Finite Nets as a Measure of Small-Scale Size
A set can be bounded without being coverable by finitely many small metric balls. Total boundedness captures the latter property: at every chosen scale, finitely many points suffice to approximate every point of the set. In finite-dimensional Euclidean space, boundedness does guarantee this finite approximation property. The construction uses a finite grid and works even when the set is neither closed nor made of grid points.
Let \((X,d)\) be a metric space, with its metric topology, and let \(A\subseteq X\). For \(a\in X\) and \(r>0\), write \(B_d(a,r)=\{x\in X:d(x,a)<r\}\). The inequality in this definition is strict: a point at distance exactly \(r\) is not in the open ball.
Some conventions allow the centres of a net to lie anywhere in \(X\), rather than requiring them to belong to \(A\). We use centres in \(A\). This convention makes a net a finite selection of points from the set being approximated and gives the direct equivalence below.
Nets and Finite Ball Covers
Proof. Suppose \(N\subseteq A\) is a finite \(\varepsilon\)-net. By the definition of a net, for each \(x\in A\) there is \(n\in N\) with \(d(x,n)<\varepsilon\), which says precisely that \(x\in B_d(n,\varepsilon)\). Hence the balls centred at the finitely many members of \(N\) cover \(A\). Conversely, suppose \(A\subseteq\bigcup_{i=1}^m B_d(a_i,\varepsilon)\) with each \(a_i\in A\). For every \(x\in A\), membership in this union gives some \(i\) such that \(d(x,a_i)<\varepsilon\). Thus the finite set \(\{a_1,\ldots,a_m\}\) is an \(\varepsilon\)-net. The assertion about total boundedness follows by applying this equivalence for each positive \(\varepsilon\). If \(A=\varnothing\), the empty family of balls covers \(A\), and the empty set is its net, so the equivalence includes this case. \(\square\)
The radius is part of the requirement: a finite cover at one radius says nothing by itself about covers at every smaller radius. Also, a net does not assert that its centres are the only points of \(A\); it asserts that every point is within the specified strict distance of some centre.
Worked Example: A Net for a Half-Open Interval
Take \(A=[0,1)\subseteq\mathbb{R}\) with the standard metric and choose \(\varepsilon=\frac14\). Consider the four centres \(N=\{\frac18,\frac38,\frac58,\frac78\}\). They all belong to \(A\). Divide \(A\) into the four half-open intervals \([0,\frac14)\), \([\frac14,\frac12)\), \([\frac12,\frac34)\), and \([\frac34,1)\). In each interval, choose its midpoint as the corresponding centre.
For any \(x\) in one of these intervals, its distance from that interval’s midpoint is at most \(\frac18\). For example, if \(x\in[\frac14,\frac12)\), then \(\frac14\le x<\frac12\), so \(|x-\frac38|\le\frac18\). The same endpoint calculation applies to the other three intervals. Since \(\frac18<\frac14\), every such \(x\) is in the open ball of radius \(\frac14\) about its centre. Thus \(N\) is a finite \(\frac14\)-net.
Check: the endpoints \(0\) and points arbitrarily close to \(1\) are included in this verification: \(|0-\frac18|=\frac18<\frac14\), and for \(\frac34\le x<1\), \(|x-\frac78|\le\frac18<\frac14\). The missing endpoint \(1\) needs no coverage because \(1\notin A\).
Constructing Nets in Euclidean Space
The grid construction works because a bounded region in \(\mathbb{R}^n\) meets only finitely many grid cells, and each cell can be made to have diameter less than the requested radius. The centres need not be the grid vertices: in each cell that meets \(A\), select a point of \(A\) in that cell. This ensures the centres belong to the set being covered.
Proof. If \(A=\varnothing\), it has the empty finite \(\varepsilon\)-net for every \(\varepsilon>0\). Now assume \(A\ne\varnothing\), fix \(\varepsilon>0\), and let \(n\ge1\). Since \(A\) is bounded, there is \(M>0\) such that \(A\subseteq[-M,M]^n\). Set \(\delta=\varepsilon/(2\sqrt n)>0\). For each integer \(k\), consider the half-open interval \([k\delta,(k+1)\delta)\). Every coordinate of every point in \(A\) lies in one of these intervals, and only finitely many indices can occur: if \(-M\le t\le M\) and \(t\in[k\delta,(k+1)\delta)\), then \(k\delta\le M\) and \((k+1)\delta>-M\), which bound \(k\) above and below.
Taking products of these intervals in the \(n\) coordinates gives finitely many cells meeting \(A\). For each such cell, select one point of its intersection with \(A\), and let \(N\) be the finite set of selected points. To check the radius, take any \(x\in A\) and the selected point \(a\in N\) from the same cell. In each coordinate, \(x\) and \(a\) lie in the same interval of length \(\delta\), so their coordinate difference has absolute value less than \(\delta\). Therefore
Since both sides are nonnegative, this gives \(\|x-a\|<\varepsilon/2<\varepsilon\). Every \(x\in A\) is thus within \(\varepsilon\) of a point of \(N\). The set \(N\subseteq A\) is finite, so it is a finite \(\varepsilon\)-net. Since \(\varepsilon>0\) was arbitrary, \(A\) is totally bounded. \(\square\)
Worked Example: A Net for the Unit Square
Let \(A=[0,1]^2\subseteq\mathbb{R}^2\), and choose \(\varepsilon=\frac12\). Use the 16 centres \((i/3,j/3)\) for integers \(i,j\in\{0,1,2,3\}\). Each centre lies in \(A\).
