Small Scales Do Not Supply Missing Limits
Total boundedness says that a metric space can be covered by finitely many small balls, no matter how small the chosen radius is. Completeness asks a different question: must every Cauchy sequence in the space converge to a point that still belongs to it? A space can satisfy the first condition and fail the second. Dense subsets of complete spaces provide a systematic way to see this distinction: they inherit total boundedness from a totally bounded ambient space, but a missing limit can prevent them from being complete.
We use the metric inherited by a subset \(A\) of a metric space \((X,d)\): for \(a,b\in A\), their distance in \(A\) is \(d(a,b)\). A sequence is Cauchy if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). It converges in a space if it has a limit belonging to that space.
In the definition of an \(\varepsilon\)-net, the centers must belong to the space being covered. This matters for subsets: a net for \(X\) may have centers outside \(A\), so it is not automatically a net for \(A\). The following result handles that issue.
Proof. If \(A=\varnothing\), the empty set is a finite \(\varepsilon\)-net for \(A\), for every \(\varepsilon>0\). Now suppose \(A\ne\varnothing\), and fix \(\varepsilon>0\). Since \(X\) is totally bounded, it has a finite \(\varepsilon/2\)-net \(F\subseteq X\). For each \(p\in F\) whose ball \(B(p,\varepsilon/2)\) meets \(A\), choose one point \(a_p\in A\cap B(p,\varepsilon/2)\). There are only finitely many such points \(a_p\). Given \(a\in A\), some \(p\in F\) satisfies \(d(a,p)<\varepsilon/2\). This ball meets \(A\), and its chosen point \(a_p\) satisfies \(d(a_p,p)<\varepsilon/2\). The triangle inequality gives \(d(a,a_p)\le d(a,p)+d(p,a_p)<\varepsilon\). Thus the chosen points form a finite \(\varepsilon\)-net for \(A\). \(\square\)
Completeness Inside a Complete Space
For a subset \(A\subseteq X\), being closed means that its complement is open in \(X\). Equivalently, \(A\) contains all its limit points in \(X\); in metric terms, any convergent sequence of points of \(A\) has its limit in \(A\). The next proposition explains exactly how this condition relates to completeness when the ambient metric space is complete.
Proof. Suppose first that \(A\) is closed, and let \((a_n)\) be a Cauchy sequence in \(A\). It is also Cauchy in \(X\), because the distances are unchanged. Completeness of \(X\) gives a limit \(x\in X\). Since \(A\) is closed and every \(a_n\) belongs to \(A\), the limit \(x\) belongs to \(A\). Thus every Cauchy sequence in \(A\) converges in \(A\), so \(A\) is complete.
Conversely, suppose \(A\) is complete. Let \(x\) be in the closure of \(A\) in \(X\). The neighborhood characterization of closure says that every open ball about \(x\) meets \(A\). For each \(n\in\mathbb{N}\), choose \(a_n\in A\) with \(d(a_n,x)<1/n\). For \(m,n\ge N\), the triangle inequality gives \(d(a_m,a_n)\le d(a_m,x)+d(x,a_n)<1/m+1/n\le 2/N\). Given \(\varepsilon>0\), choose \(N\) with \(2/N<\varepsilon\); then \((a_n)\) is Cauchy in \(A\). By completeness of \(A\), it converges to some \(a\in A\). It also converges to \(x\) in \(X\). Limits in a metric space are unique: if a sequence converges to both \(a\) and \(x\), then \(d(a,x)\le d(a,a_n)+d(a_n,x)\) for every \(n\), and letting \(n\) grow makes the right side arbitrarily small, so \(d(a,x)=0\) and \(a=x\). Therefore \(x=a\in A\). Every point in the closure of \(A\) lies in \(A\), so \(A\) is closed. \(\square\)
In particular, a dense proper subset of a complete metric space is not complete: density means its closure is the whole ambient space, and being proper means it is not that whole space, so it is not closed. If the ambient space is also totally bounded, the proposition above shows that this dense subset is totally bounded nonetheless.
Worked Example: The Rationals in the Unit Interval
Let \(I=[0,1]\) have the usual metric, and let \(A=\mathbb{Q}\cap I\) have the inherited metric. The interval \(I\) is complete, since it is closed in the complete metric space \(\mathbb{R}\). We verify directly that \(A\) is totally bounded, then exhibit a Cauchy sequence in \(A\) that does not converge in \(A\).
Finite nets. Fix \(\varepsilon>0\), and choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). Use the finite set \(F=\{j/N: j=0,1,\ldots,N\}\). Its points are rational and belong to \(I\), so \(F\subseteq A\). For any \(x\in A\), let \(j\) be the greatest integer not exceeding \(Nx\). Then \(0\le j\le N\) and \(0\le x-j/N<1/N<\varepsilon\). Hence \(d(x,j/N)<\varepsilon\), and \(F\) is a finite \(\varepsilon\)-net for \(A\). This works for every positive \(\varepsilon\).
A missing limit. Set \(\alpha=\sqrt{2}/2\), and for each \(n\in\mathbb{N}\) define \(q_n=\lfloor 10^n\alpha\rfloor/10^n\), where \(\lfloor t\rfloor\) is the greatest integer less than or equal to \(t\). Since \(0<\alpha<1\), each \(q_n\) is rational and lies in \(A\). The first values are \(q_1=7/10\), because \(7^2=49<50<64=8^2\); \(q_2=70/100\), because \(70^2=4900<5000<5041=71^2\); and \(q_3=707/1000\), because \(707^2=499849<500000<501264=708^2\). These comparisons determine the floors exactly.
