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Completing a Metric Space COMPLETING A METRIC SPACE A completion of a metric space X is an isometry from X to a complete metric space, 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, jf(x)jZ ≤ jxjX for all x 2 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 Cauchy sequence f ι(xi) g 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 f yi g in Y converges in Y . Since each yi is the limit of a Cauchy sequence i in ι(X), there exists a sequence f xi g in X such that dY (ι(xi); yi) < 1=2 for each i. To show that f yi g converges in Y , we show that: • Because f yi g is Cauchy in Y , also f ι(xi) g is Cauchy in Xe, where it has a limit y. This limit lies in Y by definition of Y . • Because f ι(xi) g converges to y in Y , so does f yi g. io For the first bullet, given " > 0, there exists io such that 1=2 < " and dY (yi; yj) < " for all i; j ≥ io. Thus f ι(xi) g is Cauchy because for all i; j ≥ io, dY (ι(xi); ι(xj)) ≤ dY (ι(xi); yi) + dY (yi; yj) + dY (yj; ι(xj)) < 3": io Similarly for the second bullet, given " > 0 there exists io such that 1=2 < " and dY (ι(xi); y) < " for all i ≥ io. Thus limi yi = y because for all i ≥ io, dY (yi; y) ≤ dY (yi; ι(xi)) + dY (ι(xi); y) < 2": 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, jF (~x)jZ = jF (lim ι(xi))jZ = j lim F (ι(xi))jZ = j lim f(xi)jZ i i i = lim jf(xi)jZ = lim jxijX = lim jι(xi)j = j lim ι(xi)j = jx~j : 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 = f xi g. 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 " > 0 there exists io(c) such that dX (xi; xj) < " for all i; j ≥ io(c), and similarly forc ~. Let io = maxf io(c); io(~c) g. For any indices i; j ≥ io, ordered so that dX (xi; x~i) ≤ dX (xj; x~j), the triangle inequality gives 0 ≤ dX (xj; x~j) − dX (xi; x~i) ≤ dX (xj; xi) + dX (~xi; x~j) < 2": This shows that the real sequence f dX (xi; x~i) g is Cauchy, and hence it has a limit as claimed. Next, given sequences c; c;~ c0 2 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 f ci g be a Cauchy sequence in C. Thus each ci = f xi;j g is a Cauchy sequence in X. Let " > 0 be given. Since f ci g is Cauchy in C, 0 9 i("): dC (ci; ci0 ) < " 8 i; i ≥ i("); and so, since dC (ci; ci0 ) = limj dX (xi0;j; xi;j) by definition, 0 0 0 (∗) 8 i; i ≥ i(") 9 j(i; i ): dX (xi0;j; xi;j) < " 8 j ≥ j(i; i ): Since each ci is Cauchy in X, 0 (∗∗) 8 i; 9 j(i): dX (xi;j; xi;j0 ) < " 8 j; j ≥ j(i): We may take j(i0) ≥ j(i); j(i; i0) for all i0 > i ≥ i("). Define a diagonal-like sequence, c = f xi g where xi = xi;j(i): To see that c is Cauchy, note that for any i0 > i ≥ i("), 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 " by (∗) because j(i0) ≥ j(i; i0). The second term is less 0 than " by (∗∗) because j(i); j(i ) ≥ j(i). To see that limi ci = c, take any i ≥ i(") 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 " by (∗) because j(i0) ≥ j(i; i0). The second term is less than " 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 = f xi g: i The limit exists because the sequence f f(xi) g is Cauchy in Z in consequence of the sequence f xi g being Cauchy in X and f being a subisometry, and Z is complete.
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