What a Completion Adds
A metric space may contain Cauchy sequences whose points get arbitrarily close to one another but have no limit in the space. A completion provides a larger space in which those missing limits are present, while leaving every distance between original points unchanged. The key example is the inclusion of the rational numbers in the real line: the rationals are dense in the reals, the reals are complete, and the inclusion preserves distance.
A sequence \((x_n)\) in a metric space \((X,d_X)\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d_X(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). The space is complete if every Cauchy sequence in it converges to a point of that space. A subset \(D\) of a metric space \(Y\) is dense in \(Y\) if every open ball about every point of \(Y\) intersects \(D\).
The map \(j\) need not be an inclusion. Its distance-preserving property implies it is injective: if \(j(x)=j(x')\), then \(d_X(x,x')=d_Y(j(x),j(x'))=0\), so \(x=x'\). Thus we can identify \(X\) with its image when convenient, but the map \(j\) records precisely how the original space sits inside the extension.
The three requirements do different jobs. Completeness says Cauchy sequences in \(Y\) have limits in \(Y\). Density says the extension has no points separated by a positive distance from all of \(j(X)\). Isometry says distances among original points have not been altered. A larger complete space by itself is not necessarily a completion: it must also contain a dense, distance-preserving copy of \(X\).
The Rationals Inside the Reals
Give \(\mathbb{Q}\) and \(\mathbb{R}\) their usual metrics, \(d(s,t)=|s-t|\). We first verify density, including at the two endpoints of the interval \([0,1]\), where an argument using only rational numbers near an interior point would not apply.
Worked Example: Rational Points Are Dense in the Unit Interval
Let \(D=\mathbb{Q}\cap(0,1)\), with the metric inherited from \(\mathbb{R}\). We show that every open ball in \([0,1]\), with its inherited metric, meets \(D\). Equivalently, for every \(y\in[0,1]\) and every \(\varepsilon>0\), we find \(q\in D\) with \(|q-y|<\varepsilon\).
If \(y=0\), choose \(n\in\mathbb{N}\) with \(n\ge2\) and \(1/n<\varepsilon\), possible by the Archimedean property. Then \(q=1/n\) is rational, \(0<1/n<1\), and \(|q-0|=1/n<\varepsilon\).
If \(y=1\), choose \(n\in\mathbb{N}\) with \(n\ge2\) and \(1/n<\varepsilon\). Then \(q=1-1/n\) is rational and satisfies \(0<q<1\). Its distance from \(1\) is \(|q-1|=1/n<\varepsilon\).
If \(0<y<1\), let \(\delta=\min\{\varepsilon,y,1-y\}/2\), which is positive. The rationals are dense in \(\mathbb{R}\): to see this, choose \(n\in\mathbb{N}\) with \(1/n<\delta\), and choose an integer \(m\) with \(m\le ny<m+1\). Then \(m/n\in\mathbb{Q}\) and \(0\le y-m/n<1/n<\delta\). Set \(q=m/n\). Since \(|q-y|<\delta\le y/2\), we have \(q>y-y/2>0\); since \(|q-y|<\delta\le(1-y)/2\), we have \(q<y+(1-y)/2<1\). Also \(|q-y|<\delta\le\varepsilon/2<\varepsilon\). Thus \(q\in D\) and lies within the required distance of \(y\).
All three cases give a point of \(D\) in every ball about \(y\), so \(D\) is dense in \([0,1]\). In particular, the same construction, or the rational approximation argument for an arbitrary real \(y\), proves that \(\mathbb{Q}\) is dense in \(\mathbb{R}\).
For arbitrary \(y\in\mathbb{R}\), the rational approximation used above has no restriction \(0<q<1\): choose \(n\) with \(1/n<\varepsilon\), and an integer \(m\) with \(m\le ny<m+1\). Then \(q=m/n\) satisfies \(0\le y-q<1/n<\varepsilon\). This proves directly that every open ball in \(\mathbb{R}\) meets \(\mathbb{Q}\).
