Proving the completion of a metric space is complete

[tbrown122387]

Homework Statement

Having a little trouble on number 24 of Chapter 3 in Rudin's Principles of Mathematical Analysis. How do I prove that the completion of a metric space is complete?

Homework Equations

X is the original metric space, X^* is the completion, or the set of equivalence classes generated by the metric \bigtriangleup(P,Q) = \lim_{n \to \infty} d(p_n, q_n) where P,Q \in X^* and \{p_n\} \in P and \{q_n\} \in Q.

The Attempt at a Solution

I guess the thing that's confusing me is thinking about Cauchy sequences of equivalence classes. Every time you compare two new equivalence classes, you compare the limit of two real number sequences I guess. Am I thinking about this correctly?

My gut instinct is to use Baire's theorem for this. Maybe construct some shrinking neighborhoods and show the infinite intersection is nonempty? If this is a good path to take, I'll have two questions:
1. How to construct the neighborhoods (maybe let N_{r_n}p_n be the smallest neighborhood containing the previous point, centered at p_n)?

2. How to show these neighborhoods are dense in X^*? How can an equivalence class be a limit or a point of a neighborhood?

 

[tbrown122387]

Tell me if you think this works:

Assume \{P_n\} is Cauchy in X^* . Fix \epsilon > 0 and pick N large enough so that n \ge N implies \bigtriangleup (P_n, P_m) < \epsilon . This means that \lim_{n \to \infty} d(p_n, p_m) < \epsilon for \{p_n\}, \{p_m\} in P_n, P_m respectively. Since \epsilon was arbitrary, \{p_n\}, \{p_m\} are equivalent, meaning they're in the same equivalence class. Name this class P. Clearly \bigtriangleup (P_n, P) = 0 since for any \{p_j\} \in P, \lim d(p_n, p_j) = 0 (by 24(b)). So, Cauchy sequences of equivalence classes converge to some specific equivalence class.

Note: in 24(b) we showed that it if you substitute in different sequences from the same equivalence class, \bigtriangleup stays the same.

 

[tbrown122387]

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