeness. Their interval form makes calculations on \(\mathbb{R}\) explicit, while the containment and intersection criteria turn geometric questions into inequalities. The smaller-ball property explains why strict distance bounds are useful: every point inside a ball has some positive tolerance before it can leave that ball.