Why Closedness and Completeness Are Connected
Completeness concerns Cauchy sequences: do they converge to points of the space? Closedness concerns which limits a subset retains from its ambient space. In a metric space, these ideas meet because every Cauchy sequence in a complete ambient space has an ambient limit, and closedness ensures that this limit remains in the subset. In the other direction, a point in the closure of a complete subspace can be approached by a sequence from that subspace; completeness supplies a limit inside it.
Throughout, \((X,d)\) is a metric space, and each subset \(A\subseteq X\) is given the subspace metric, which is the restriction of \(d\) to pairs of points in \(A\). A sequence \((x_n)\) in a metric space is Cauchy if, for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). A metric space is complete if every Cauchy sequence in it converges to a point of that space. Convergence to \(x\) means that for every \(\varepsilon>0\), eventually \(d(x_n,x)<\varepsilon\).
The distinction in the last sentence matters. The same set with a different metric can have different completeness behavior, and a subset can be closed in one ambient space but not another. The results below always use the inherited metric and explicitly state what is assumed about the ambient space.
Closed Subspaces of Complete Spaces
Proof. Let \((a_n)\) be a Cauchy sequence in \(A\). The distance between any two of its terms is the same whether computed in \(A\) or in \(X\), since \(A\) has the subspace metric. Thus \((a_n)\) is Cauchy in \(X\). Completeness of \(X\) gives a point \(x\in X\) such that \(a_n\to x\) in \(X\).
We show that \(x\in A\). If \(x\notin A\), then \(x\in X\setminus A\). Since \(A\) is closed, \(X\setminus A\) is open. Therefore some \(r>0\) satisfies \(B_d(x,r)\subseteq X\setminus A\). But \(a_n\to x\), so for all sufficiently large \(n\), \(a_n\in B_d(x,r)\). This contradicts \(a_n\in A\) and \(B_d(x,r)\subseteq X\setminus A\). Hence \(x\in A\). The sequence converges to a point of \(A\), proving that \(A\) is complete. If \(A=\varnothing\), there are no sequences in \(A\), so the definition of completeness holds vacuously. \(\square\)
The proof uses closedness precisely at the point where it rules out an ambient limit outside \(A\). Completeness of \(X\) alone does not say that the limit of a sequence from an arbitrary subset stays in that subset.
Worked Example: The Closed Interval \([1,2]\)
Consider \(A=[1,2]\subseteq\mathbb{R}\), where \(\mathbb{R}\) has its standard metric \(d(x,y)=|x-y|\). The real line is complete, by the completeness theorem for the real numbers. The complement of \(A\) is \((-\infty,1)\cup(2,\infty)\), which is open in \(\mathbb{R}\): each of its points lies in an open interval contained in the complement. Thus \(A\) is closed, and Theorem 1 implies that \([1,2]\) is complete.
For a direct check on a particular sequence, let \(x_n=1+1/n\). For every \(n\ge1\), \(1\le x_n\le2\), so \(x_n\in A\). For \(m,n\ge N\),
Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>2/\varepsilon\). Then \(2/N<\varepsilon\), so the sequence is Cauchy. Its proposed limit is \(1\), which belongs to \([1,2]\), and the distance check is \(|x_n-1|=1/n\to0\). For instance, \(x_1=2\), \(x_2=3/2\), and \(x_3=4/3\); the formula gives \(|x_3-1|=1/3\), as required. This calculation illustrates the theorem, while the theorem establishes completeness for every Cauchy sequence in the interval.
Complete Subspaces Are Closed
The converse direction does not require the ambient space to be complete. The key is that a point in the closure can be approximated by points of the subset at arbitrarily small distances. We include that construction in the proof.
Proof. If \(A=\varnothing\), then \(A\) is closed because \(X\setminus A=X\) is open. Now suppose \(A\ne\varnothing\), and let \(x\in X\) be in the closure of \(A\). By the neighborhood test for closure, every open neighborhood of \(x\) meets \(A\). In particular, for each \(n\in\mathbb{N}\), the ball \(B_d(x,1/n)\) meets \(A\). Choose \(a_n\in A\cap B_d(x,1/n)\). Then \(d(a_n,x)<1/n\).
For \(m,n\in\mathbb{N}\), the triangle inequality gives \(d(a_m,a_n)\le d(a_m,x)+d(x,a_n)<1/m+1/n\). Given \(\varepsilon>0\), choose \(N\) with \(2/N<\varepsilon\). If \(m,n\ge N\), then \(d(a_m,a_n)<2/N<\varepsilon\), so \((a_n)\) is Cauchy in \(A\). Completeness of \(A\) gives \(a\in A\) with \(a_n\to a\) in \(A\), and hence also in \(X\), because the subspace metric is inherited from \(X\).
We also have \(a_n\to x\) in \(X\), since \(d(a_n,x)<1/n\). Limits in a metric space are unique: if \(a\ne x\), let \(\delta=d(a,x)>0\). For sufficiently large \(n\), both \(d(a_n,a)<\delta/3\) and \(d(a_n,x)<\delta/3\); the triangle inequality would give \(\delta\le d(a,a_n)+d(a_n,x)<2\delta/3\), a contradiction. Thus \(x=a\in A\). We have shown that every point in the closure of \(A\) belongs to \(A\), so \(A\) is closed. \(\square\)
Together the two theorems give a useful equivalence when the ambient space is complete: a subset, with its inherited metric, is complete if and only if it is closed. The forward implication is Theorem 2, and the reverse implication is Theorem 1. Without completeness of the ambient space, only the forward implication is guaranteed.
