Let (X, dX) be a complete metric space and Y be a subset of X

1. Completion of a metric space

A metric space need not be complete. For example, let B = {(x, y) ∈ R² : x²+ y² < 1}

be the open ball in R². The metric subspace (B, dB) of R² is not a complete metric space.

Proposition 1.1. Let (X, d_X) be a complete metric space and Y be a subset of X. Then (Y, dY) is complete if and only if Y is a closed subset of X.

Proof. This is left to the reader as an exercise. 

This proposition allows us to construct many examples of metric spaces which are not complete.

Let D = B be the closure of B in R². Proposition 1.1 implies that (D, d_D) is a complete metric space. Although (B, dB) is not complete, B is a “dense” subset of a complete metric space (D, dD).

Definition 1.1. Let (M, d) be a metric space and D be a subset of M. We say that D is a dense subset of M if D = M.

Example 1.1. The set of rational numbers Q is a dense subset of R.

These observations lead to the notion of completion of a metric space.

Definition 1.2. Let (X, d) be a metric space. A completion of (X, d) is a complete metric space ( bX, bd) together with a function j : X → X such that

d(j(x), j(y)) = d(x, y),b for any x, y ∈ X and j(X) is dense in bX, i.e. j(X) = bX.

A function f : (X, d) → (Y, ρ) is called an isometry if ρ(f (x), f (x⁰)) = d(x, x⁰) for any x, x⁰ ∈ X. An isometry is always injective.

Definition 1.3. A function f : (X, d) → (Y, ρ) is an isomorphism if f is an isometry and f is surjective. Two metric spaces are called isomorphic if there is an isomorphism between them.

Remark. If f : X → Y is an isometry, then X is always isomorphic to f (X).

Recall that B(S) is a Banach space. Denote d∞(f, g) = kf − gk∞.

Theorem 1.1. Any metric space (X, d) is isomorphic to a metric subspace of (B(X), d∞).

Proof. Let (X, d) be a metric space. For each x ∈ X, we define a function f_x : X → R by fx(t) = d(x, t). Let us fixed x0 ∈ X. Observe that

|f_x(t) − fx0(t)| = |d(x, t) − d(x0, t)| ≤ d(x, x0) for any t ∈ X.

Hence if we define another function j_x : X → R by jx = f_x− f_x0. Then j_x is a real valued bounded function on X, i.e. jx ∈ B(X). Moreover,

|j_x(t) − jy(t)| = |(fx(t) − fx0(t)) − (fy(t) − fx0(t))| = |fx(t) − fy(t)| ≤ d(x, y) for any t ∈ X. Hence kjx− j_yk_∞≤ d(x, y). In fact,

|j_x(y) − j_y(y)| = |f_x(y) − f_y(y)| = |d(x, y)| = d(x, y).

We see that kj_x− j_yk_∞ = d(x, y), i.e. d∞(j_x, j_y) = d(x, y) for any x, y ∈ X. We obtain an isometry

j : X → B(X), x 7→ j_x.

Let j(X) = Y and (Y, d_Y) be the metric subspace of (B(X), d∞) associated with Y. Since j is an isometry and j(X) = Y, j : X → Y is an isomorphism of metric spaces. 

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Corollary 1.1. Every metric space has a completion.

Proof. Let j : X → B(X) as above and bX = j(X) and ( bX, bd) be the associated metric subspace of (B(X), d∞). Since bX is closed in B(X) and B(X) is complete, ( bX, bd) is complete

by Proposition 1.1. We obtain a completion of (X, d). 

Theorem 1.2. Let (X, d) be a metric space. Suppose (X₁, d₁, j₁) and (X₂, d₂, j₂) are two completions of (X, d). Then (X1, d1) and (X2, d2) are isomorphic.

Before proving this theorem, we need the following Lemma.

Lemma 1.1. Let (X, d) and (Y, ρ) be metric spaces and D be a dense subset of X. Suppose Y is complete and f : D → Y is an isometry. Then there exists a unique isometry F : X → Y so that F |D = f.

Proof. When x ∈ D, we define F (x) = f (x). Now, we want to define F (x) for x ∈ X \ D.

Since D is dense in X, for any x ∈ X, we can choose a sequence (x_n) in D so that (x_n) is convergent to x in X. Since (xn) is convergent to x in X, it is a Cauchy sequence in X. For any  > 0, there exists N_ ∈ N so that d(xn, x_m) <  whenever n, m ≥ N_. Since f is an isometry,

ρ(f (xn), f (xm)) = d(xn, xm) < 

whenever n, m ≥ N_. This shows that (f (x_n)) is a Cauchy sequence in Y. Since Y is complete, (f (x_n)) is convergent. Denote the limit of (f (x_n)) by y. We may define F (x) by y.

We need to make sure that y is uniquely determined by x. Let (x⁰_n) be another sequence in X which converges to x. Claim (f (x⁰_n)) is also convergent to y. Let y⁰ be the limit of (f (x⁰_n)). By triangle inequality,

ρ(y, y⁰) ≤ ρ(y, f (xn)) + ρ(f (xn), f (x⁰_n)) + ρ(f (x⁰_n), y⁰)

≤ ρ(y, f (x_n)) + d(xn, x⁰_n) + ρ(f (x⁰_n), y⁰)

≤ ρ(y, f (x_n)) + d(x_n, x) + d(x⁰_n, x) + ρ(f (x⁰_n), y⁰).