Proof. For \(q,r\in\mathbb{Q}\), the distance in \(\mathbb{R}\) between \(j(q)\) and \(j(r)\) is \(|q-r|\), which is exactly the distance in \(\mathbb{Q}\). Thus \(j\) is an isometric embedding. The rational approximation argument shows that \(j(\mathbb{Q})=\mathbb{Q}\) is dense in \(\mathbb{R}\). Finally, the standard completeness theorem for the real numbers says that every Cauchy sequence of real numbers converges to a real number; it follows from the least-upper-bound property of \(\mathbb{R}\). Hence \(\mathbb{R}\) is complete, and all three requirements in the definition hold. \(\square\)
What the Extension Does to Cauchy Sequences
A dense complete extension has a direct consequence for sequences from the original space: their images acquire limits in the extension. In the other direction, density lets us approximate every point of the extension by points from the original space.
Proof. For (1), let \((x_n)\) be Cauchy in \(X\). Given \(\varepsilon>0\), choose \(N\) so that \(d_X(x_m,x_n)<\varepsilon\) for \(m,n\ge N\). Since \(j\) is isometric, \(d_Y(j(x_m),j(x_n))=d_X(x_m,x_n)<\varepsilon\) for those indices. Thus \((j(x_n))\) is Cauchy in \(Y\). Completeness of \(Y\) gives a point \(y\in Y\) with \(j(x_n)\to y\).
For (2), fix \(y\in Y\). For each \(n\in\mathbb{N}\), density gives \(x_n\in X\) such that \(d_Y(j(x_n),y)<1/n\). Given \(\varepsilon>0\), choose \(N\) with \(2/N<\varepsilon\). For \(m,n\ge N\), the triangle inequality gives \(d_X(x_m,x_n)=d_Y(j(x_m),j(x_n))\le d_Y(j(x_m),y)+d_Y(y,j(x_n))<1/m+1/n\le2/N<\varepsilon\). So \((x_n)\) is Cauchy in \(X\). The choice also gives \(d_Y(j(x_n),y)<1/n\), which tends to zero, hence \(j(x_n)\to y\). \(\square\)
This theorem explains the role of density and completeness without prescribing a construction of \(Y\). Density supplies approximating points from \(X\); completeness supplies limits for Cauchy sequences in \(Y\). In a completion of \(\mathbb{Q}\), for example, rational sequences can converge to points that are not rational.
Worked Example: The Rationals Are Not Complete
We exhibit a Cauchy sequence of rationals that cannot converge to a rational. For \(n\in\mathbb{N}\), let \(m_n\) be the greatest integer not exceeding \(10^n\sqrt{2}\), and put \(q_n=m_n/10^n\). Each \(q_n\) is rational. The defining inequalities for \(m_n\) give \(m_n\le10^n\sqrt{2}<m_n+1\), and division by \(10^n>0\) gives \(0\le\sqrt{2}-q_n<10^{-n}\).
For \(m,n\ge N\), the triangle inequality yields \(|q_m-q_n|\le|q_m-\sqrt{2}|+|\sqrt{2}-q_n|<10^{-m}+10^{-n}\le2\cdot10^{-N}\). Given \(\varepsilon>0\), choose \(N\) large enough that \(2\cdot10^{-N}<\varepsilon\). Then \(|q_m-q_n|<\varepsilon\) for all \(m,n\ge N\), so \((q_n)\) is Cauchy in \(\mathbb{Q}\).
The same error estimate \(0\le\sqrt{2}-q_n<10^{-n}\) shows \(q_n\to\sqrt{2}\) in \(\mathbb{R}\). If \((q_n)\) converged in \(\mathbb{Q}\) to some \(q\in\mathbb{Q}\), it would also converge to \(q\) in \(\mathbb{R}\). The triangle inequality would then imply \(|q-\sqrt{2}|\le|q-q_n|+|q_n-\sqrt{2}|\to0\), so \(q=\sqrt{2}\).
But \(\sqrt{2}\notin\mathbb{Q}\). Indeed, if \(\sqrt{2}=a/b\) for relatively prime positive integers \(a,b\), then \(a^2=2b^2\), so \(a\) is even. Write \(a=2k\); then \(4k^2=2b^2\), hence \(b^2=2k^2\), so \(b\) is even as well. This contradicts that \(a,b\) are relatively prime. Therefore the Cauchy sequence \((q_n)\) has no limit in \(\mathbb{Q}\), and \(\mathbb{Q}\) is not complete.
