Boundedness Is Not Total Boundedness

Two Different Kinds of Size Control

Boundedness and total boundedness both constrain how large a metric space is, but they ask different questions. Boundedness asks whether all points stay within one fixed distance of a chosen point. Total boundedness asks whether, at every prescribed small scale, finitely many balls suffice to cover the space. The difference is important: an infinite space can have all its points at distance at most one from each other and still require infinitely many balls of radius one-half.

Throughout, $(X,d)$ denotes a metric space. An open ball of radius $r>0$ about $x\in X$ is $B(x,r)=\{y\in X:d(x,y)<r\}$. In statements about a subset $A$ of a metric space, we give $A$ the subspace metric: the distance between $x,y\in A$ is the same $d(x,y)$ as in the ambient space. Thus the centers of balls covering $A$ can be taken in $A$.

Definition: A metric space $X$ is bounded if there are $x_0\in X$ and $R>0$ such that $d(x,x_0)\le R$ for every $x\in X$. It is totally bounded if, for every $\varepsilon>0$, there are finitely many points $x_1,\ldots,x_N\in X$ such that $X\subseteq\bigcup_{i=1}^N B(x_i,\varepsilon)$. Such a finite set of centers is called a finite $\varepsilon$-net. The empty space is both bounded and totally bounded, with the covering condition satisfied by the empty family of balls.

The strict inequality in the definition of an open ball matters: membership in $B(x,\varepsilon)$ means distance strictly less than $\varepsilon$. Boundedness uses a single fixed radius, whereas total boundedness must work for every positive $\varepsilon$, however small.

Proposition: Every nonempty metric space is bounded if and only if its distances are bounded above; that is, there is a finite $D$ such that $d(x,y)\le D$ for all $x,y\in X$.

Proof. Suppose $X$ is bounded, with center $x_0$ and radius $R$. For any $x,y\in X$, the triangle inequality gives $d(x,y)\le d(x,x_0)+d(x_0,y)\le R+R=2R$. Thus all pairwise distances have a finite upper bound. Conversely, suppose $d(x,y)\le D$ for all $x,y\in X$, where $X\ne\varnothing$. Choose $x_0\in X$. Then $d(x,x_0)\le D$ for every $x\in X$, so $X$ is bounded (if necessary, take any positive radius greater than or equal to $D$). $\square$

A Bounded Space That Is Not Totally Bounded

The discrete metric makes the distinction exact. Each pair of different points is separated by distance one, so the space has uniformly limited diameter. At any radius less than or equal to one, however, an open ball contains only its center.

Worked Example: The Infinite Discrete Metric

Let $X=\mathbb{N}=\{1,2,\ldots\}$ and define $d(m,n)=0$ if $m=n$, and $d(m,n)=1$ if $m\ne n$. This is a metric: it is nonnegative and symmetric, and distance zero occurs exactly when the points agree. For the triangle inequality, if $m=n$, then $d(m,n)=0\le d(m,k)+d(k,n)$. If $m\ne n$, the left side is 1; for any $k$, at least one of $m\ne k$ or $k\ne n$ holds, so the right side is at least 1.

Boundedness. Fix $x_0=1$. For every $n\in X$, $d(n,1)=0$ when $n=1$, and $d(n,1)=1$ when $n\ne1$. Therefore $d(n,1)\le1$ for every $n$, and $X$ is bounded. In fact, every distance in $X$ is either 0 or 1.

Failure of total boundedness. Take $\varepsilon=\tfrac12$. For any center $m\in X$, the condition $d(m,n)<\tfrac12$ holds exactly when $n=m$: the distance is 0 for $n=m$, and it is 1, which is not less than $\tfrac12$, for $n\ne m$. Hence $B(m,\tfrac12)=\{m\}$. A union of finitely many such balls contains only finitely many natural numbers, so it cannot cover $X$. Thus $X$ is not totally bounded.

Check: any finite collection of $N$ centers covers at most $N$ points at radius $\tfrac12$, while $X$ has infinitely many points. This verifies failure using the finite-net definition, not just the appearance of the metric.

This example disproves the general implication “bounded implies totally bounded.” Boundedness limits distances between points but places no limit on how many points may be separated from one another by a fixed positive distance.

Finite Nets in Euclidean Space

In $\mathbb{R}^n$ with its Euclidean metric, bounded sets do have the stronger property. The reason is that a bounded region can be divided into finitely many small boxes. From each box that meets the set, choose one point of the set; those points form a finite net. The next result proves this construction, including the case of a set that does not fill its surrounding box.

Theorem: Every bounded subset $A\subseteq\mathbb{R}^n$, with the Euclidean metric, is totally bounded.

