COMPLETING A SPACE

A completion of a X is an isometry from X to a , ι : X −→ X,e Xe a complete metric space and ι an isometry, that satisfies a mapping property encoded in the diagram

XeO M M M F ι M M M f M X &/ Z Here f : X −→ Z is any subisometry from X to an auxiliary complete metric space,

|f(x)|Z ≤ |x|X for all x ∈ X,Z complete, and the induced map F is unique, and F is understood to be a subisometry as well.

1. Density Granting a completion ι : X −→ Xe, we show that X is dense in Xe. Literally the statement is that ι(X) is dense in Xe, i.e., any point in Xe is the limit of a { ι(xi) } in Xe. Indeed, let Y ⊂ Xe be the subspace of such limits, which clearly includes ι(X). To show that Y is complete, we must show that any Cauchy sequence { yi } in Y converges in Y . Since each yi is the limit of a Cauchy sequence i in ι(X), there exists a sequence { xi } in X such that dY (ι(xi), yi) < 1/2 for each i. To show that { yi } converges in Y , we show that:

• Because { yi } is Cauchy in Y , also { ι(xi) } is Cauchy in Xe, where it has a limit y. This limit lies in Y by definition of Y . • Because { ι(xi) } converges to y in Y , so does { yi }.

io For the first bullet, given ε > 0, there exists io such that 1/2 < ε and dY (yi, yj) < ε for all i, j ≥ io. Thus { ι(xi) } is Cauchy because for all i, j ≥ io,

dY (ι(xi), ι(xj)) ≤ dY (ι(xi), yi) + dY (yi, yj) + dY (yj, ι(xj)) < 3ε.

io Similarly for the second bullet, given ε > 0 there exists io such that 1/2 < ε and dY (ι(xi), y) < ε for all i ≥ io. Thus limi yi = y because for all i ≥ io,

dY (yi, y) ≤ dY (yi, ι(xi)) + dY (ι(xi), y) < 2ε.

Next we show that ιo : X −→ Y , where ι0 is ι with its codomain restricted from Xe to Y , satisfies the completion mapping property, so that in fact Y is all of Xe. Indeed, letting inc denote the inclusion map, the diagram inc / Y _ XeO ?? M M ?? M F ?? ι M ιo ?? M M f M X &/ Z 1 2 COMPLETING A METRIC SPACE shows that f induces the map G = F ◦inc : Y −→ Z, which inherits the subisometry property from F because the inclusion is an isometry. Furthermore, any induced map G out of Y is determined by its values on ιo(X), because ιo(X) is dense in Y by the construction of Y and because the isometry is continuous. More specifically, every point y of Y takes the form y = limi ι(xi), and so

G(y) = G(lim ι(xi)) = lim G(ι(xi)). i i

Thus G is uniquely determined by f because G ◦ ιo = f, and so ιo : X −→ Y satisfies the completion mapping property. This shows that Y is all of Xe. Although the following observation isn’t needed, still we note that because X is dense in Xe, if a map f : X −→ Z from X to a complete metric space is an isometry, rather than just a subisometry, then the resulting map F : Xe −→ Z is an isometry as well. This is because F is an isometry on the dense subspace ι(X) and F is continuous. Belaboring the matter and carefully not taking the limit of a Cauchy sequence in X,

|F (˜x)|Z = |F (lim ι(xi))|Z = | lim F (ι(xi))|Z = | lim f(xi)|Z i i i

= lim |f(xi)|Z = lim |xi|X = lim |ι(xi)| = | lim ι(xi)| = |x˜| . i i i Xe i Xe Xe

2. Uniqueness Granting a completion, a standard argument shows that it is determined up to unique isometric isomorphism, as follows. First, letting ι : X −→ Xe take the role of f : X −→ Z in the characterizing property gives a unique F in the commutative diagram

XeO M M M F ι M M M ι M& X / Xe Of course the identity map works, but the characterizing property says that only the identity map of Xe extends the isometry of X into Xe to a self-subisometry of Xe. Second, if ιj : X −→ Xej for j = 1, 2 are completions of X then in the diagram

Xe1 f Xe2 8 Xe1 NNN O pp NNN ppp NN ι2 pp ι1 NN pp ι1 NNN ppp X p we let ι1 : X −→ Xe1 and ι2 : X −→ Xe2 respectively take the roles of the completion and the test subisometry, and then exchange their roles, to get unique subisometries F21,F1,2 in the commutative diagram

F21 / F12 / Xe1 f Xe2 8 Xe1 NNN O pp NNN ppp NN ι2 pp ι1 NN pp ι1 NNN ppp X p COMPLETING A METRIC SPACE 3

Thus F12 ◦F21 subisometrically extends the identity map of X to an isometry of Xe1, making it the identity map of Xe1 as noted above. And the same argument with 1 and 2 exchanged, shows that F21 ◦ F12 is the identity map of Xe2. Further, the identity map is a true isometry. Altogether, then, F21 is an isometric isomorphism. This shows that any two completion spaces Xe1 and Xe2 of X are isomorphic by a unique isometry, which is to say that the completion is determined up to unique isometric isomorphism.

3. Construction 3.1. Pseudometric Space of Cauchy Sequences. Let C be the space of Cauchy sequences in X. Let a typical element of C be c = { xi }. We show that the function

d : C × C −→ ≥0, dC (c, c˜) = lim dX (xi, x˜i) R i is meaningful, i.e., the limit in the previous display exists. Given ε > 0 there exists io(c) such that dX (xi, xj) < ε for all i, j ≥ io(c), and similarly forc ˜. Let io = max{ io(c), io(˜c) }. For any indices i, j ≥ io, ordered so that dX (xi, x˜i) ≤ dX (xj, x˜j), the gives

0 ≤ dX (xj, x˜j) − dX (xi, x˜i) ≤ dX (xj, xi) + dX (˜xi, x˜j) < 2ε.

