# Confusion about Completion of Metric Space Proof

I have started studying Functional Analysis following "Introduction to Functional Analysis with Applications". In chapter 1-6 there is the following proof

For any metric space $X$, there is a complete metric space $\hat{X}$ which has a subspace $W$ that is isometric with $X$ and is dense in $\hat{X}$

(Page 1 & 2) http://i.imgur.com/CRXjh.png

(Page 3 & 4) http://i.imgur.com/PogqC.png

I think I understand parts (a) and (b). At the top of page 3, section (c) where it is proving $\hat{X}$ is complete it states:

Let $(\hat{x_{n}})$ be any Cauchy Sequence in $\hat{X}$. Since $W$ is dense in $\hat{X}$, for every $\hat{x_{n}}$, there is a $\hat{z_{n}}\varepsilon W$ such that $\hat{d}(\hat{x_{n}},\hat{z_{n}}) < \frac{1}{n}$

I do not understand why we choose $\frac{1}{n}$. Would some ε > 0, for each $\hat{x_{n}}$, not suffice? I assume it must not, but I don't see why, so I must not understand this proof. Maybe i'm not sure on what n is referring to because of the subscripts n on the lefthand side. Is n the index of the Cauchy sequence in $\hat{X}$, $(\hat{x_{n}})$? Is it the index of the Cauchy sequence in X, $\hat{x_{n}}$?

Any help would be greatly appreciated, i am pretty dumb and this has puzzled me for a couple days.

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As far as I can tell the author, at that place, is just in need of a (any) sequence $\epsilon_n$ tending to zero as $n\rightarrow \infty$. You could choose any $\epsilon_n$ instead with the property that $\epsilon_n \rightarrow 0$, but $1/n$ is easy to write down and handle. Note though the dependency between the index of the elements of the sequence and the sequence tending to zero. You cannot just take any fixed (independent of $n$) $\epsilon >0$ here.
What I don't understand is, it seems like if W is dense in $\hat{X}$, for any $\hat{x_{n}}$ we should be able to find a $\hat{z_{n}} \varepsilon W$ as close as we like to $\hat{x_{n}}$. Why do the $\hat{x_{n}}$ get closer to the $\hat{z_{n}}$ as $n \to \infty$, instead of the $\hat{z_{n}}$ always being chosen infinitesimally close to the $\hat{x_{n}}$. – Petrarch Jul 1 '12 at 20:43
You can indeed find $\hat{z}_n$ as close as you like to $\hat{x}_n$, but this simply means that for any $\epsilon$ there exists $\hat{z}_n$ such that the distance to $\hat{x}_n$ is smaller than $\epsilon$. You cannot choose a fixed element $\hat{z}_n$ that is 'infinitesimally' close to $\hat{x}_n$ (such elements do not exist, any element that is not $\hat{x}_n$ has some positive distance to $\hat{x}_n$). What you can do is choose the $\hat{z}_n$ such that they get ever closer to the $\hat{x}_n$ as $n$ increases. – Eepzy Jul 1 '12 at 20:52
I think I get it, maybe...Normally they write that For Every $\varepsilon > 0$ there exists an $N$ such that $d(x_{n},y_{n}) < \varepsilon$ for $n>N$. If we want to say something about all n instead of just $n>N$, we can say $d(x_{n},y_{n}) < \frac{1}{n}$ or any function that goes to zero as $n \to \infty$. Is that right at all? Thank you both Thomas and Eepzy for your help and patience. – Petrarch Jul 1 '12 at 21:11
@AdamRubinson no. As shown in the OPs text, you can write down explicitly $1/n$. – user20266 Jul 2 '12 at 14:27