Since (x_n) and (x⁰_n) are convergent to x and (f (x_n)) is convergent to y and (f (x⁰_n)) is convergent to y⁰, for any  > 0, we choose N ∈ N so that d(xn, x) and d(x⁰_n, x) and ρ(f (x_n), y) and ρ(f (x⁰_n), y⁰) are all less than /4. This implies that ρ(y, y⁰) <  for any

 > 0. This implies y = y⁰. Thus we obtain a function F : X → Y defined by

F (x) =

(f (x) if x ∈ D

n→∞lim f (x_n) if x ∈ X \ D and x = lim

n→∞x_n.

Let us prove that F is an isometry. When x, y ∈ D, F (x) = f (x) and F (y) = f (y). In this case, ρ(F (x), F (y)) = ρ(f (x), f (y)) = d(x, y). Let us prove the case when x ∈ X \ D and y ∈ D. Then ρ(F (x), F (y)) = ρ(F (x), f (y)). We want to show ρ(F (x), f (y)) = d(x, y).

Choose a sequence (x_n) which converges to x in X. By triangle inequality,

|d(x_n, y) − d(x, y)| ≤ d(xn, x) we find lim

n→∞d(xn, x) = d(x, y). By triangle inequality,

|ρ(F (x), f (y)) − ρ(f (x_n), f (y))| ≤ d(f (x_n), F (x)), we find lim

n→∞ρ(f (xn), f (y)) = ρ(F (x), f (y)). Since xn, y ∈ D, ρ(f (xn), f (y)) = d(xn, y).

This implies that

ρ(F (x), f (y)) = lim

n→∞ρ(f (x_n), f (y)) = lim

n→∞d(x_n, y) = d(x, y).

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Finally, let us show that for x, y ∈ X \D, ρ(F (x), F (y)) = d(x, y). Choose a sequence (yn) in X convergent to y. Then ρ(F (x), F (y_n)) = ρ(F (x), f (y_n)) = d(x, y_n). By triangle inequality as above, lim

n→∞d(x, yn) = d(x, y) and lim

n→∞ρ(F (x), f (yn)) = ρ(F (x), F (y)). We prove that ρ(F (x), F (y)) = d(x, y) when x, y ∈ X \ D.

Let us prove that such F is unique. Let F⁰ : X → Y be an isometry so that F⁰|_D = f.

For x ∈ D, F⁰(x) = f (x) = F (x). For x ∈ X \ D, choose a sequence (xn) in D such that (x_n) is convergent to x. Then

ρ(F (x), F⁰(x)) ≤ ρ(F (x), F (x_n))+ρ(F (x_n), F⁰(x_n))+ρ(F⁰(x_n), F⁰(x)) = ρ(F (x), f (x_n))+d(x_n, x).

By taking n → ∞, we find ρ(F (x), F⁰(x)) = 0. Thus F (x) = F⁰(x) when x ∈ X \ D. We conclude that F (x) = F⁰(x) for all x ∈ X. This proves our assertion.  Now let us go back to prove that any two completions of a metric space are isomorphic.

Proof. Let f = j2◦j₁⁻¹: j1(X) → X2. Then f is an isometry. Since X2is complete, f can be extended to an isometry F : X₁→ X₂. Let us prove that F is surjective. Let y ∈ X₂. Choose a sequence (yn) in j2(X) so that (yn) is convergent to y. Since (yn) in j2(X), we can choose (x_n) in X so that y_n = j₂(x_n). Take z_n = j₁(x_n) for any n ≥ 1. Since j₁, j₂ are isometry, (zn) is a Cauchy sequence in X1. By completeness of X1, we choose z = lim

n→∞zn. Let us show that F (z) = y. By triangle inequality, we can show that d2(F (z), y) = lim

n→∞d2(F (zn), y) :

|d₂(F (z), y) − d2(F (zn), y)| ≤ d2(F (zn), F (z)) = d1(zn, z).

Moreover, it follows from the definition that

n→∞lim d₂(F (j₁(x_n)), y) = lim

n→∞d₂(j₂(x_n), y) = lim

n→∞d₂(y_n, y) = 0.

This implies that d₂(F (z), y) = 0. Hence y = F (z). We prove that F is surjective.

 This theorem implies that the completion of a metric space is unique up to isomorphisms.

Corollary 1.2. (Universal property of completion of a metric space) Let (X, d) be a metric space. Given any isometry f : X → Y into a complete metric space Y and any completion ( bX, bd, j) of (X, d) there is a unique isometry F : bX → Y such that f = F ◦ j.

Proof. Let Z = j(X). Then j : X → Z is an isomorphism. Define g : Z → Y by f ◦ j⁻¹. Since Z is dense in bX and g is an isometry, by Lemma 1.1, there is a unique isometry F : bX → Y so that F |Z = g. This is equivalent to say that F ◦ j = f. We prove our assertion.