Proving Y is Complete in X's Isometric Embedding

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The discussion focuses on proving the completeness of the closure Y of the isometric embedding of a metric space X into the space of continuous functions C(X). It establishes that for any point p in X, the function f_p is continuous and that the distance between functions f_p and f_q corresponds to the distance between points p and q in X. The challenge lies in demonstrating that a Cauchy sequence in Y converges within Y, leveraging the fact that C(X) is complete and that Cauchy sequences in Y correspond to Cauchy sequences in X. The participants are seeking assistance in bridging the gap between the Cauchy sequence in Y and its convergence. The conclusion emphasizes the isometric relationship between X and the dense subset Y within the complete metric space C(X).
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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.
 
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