Worked Example: A Dense-Incompleteness Test in \(\mathbb{Q}\)
Give \(\mathbb{Q}\) the metric inherited from \(\mathbb{R}\), and let \(A=\mathbb{Q}\cap(0,1)\). Define \(\alpha=\sqrt{2}/2\), and for \(n\in\mathbb{N}\) set
Here \(\lfloor t\rfloor\) denotes the greatest integer less than or equal to \(t\). Each \(a_n\) is rational. Since \(1<\sqrt{2}<2\), we have \(0<a_n\le\alpha<1\), so \(a_n\in A\). The defining property of the floor gives \(0\le \sqrt{2}-\lfloor10^n\sqrt{2}\rfloor/10^n<10^{-n}\), and therefore \(0\le\alpha-a_n<1/(2\cdot10^n)\).
The first two terms can be checked explicitly: \(10\sqrt{2}\) is between \(14\) and \(15\), so \(a_1=14/20=7/10\); \(100\sqrt{2}\) is between \(141\) and \(142\), so \(a_2=141/200\). For any \(m,n\),
If \(m,n\ge N\), the right side is at most \(10^{-N}\). Given \(\varepsilon>0\), choose \(N\) so that \(10^{-N}<\varepsilon\); then the sequence is Cauchy. In \(\mathbb{R}\), it converges to \(\alpha\), as the error bound \(0\le\alpha-a_n<1/(2\cdot10^n)\) tends to zero.
There is no limit for this sequence in \(A\). If it converged to some \(q\in A\), it would also converge to \(q\) in \(\mathbb{R}\); uniqueness of limits would force \(q=\alpha\). But \(\alpha\) is irrational: if \(\sqrt{2}/2\) were rational, then \(\sqrt{2}\) would be rational, which is impossible. Indeed, if \(\sqrt{2}=u/v\) for coprime positive integers \(u,v\), then \(u^2=2v^2\) makes \(u\) even; writing \(u=2k\) then makes \(v\) even, contradicting coprimality. Thus \(A\) is not complete.
This same sequence also shows why the conclusion is about \(A\), not about \(\mathbb{Q}\): its terms converge in \(\mathbb{Q}\) to \(0\) only if the sequence tends to \(0\), which it does not; instead they converge in \(\mathbb{R}\) to the irrational \(\alpha\), so they have no limit in \(\mathbb{Q}\) either. A simpler sequence, \(1/(n+1)\), does converge in \(\mathbb{Q}\) to \(0\), so it does not witness incompleteness of \(\mathbb{Q}\). It does witness incompleteness of \(A\), because its terms lie in \(A\) and its limit \(0\) is not in \(A\).
Finally, \(A\) is not closed in \(\mathbb{Q}\). The point \(0\) belongs to \(\mathbb{Q}\setminus A\), but every open neighborhood of \(0\) in \(\mathbb{Q}\) contains \(B_{\mathbb{Q}}(0,\varepsilon)=\{q\in\mathbb{Q}:|q|<\varepsilon\}\) for some \(\varepsilon>0\). Choose an integer \(k\ge2\) with \(k>1/\varepsilon\). Then \(0<1/k<\varepsilon\) and \(1/k\in A\). Thus no such neighborhood is contained in \(\mathbb{Q}\setminus A\), so the complement is not open. There is no conflict with Theorem 1: the ambient space \(\mathbb{Q}\) is not complete.
Worked Example: The Integers in the Real Line
Let \(A=\mathbb{Z}\subseteq\mathbb{R}\) with the usual metric. We verify directly that \(A\) is closed. If \(x\in\mathbb{R}\setminus\mathbb{Z}\), there is an integer \(k\) with \(k<x<k+1\). Put \(r=\min(x-k,k+1-x)/2\), which is positive. Every integer is at most \(k\) or at least \(k+1\), so its distance from \(x\) is at least \(\min(x-k,k+1-x)=2r\). Consequently \(B(x,r)\) contains no integer. Each point of the complement therefore has an open neighborhood in the complement, proving that \(\mathbb{Z}\) is closed.
Since \(\mathbb{R}\) is complete, Theorem 1 now shows that \(\mathbb{Z}\) is complete with its inherited metric. As a check on the distance estimate, at \(x=1/3\) we may take \(k=0\), giving \(r=\min(1/3,2/3)/2=1/6\). The ball \((1/6,1/2)\) contains no integer. The theorem, rather than this single check, applies to every noninteger real number.
A Common Pitfall: Closed in an Incomplete Space
Do not reverse Theorem 1 without its ambient completeness hypothesis. For example, \(\mathbb{Q}\) is closed as a subset of itself: its complement in \(\mathbb{Q}\) is empty, and the empty set is open. But the sequence in the preceding example is Cauchy in \(\mathbb{Q}\) and has no rational limit. Thus \(\mathbb{Q}\) is not complete. The mistaken argument “\(\mathbb{Q}\) is closed, so it is complete” omits the required assumption that the ambient metric space is complete. The correct statement is: if the ambient space is complete, then its closed subspaces are complete; in any metric space, complete subspaces are closed.
Check Your Understanding
For each question, keep track of both the ambient space and the metric on the subset.
- Which hypothesis in Theorem 1 ensures that a Cauchy sequence in a closed subset has an ambient limit?
- In the proof of Theorem 2, why can a point in the closure be approximated by a sequence whose \(n\)th term is within \(1/n\)?
- Is \((0,1)\) complete with its inherited metric from \(\mathbb{R}\)? Test the sequence \(1/(n+1)\).
- Explain why the fact that \(\mathbb{Q}\) is closed in itself does not imply that \(\mathbb{Q}\) is complete.
- State the equivalence between closedness and completeness that holds for subsets of a complete metric space.