This shows that the real sequence { dX (xi, x˜i) } is Cauchy, and hence it has a limit as claimed. Next, given sequences c, c,˜ c0 ∈ C and any index i, the triangle inequality 0 0 in X gives dX (xi, x˜i) ≤ dX (xi, xi) + dX (xi, x˜i), and so passing to the limit gives the triangle inequality in C,

0 0 dC (c, c˜) ≤ dC (c, c ) + dC (c , c).

Also, dC is obviously symmetric and nonnegative. However, it is a pseudometric rather than a metric because it is not strictly positive. We will return to this point later.

3.2. Completeness of C. The proof that the pseudometric space C is complete is the main point of interest in all this. The argument is Cantor diagonalization. Let { ci } be a Cauchy sequence in C. Thus each ci = { xi,j } is a Cauchy sequence in X. Let ε > 0 be given. Since { ci } is Cauchy in C, 0 ∃ i(ε): dC (ci, ci0 ) < ε ∀ i, i ≥ i(ε), and so, since dC (ci, ci0 ) = limj dX (xi0,j, xi,j) by definition, 0 0 0 (∗) ∀ i, i ≥ i(ε) ∃ j(i, i ): dX (xi0,j, xi,j) < ε ∀ j ≥ j(i, i ).

Since each ci is Cauchy in X, 0 (∗∗) ∀ i, ∃ j(i): dX (xi,j, xi,j0 ) < ε ∀ j, j ≥ j(i). We may take j(i0) ≥ j(i), j(i, i0) for all i0 > i ≥ i(ε). Define a diagonal-like sequence,

c = { xi } where xi = xi,j(i). To see that c is Cauchy, note that for any i0 > i ≥ i(ε),

dX (xi0 , xi) = dX (xi0,j(i0), xi,j(i)) ≤ dX (xi0,j(i0), xi,j(i0)) + dX (xi,j(i0), xi,j(i)). 4 COMPLETING A METRIC SPACE

The first term is less than ε by (∗) because j(i0) ≥ j(i, i0). The second term is less 0 than ε by (∗∗) because j(i), j(i ) ≥ j(i). To see that limi ci = c, take any i ≥ i(ε) and note that for any i0 ≥ i, j(i),

dX (xi0 , xi,i0 ) = dX (xi0,j(i0), xi,i0 ) ≤ dX (xi0,j(i0), xi,j(i0)) + dX (xi,j(i0), xi,i0 ). Again the first term is less than ε by (∗) because j(i0) ≥ j(i, i0). The second term is less than ε by (∗∗) because j(i0), i0 ≥ j(i).

3.3. Mapping Property of C. Now we show that the pseudometric space C satisfies the mapping property of the completion other than not being a true metric space. The isometry from X to C takes each element to the corresponding constant sequence. Given a subisometry f : X −→ Z where Z is complete, the only possible compatible induced isometry is

F : C −→ Z,F (c) = lim f(xi) where c = { xi }. i

The limit exists because the sequence { f(xi) } is Cauchy in Z in consequence of the sequence { xi } being Cauchy in X and f being a subisometry, and Z is complete. With F now well defined, also it is a subisometry because for any c, c˜ ∈ C we have

dZ (F (c),F (˜c)) = dZ (lim f(xi), lim f(˜xi)) where c = { xi } andc ˜ = { x˜i } i i

= lim dZ (f(xi), f(˜xi)) because dZ is continuous i

≤ lim dX (xi, x˜i) because f is a subisometry i = dC (c, c˜) by definition of dC .

3.4. True Metric Space Xe = C/ ∼, Its Mapping Property. Finally, to repair the problem that C is only a pseudometric space, construct its quotient space

Xe = C/ ∼, where the equivalence relation is

c ∼ c˜ ⇐⇒ dC (c, c˜) = 0, and the elements of the quotient space are equivalence classes,

[c] = { c˜ ∈ C :c ˜ ∼ c }.

0 We show that dC is defined on pairs of classes. Suppose that c ∼ c in C and take 0 anyc ˜ in C. Let c = { xi } and similarly for c andc ˜. Thus

0 0 dC (c , c˜) ≤ dC (c , c) + dC (c, c˜) = dC (c, c˜),

0 0 and similarly dC (c, c˜) ≤ dC (c , c˜). Since dC is symmetric, also dC (c, c˜) = dC (c, c˜ ) for allc ˜0 ∼ c˜. Thus d ([c], [˜c]) is well defined as the common value of d (c0, c˜0) Xe C over all c0 ∈ [c] andc ˜0 ∈ [˜c]. And if d ([c], [˜c]) = 0 then [c] = [˜c], so d C is a true Xe Xe metric. The projection map π : C −→ Xe taking each c ∈ C to [c] is an isometry. To see that the mapping property of the completion holds for Xe, we check that any induced map out of C is constant on equivalence classes and therefore factors COMPLETING A METRIC SPACE 5 uniquely through Xe,

c ∼ c˜ ⇐⇒ dC (c, c˜) = 0

⇐⇒ lim dX (xi, x˜i) = 0 i =⇒ lim dZ (f(xi), f(˜xi)) = 0 because f is a subisometry i

⇐⇒ dZ (lim f(xi), lim f(˜xi)) = 0 i i

⇐⇒ dZ (F (c),F (˜c)) = 0 ⇐⇒ F (c) = F (˜c). Thus we have X eO G G π G G G F C G O SSSS G SSS F G ι SSSS G SSSS G f SSSG# X S/) Z The map F : Xe −→ Z is a subisometry in consequence of F : C −→ Z being a subisometry, because the projection π : C −→ Xe is an isometry.