Proving Y is Complete in X's Isometric Embedding

[ehrenfest]

Homework Statement

Let X be a metric space, with metric d. Fix a point a in X. Assign to each p in X the function f_p defined by

$$f_p(x) = d(x,p)-d(x,a)$$

where x in X.

Prove that $|f_p(x)|\leq d(a,p)$ for all x in X, and that therefore, f_p in C(X).

Prove that $||f_p-f_q|| = d(p,q)$ for all p,q in X.

If $\phi(p) = f_p$ it follows that \phi is an isometry (a distance-preserving mapping) of X onto $\phi(X) \subset C(X)$.

Let Y be the closure of \phi(X) in C(X). Show that Y is complete.
(Conclusion: X is isometric to a dense subset of a complete metric space Y)

Note: Rudin uses C(X) as the set of complex-valued, continuous, bounded functions with domain X.

Homework Equations

The Attempt at a Solution

I can do everything but show that Y is complete. Of course we know that C(X) is complete. We also know that if {f_p_i} is a Cauchy sequence in \phi(X), then {p_i} is a Cauchy sequence in X.

If {g_n} is a Cauchy sequence in Y, then
$${g_n} = {\lim_{i \to \infty} f_{n,p_i}}$$
but I don't know how to use that to prove that g_n converges.

 

[ehrenfest]

anyone?