For any coordinate \(t\in[0,1]\), one of \(0,\frac13,\frac23,1\) is within \(\frac16\) of \(t\): these values divide \([0,1]\) into intervals of length \(\frac13\), and each interval’s endpoints are \(\frac13\) apart. In each coordinate of a point \(x=(x_1,x_2)\), choose such a centre coordinate. The resulting grid point \(a\) satisfies
Thus \(\|x-a\|<\frac12\), and the 16 grid points form a finite \(\frac12\)-net. The strict final inequality matters because the net is defined using open balls.
Check: at a corner such as \((0,0)\), the centre \((0,0)\) gives distance \(0<\frac12\). The distance bound also covers points on the edges; no separate assumption that \(x\) is an interior point was used.
Worked Example: A Net for an Infinite Bounded Set
In \(\mathbb{R}\), let \(A=\{1/m:m\in\mathbb{N}\}\) and choose \(\varepsilon=\frac13\). This set has infinitely many points, so we will cover its tail with one ball and handle its first four points with individual centres. Take \(N=\{1,\frac12,\frac13,\frac14,\frac15\}\subseteq A\).
The points \(1,\frac12,\frac13,\frac14\) are themselves centres, so each is at distance \(0<\frac13\) from \(N\). For every integer \(m\ge5\), we have \(0<1/m\le1/5\), and therefore
Hence every remaining point is within \(\frac13\) of the centre \(\frac15\), and \(N\) is a finite \(\frac13\)-net. The displayed calculation also checks the tail case \(m=5\), where the distance is \(0\).
Boundedness Is Not the Same as Total Boundedness
In Euclidean space, boundedness is enough to produce finite nets because the finite grid construction is available. This implication is not true for arbitrary metric spaces. Conversely, total boundedness does imply boundedness for every nonempty metric space, as the next proposition shows.
Proof. Let \(A\ne\varnothing\) be totally bounded, and choose \(a_0\in A\). Because \(A\) is totally bounded, it has a finite \(1\)-net \(N=\{a_1,\ldots,a_m\}\subseteq A\). Since \(A\ne\varnothing\), a net covering \(A\) cannot be empty, so \(m\ge1\). For every \(x\in A\), some \(a_i\) satisfies \(d(x,a_i)<1\). The triangle inequality gives \(d(x,a_0)\le d(x,a_i)+d(a_i,a_0)<1+d(a_i,a_0)\). The finitely many distances \(d(a_i,a_0)\) have a maximum \(D\), so \(d(x,a_0)<1+D\) for every \(x\in A\). This is a uniform bound on the distance from \(a_0\), proving that \(A\) is bounded. \(\square\)
Common pitfall. A bounded set need not be totally bounded in an arbitrary metric space. Let \(X\) be an infinite set with the discrete metric \(d(x,y)=0\) if \(x=y\) and \(d(x,y)=1\) if \(x\ne y\), and take \(A=X\). The triangle inequality holds: if \(x\ne z\), then at least one of \(x\ne y\) or \(y\ne z\) holds, so \(d(x,y)+d(y,z)\ge1=d(x,z)\); if \(x=z\), the triangle inequality follows from nonnegativity. The set \(A\) is bounded because all distances are at most \(1\). But for \(\varepsilon=\frac12\), each open ball \(B_d(x,\frac12)\) is just \(\{x\}\): equal points have distance \(0<\frac12\), and distinct points have distance \(1\not<\frac12\). Finitely many such balls cover only finitely many points, so they cannot cover the infinite set \(X\). Thus \(A\) is not totally bounded. The finite-grid theorem depends on Euclidean structure, not merely on boundedness.
Total boundedness is useful as a quantitative form of compactness: it says that at every positive scale there is a finite description of the set’s location. The grid proof makes that description explicit in finite-dimensional Euclidean space. It does not by itself assert that a set is compact; compactness involves an open-cover condition, and further results connect these notions under additional hypotheses.
Check Your Understanding
For each question, keep track of the strict inequality in the definition of an open ball.
- For \(A=[0,2]\subseteq\mathbb{R}\) and \(\varepsilon=\frac12\), give a finite \(\varepsilon\)-net whose centres lie in \(A\), and verify coverage.
- Explain why allowing centres outside \(A\) is a different convention, and show how to obtain centres in \(A\) when \(A\ne\varnothing\) and an ambient-centred cover of radius \(\varepsilon/2\) is given.
- Prove directly from the definition that every finite subset of a metric space is totally bounded.
- In the infinite discrete metric space, determine the open balls of radius \(2\). Does the whole space have a finite \(2\)-net?
- In the grid proof for \(\mathbb{R}^n\), where is boundedness used, and where is the Euclidean norm used?