Proof. If $A=\varnothing$, it is totally bounded by the empty covering. Suppose $A\ne\varnothing$. Since $A$ is bounded, it lies in some cube $[-M,M]^n$ for a finite $M\ge0$. If $M=0$, then $A\subseteq\{(0,\ldots,0)\}$, so one ball of any positive radius centered at its point covers $A$. Now suppose $M>0$, and fix $\varepsilon>0$. Choose $m\in\mathbb{N}$ so large that $2M\sqrt n/m<\varepsilon$. Divide each interval $[-M,M]$ into $m$ subintervals of length $2M/m$, assigning shared endpoints to just one adjacent subinterval so that the resulting boxes partition the cube. There are $m^n$ boxes. For each box $C$ that meets $A$, choose one point $a_C\in A\cap C$. This requires choices from only finitely many nonempty sets. If $x\in A\cap C$, then in each coordinate the difference between $x$ and $a_C$ is at most $2M/m$. Therefore $$ \|x-a_C\|\le \sqrt{n(2M/m)^2}=\frac{2M\sqrt n}{m}<\varepsilon. $$ Thus every $x\in A$ lies in the open ball $B(a_C,\varepsilon)$ for its box $C$. The finitely many selected centers cover $A$, proving total boundedness. $\square$

Worked Example: A Bounded Interval

Consider $A=(0,1)\subseteq\mathbb{R}$ with the usual metric. First, it is bounded: for every $x\in(0,1)$, $0<x<1$, so $|x-\tfrac12|<\tfrac12$. Now exhibit a finite net at the specific scale $\varepsilon=\tfrac14$.

Use the three centers $\tfrac16,\tfrac12,\tfrac56$, all of which belong to $A$. They correspond to the three intervals $(0,\tfrac13]$, $(\tfrac13,\tfrac23]$, and $(\tfrac23,1)$, with endpoint assignment chosen to avoid overlap concerns. For a point $x$ in any one of these intervals, its distance from that interval’s center is at most $\tfrac16$. Since $\tfrac16<\tfrac14$, the point lies in the open ball of radius $\tfrac14$ about that center. Consequently these three balls cover $A$.

Check: the largest distance needed is $\tfrac16$, and $\tfrac16<\tfrac14$ because $2<3$ after multiplying both sides by 12. The inequality is strict, as required for membership in the open balls. This gives an explicit finite net at the stated scale; the theorem supplies such a net for every positive scale.

Worked Example: A Rectangle in the Plane

Let $Q=[-1,1]\times[0,1]\subseteq\mathbb{R}^2$, with Euclidean distance, and use $\varepsilon=\tfrac12$. Divide the horizontal interval $[-1,1]$ into four intervals of width $\tfrac12$, and the vertical interval $[0,1]$ into two intervals of width $\tfrac12$. Assign each shared boundary to one of the adjoining intervals, so every point belongs to one of the resulting eight rectangular cells.

Choose the midpoint of each cell as a center. The four horizontal midpoint coordinates are $-\tfrac34,-\tfrac14,\tfrac14,\tfrac34$, and the two vertical midpoint coordinates are $\tfrac14,\tfrac34$. Thus there are $4\cdot2=8$ centers, all in $Q$. For any point in a cell, its horizontal and vertical coordinate differences from that cell’s center are each at most $\tfrac14$. Its squared distance from the center is therefore at most $(\tfrac14)^2+(\tfrac14)^2=\tfrac18$. Since $\tfrac18<\tfrac14=(\tfrac12)^2$, its distance is strictly less than $\tfrac12$. The eight open balls of radius $\tfrac12$ centered at these points cover $Q$.

Check: the distance bound is $\sqrt{\tfrac18}=\tfrac{\sqrt2}{4}$, and $\tfrac{\sqrt2}{4}<\tfrac12$ because squaring the positive sides gives $\tfrac18<\tfrac14$. The count is eight cells, obtained from four horizontal pieces and two vertical pieces. This verifies both the covering radius and the number of centers used.

A Common Mistake: Confusing the Radius Conditions

A common error is to use a diameter bound as if it automatically gave a finite cover at a chosen radius. For the infinite discrete metric above, every two points have distance at most 1. A mistaken argument might say that the ball of radius 1 around $1$ covers the whole space because all distances are at most 1.

Wrong calculation: for $n\ne1$, $d(n,1)=1$, and “$1\le1$, so $n\in B(1,1)$.” The last step is false: the ball condition is $d(n,1)<1$, and $1<1$ is false. In fact $B(1,1)=\{1\}$. More generally, even knowing that a space has a finite diameter gives no finite covering by small-radius balls: the radius $\tfrac12$ balls in this example are all singletons.

The Euclidean theorem works for a different reason than boundedness alone. A bounded Euclidean set fits into a finite-dimensional box, and that box can be subdivided into finitely many cells whose diameters are as small as required. For the infinite discrete metric space, no matter how many distinct centers are selected, a ball of radius $\tfrac12$ covers only its own center.

Takeaway: Total boundedness always implies boundedness: for a nonempty space, take a finite cover by balls of radius $1$, and use the triangle inequality and the finite distances among their centers to obtain a finite bound on all pairwise distances. The reverse implication fails for general metric spaces, as the infinite discrete metric demonstrates. In Euclidean space, bounded subsets are totally bounded because finite grids provide covers at every positive scale.

Check Your Understanding

For each question, keep the strict inequality in the definition of an open ball in view.

In the infinite discrete metric, compute $B(m,\varepsilon)$ for $0<\varepsilon\le1$, and explain why no finite collection of these balls covers $\mathbb{N}$.
Is every finite metric space totally bounded? Give a finite $\varepsilon$-net for an arbitrary positive $\varepsilon$.
For a bounded subset of $\mathbb{R}^n$, why must the number of grid cells used in the proof be finite?
Give an example of a bounded subset of $\mathbb{R}$ and construct a finite net for one specified positive radius.
Explain why the statement $d(x,y)\le R$ does not, by itself, imply $y\in B(x,R)$.

Boundedness Is Not Total Boundedness

What You'll Learn