Check: Its limit in \(\mathbb{R}\) is the irrational number \(\sqrt{2}\), while every term is rational. The sequence therefore demonstrates exactly why completeness of \(\mathbb{R}\) does not imply completeness of its dense subspace \(\mathbb{Q}\).
Other Simple Dense Complete Extensions
Worked Example: Completing the Open Interval
Give \(X=(0,1)\) and \(Y=[0,1]\) the metrics inherited from \(\mathbb{R}\), and let \(j:X\to Y\) be inclusion. It preserves distance because both metrics assign distance \(|x-x'|\) to \(x,x'\in X\). The density proof in the first worked example shows that every point of \(Y\), including \(0\) and \(1\), can be approximated arbitrarily closely by points of \(X\).
To verify completeness, let \((y_n)\) be Cauchy in \(Y\). It is also Cauchy in \(\mathbb{R}\), since the distances are the same, so the completeness of \(\mathbb{R}\) gives a real limit \(y\). Because \(0\le y_n\le1\) for every \(n\), we must have \(0\le y\le1\): if \(y<0\), convergence would eventually force \(y_n<0\), and if \(y>1\), it would eventually force \(y_n>1\), each a contradiction. Thus \(y\in Y\), and the sequence converges in \(Y\). So \([0,1]\) is a completion of \((0,1)\). The added points are precisely the two endpoints, which are approached by points of \(X\) but do not belong to \(X\).
Worked Example: A Complete Space Needs No Added Points
Let \(X=\{a,b\}\) with the discrete metric \(d(a,b)=1\), \(d(a,a)=d(b,b)=0\). This space is complete. In fact, if a sequence in \(X\) is Cauchy, apply the definition with \(\varepsilon=1\). Beyond some index, every two terms have distance less than \(1\); distinct points have distance exactly \(1\), so all terms beyond that index are equal. The sequence is eventually constant and converges to that constant value.
With \(Y=X\) and \(j=\operatorname{id}_X\), the map preserves distances, and \(Y\) is complete. Its image is dense in itself: every open ball centered at a point contains that point, since the center is at distance \(0\) from itself. Hence \(X\) is a completion of itself. A completion is not required to add new points when the original metric space is already complete.
A Common Pitfall: Density Does Not Mean Membership
Common pitfall. From the fact that \(\mathbb{Q}\) is dense in \(\mathbb{R}\), one might incorrectly conclude that every real number is rational. Density says that every open ball about a real number contains at least one rational; it does not say that the center belongs to \(\mathbb{Q}\). For instance, the estimate \(0\le\sqrt{2}-q_n<10^{-n}\) gives rational points within \(10^{-n}\) of \(\sqrt{2}\) for every \(n\), but the parity argument above proves \(\sqrt{2}\notin\mathbb{Q}\).
There is also a useful endpoint check in the unit interval. A tempting but wrong choice for a rational point near \(1\) is \(1/n\), since \(|1-1/n|\) is not generally small when \(n\) is large: it equals \(1-1/n\). The correct choice is \(1-1/n\). Choose \(n\ge2\) with \(1/n<\varepsilon\); then \(1-1/n\in\mathbb{Q}\cap(0,1)\) and \(|1-(1-1/n)|=1/n<\varepsilon\). Endpoint cases need their own valid choices, rather than reusing a choice made for the other endpoint.
Check Your Understanding
For each question, keep the three requirements—complete, dense image, and isometric embedding—distinct.
- Why does the inclusion \(\mathbb{Q}\to\mathbb{R}\) preserve distances, and what argument shows its image is dense?
- In the example \(X=(0,1)\), \(Y=[0,1]\), how do you approximate \(0\) and \(1\) by points of \(X\)?
- What feature of the sequence of rational truncations of \(\sqrt{2}\) proves that \(\mathbb{Q}\) is not complete?
- Explain why a dense subset need not contain every point of the space in which it is dense.
- Show that an isometric embedding between metric spaces is injective.