By the definition of the floor, \(0\le\alpha-q_n<10^{-n}\). Thus, for \(m,n\ge N\), \( |q_m-q_n|\le |q_m-\alpha|+|\alpha-q_n|<10^{-m}+10^{-n}\le 2\cdot10^{-N}\). For any \(\varepsilon>0\), choose \(N\) with \(2\cdot10^{-N}<\varepsilon\); this proves that \((q_n)\) is Cauchy. The same estimate shows \(q_n\to\alpha\) in \(I\).
The number \(\alpha\) is irrational: if \(\sqrt{2}/2\) were rational, then \(\sqrt{2}\) would be rational; writing \(\sqrt{2}=r/s\) in lowest positive integer terms would give \(r^2=2s^2\), forcing \(r\) even and then \(s\) even, a contradiction. So \(\alpha\notin A\). If \((q_n)\) converged to some \(q\in A\), its limit in \(I\) would also be \(\alpha\), and uniqueness of limits would give \(q=\alpha\), impossible. Therefore \(A\) is totally bounded but incomplete.
Check: for every \(x\in I\) and every \(\varepsilon>0\), density of the rationals supplies a rational in \(I\) within \(\varepsilon\) of \(x\), so \(A\) is dense in \(I\). It is proper because the irrational \(\alpha\) is not in \(A\), in agreement with the theorem.
Worked Example: A Dense Half-Open Interval
Take \(A=[0,1)\) inside \(I=[0,1]\), both with the usual metric. The ambient interval \(I\) is complete and totally bounded. The subset \(A\) is totally bounded by the proposition, and it is dense in \(I\): every point of \(A\) already belongs to \(A\), while every neighborhood of \(1\) in \(I\) contains points less than \(1\).
For a direct incompleteness test, let \(x_n=1-1/n\) for \(n\ge2\). Each \(x_n\) belongs to \(A\). If \(m,n\ge N\ge2\), then \(|x_m-x_n|=|1/n-1/m|\le1/N\). Given \(\varepsilon>0\), choose an integer \(N\ge2\) with \(1/N<\varepsilon\); hence \((x_n)\) is Cauchy. In \(I\), \(x_n\to1\), since \(|x_n-1|=1/n\to0\), but \(1\notin A\). Uniqueness of limits rules out a different limit in \(A\), so \(A\) is not complete.
Check: the missing point is a limit of a sequence from \(A\), so \(A\) is not closed in \(I\). The theorem therefore predicts incompleteness, and the explicit sequence verifies it.
Worked Example: Removing an Interior Point
Again use the complete ambient interval \(I=[0,1]\), but now take \(A=I\setminus\{1/2\}\). This set is dense in \(I\): every point other than \(1/2\) belongs to \(A\), and every open neighborhood of \(1/2\) in \(I\) contains points different from \(1/2\). It is totally bounded by the subset proposition.
For \(n\in\mathbb{N}\), put \(y_n=1/2+1/(4n)\). Since \(0<1/(4n)\le1/4\), each \(y_n\) lies in \((1/2,3/4]\), so \(y_n\in A\). For \(m,n\ge N\), \(|y_m-y_n|=(1/4)|1/m-1/n|\le1/(4N)\). Choose \(N\) so that \(1/(4N)<\varepsilon\); then \((y_n)\) is Cauchy. Also \(|y_n-1/2|=1/(4n)\to0\), so its limit in \(I\) is the missing point \(1/2\). Uniqueness of limits shows that it has no limit in \(A\). Thus this dense, totally bounded subset is incomplete as well.
Check: for \(n=1\), \(y_1=1/2+1/4=3/4\in A\); as \(n\) increases the positive difference \(1/(4n)\) tends to zero, without ever equalling zero. The sequence approaches the removed point from within \(A\).
A Common Mistake and the Main Distinction
Common pitfall. One might reason: “The rationals are dense in \(I\), so their closure is \(I\); and \(I\) is complete; therefore \(\mathbb{Q}\cap I\) is complete.” The error is changing the space in which convergence is required. Completeness of \(I\) says Cauchy sequences from \(I\) have limits in \(I\); it does not say their limits belong to every subset of \(I\). In the worked example, the rational sequence \((q_n)\) is Cauchy and its limit \(\sqrt{2}/2\) lies in \(I\), but that limit is not rational. The correct criterion is the theorem: a subset of a complete metric space is complete exactly when it is closed in that ambient space.
Total boundedness concerns the existence of finite nets at each scale; it does not require that limits of Cauchy sequences stay in the space. In these examples, the finite nets control how points are distributed, while incompleteness comes from a specific limit point that has been omitted. When testing a proposed example, check these properties separately: construct finite nets with centers in the subset, then identify a Cauchy sequence whose ambient limit is not in it.
Check Your Understanding
For each question, keep track of which space supplies the metric and where the proposed limit belongs.
- In \(I=[0,1]\), explain why the set of rational numbers in \(I\) is totally bounded even though a finite net for \(I\) might have irrational centers.
- Give a Cauchy sequence in \((0,1)\) whose limit in \(I\) is not in \((0,1)\). Why does this prove incompleteness?
- State the hypotheses of the theorem characterizing complete subsets of a complete metric space, and explain why density alone is not enough to conclude that a subset is incomplete.
- Can the empty subset of a complete metric space be complete and closed? Check both definitions.
- For \(A=I\setminus\{1/2\}\), identify a point of \(I\) that witnesses nonclosedness and a sequence in \(A\) that converges to it.