Uniqueness of Metric Completions

Why Dense Copies Determine a Complete Space

A completion adds the limits missing from a metric space while preserving all its original distances. If two complete metric spaces both contain dense isometric copies of the same metric space, the points of those copies prescribe how the spaces should correspond. The main question is whether that correspondence extends to every point, and whether the extension is forced. The answer comes from approximating each point by a sequence in the dense copy and taking the limit of its image.

Let \((M,d_M)\), \((X,d_X)\), and \((Y,d_Y)\) be metric spaces. An isometric embedding \(i:M\to X\) is a map satisfying \(d_X(i(a),i(b))=d_M(a,b)\) for all \(a,b\in M\). Its image is dense in \(X\) if every open ball centered at a point of \(X\) meets \(i(M)\). A metric space is complete if every Cauchy sequence in it converges to a point of that space.

The embeddings \(i\) and \(j\) identify corresponding points: the point \(i(a)\) in \(X\) is to match \(j(a)\) in \(Y\). The theorem below says there is exactly one isometry \(F:X\to Y\) that makes all these matches. Uniqueness here means uniqueness subject to this matching condition; it does not say that the two spaces have only one isometry in total.

Proof. Since \(f\) preserves distances, \(d_Q(f(p_n),f(p))=d_P(p_n,p)\) for every \(n\). If \(p_n\to p\), the right side tends to zero, which is exactly \(f(p_n)\to f(p)\). If \((p_n)\) is Cauchy, then for every \(\varepsilon>0\) there is \(N\) such that \(d_P(p_n,p_m)<\varepsilon\) whenever \(m,n\ge N\). Distance preservation gives \(d_Q(f(p_n),f(p_m))=d_P(p_n,p_m)<\varepsilon\) for the same indices, so the image sequence is Cauchy. \(\square\)

The Uniqueness Theorem

The construction starts with a point \(x\in X\). Density lets us choose points \(a_n\in M\) such that \(i(a_n)\to x\). The sequence \(j(a_n)\) is Cauchy because the distances between its terms are exactly the distances between the corresponding points \(i(a_n)\). Completeness of \(Y\) supplies a limit, which we define to be \(F(x)\). The proof must also show that this limit does not depend on the chosen approximating sequence.

Proof. First suppose \(M\) is empty. Its image is empty, and the only dense subset of a metric space that is empty is the whole space: if \(X\) had a point \(x\), the ball \(B(x,1)\) would not meet the empty set. Thus \(X=Y=\varnothing\), and the unique map from \(X\) to \(Y\) is the required isometry. Now suppose \(M\) is nonempty.

Fix \(x\in X\). For each \(n\in\mathbb N\), density of \(i(M)\) gives \(a_n\in M\) with \(d_X(i(a_n),x)<1/n\). Thus \(i(a_n)\to x\). For \(m,n\in\mathbb N\), the triangle inequality gives \(d_X(i(a_n),i(a_m))\le d_X(i(a_n),x)+d_X(x,i(a_m))\). Both terms on the right tend to zero as \(m,n\) grow, so \((i(a_n))\) is Cauchy. Since \(i\) is an isometric embedding, \(d_Y(j(a_n),j(a_m))=d_X(i(a_n),i(a_m))\); hence \((j(a_n))\) is Cauchy. Completeness of \(Y\) gives a limit in \(Y\).

To show this limit is independent of the sequence, suppose also \(i(b_n)\to x\), where each \(b_n\in M\). Then \(d_Y(j(a_n),j(b_n))=d_X(i(a_n),i(b_n))\le d_X(i(a_n),x)+d_X(x,i(b_n))\), which tends to zero. If \(j(a_n)\to y\) and \(j(b_n)\to z\), the triangle inequality yields \(d_Y(y,z)\le d_Y(y,j(a_n))+d_Y(j(a_n),j(b_n))+d_Y(j(b_n),z)\). Each term on the right tends to zero, so \(d_Y(y,z)=0\), and the metric property gives \(y=z\). We may therefore define \(F(x)\) to be the limit of \(j(a_n)\) for any sequence with \(i(a_n)\to x\).

If \(x=i(a)\) for some \(a\in M\), the constant sequence \(a_n=a\) is an allowed approximating sequence. Its image under \(j\) is constantly \(j(a)\), so \(F(i(a))=j(a)\). To check distances, take \(x,x'\in X\), and choose \(a_n,b_n\in M\) with \(i(a_n)\to x\) and \(i(b_n)\to x'\). By the triangle inequality, \(\lvert d_X(i(a_n),i(b_n))-d_X(x,x')\rvert\le d_X(i(a_n),x)+d_X(i(b_n),x')\), so \(d_X(i(a_n),i(b_n))\to d_X(x,x')\). The same inequality in \(Y\), using \(j(a_n)\to F(x)\) and \(j(b_n)\to F(x')\), shows \(d_Y(j(a_n),j(b_n))\to d_Y(F(x),F(x'))\). But \(d_Y(j(a_n),j(b_n))=d_X(i(a_n),i(b_n))\) for every \(n\). Therefore \(d_Y(F(x),F(x'))=d_X(x,x')\), and \(F\) is an isometry. In particular, it is injective: if \(F(x)=F(x')\), then \(d_X(x,x')=d_Y(F(x),F(x'))=0\), hence \(x=x'\).

To prove surjectivity, take \(y\in Y\). Density of \(j(M)\) lets us choose \(c_n\in M\) with \(d_Y(j(c_n),y)<1/n\). Then \(j(c_n)\to y\). The sequence \((i(c_n))\) is Cauchy because \(d_X(i(c_n),i(c_m))=d_Y(j(c_n),j(c_m))\le d_Y(j(c_n),y)+d_Y(y,j(c_m))\), and the right side tends to zero as \(m,n\) grow. Completeness of \(X\) gives \(x\in X\) with \(i(c_n)\to x\). The definition of \(F\) now gives \(F(x)=\lim_{n\to\infty}j(c_n)=y\). Thus \(F\) is surjective. A distance-preserving bijection has a distance-preserving inverse, so both \(F\) and its inverse are continuous; consequently \(F\) is a homeomorphism.

Finally, let \(G:X\to Y\) be any other isometry satisfying \(G(i(a))=j(a)\) for every \(a\in M\). For \(x\in X\), choose \(a_n\) with \(i(a_n)\to x\). Because \(G\) preserves distances, \(G(i(a_n))\to G(x)\). But \(G(i(a_n))=j(a_n)\), and the definition of \(F\) gives \(j(a_n)\to F(x)\). Limits in a metric space are unique: if a sequence converges to both \(u\) and \(v\), then \(d_Y(u,v)\le d_Y(u,y_n)+d_Y(y_n,v)\to0\), so \(u=v\). Therefore \(G(x)=F(x)\) for every \(x\), and \(G=F\). This proves uniqueness. \(\square\)

Worked Applications

What the Hypotheses and the Matching Condition Do

The density hypothesis is what ensures that every point can be approximated by points whose correspondence is already known. Without density, the values on the embedded copy may leave room for multiple isometries elsewhere.

Completeness has a different role: it ensures that the Cauchy sequences of corresponding approximants actually have limits in the target space. For a concrete failure when the target is not complete, take \(M=(0,1)\) with its usual metric, \(X=[0,1]\), and \(Y=(0,1)\), using the identity embeddings into each space. The image is dense in both spaces, but \(Y\) is not complete: the sequence \(1/(n+1)\) is Cauchy and has no limit in \((0,1)\). No isometry \(F:X\to Y\) can extend the identity on \((0,1)\). Indeed, \(1/(n+1)\to0\) in \(X\), so distance preservation would imply \(F(1/(n+1))\to F(0)\) in \(Y\). The extension condition gives \(F(1/(n+1))=1/(n+1)\), which converges in \(\mathbb R\) to \(0\), not to any point of \(Y\). The missing limit is precisely the obstruction.

Common pitfall. If two approximating sequences approach the same point \(x\), their terms need not be equal at each index. For example, in \((0,1)\), let \(a_n=1/2+1/(10n)\) and \(b_n=1/2-1/(10n)\). At \(n=1\), \(a_1=3/5\), \(b_1=2/5\), so \(d(a_1,b_1)=1/5\), not zero. The incorrect step is to claim that corresponding images are equal merely because both sequences approach \(x\). The valid conclusion is that their distances tend to zero: \(d(a_n,b_n)=\lvert(1/2+1/(10n))-(1/2-1/(10n))\rvert=1/(5n)\to0\). Distance preservation gives the same estimate for their images, which proves that the two image sequences have the same limit—not that their terms agree.

The result therefore has two parts worth keeping distinct. Completeness supplies the limits needed to define the extension and to prove it is onto; density ensures the prescribed correspondence reaches every point and forces uniqueness. Once the embeddings are fixed, there is exactly one distance-preserving bijection that respects them.