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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.

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    Oh right it's because (x_n) is a Cauchy sequence. So my question is different (but I'm not going to make a thread about it): given a point x in a metric space (X,d), does there necessarily exist y in X such that d(x,y) = 3 ? I guess the answer is no because for example the unit disc with the usual metric is a counterexample. So in order to ensure the existence of such a y, we must invoke the axiom of choice. Moreover, this problem never arises for Normed Linear Space because we can simply use scaling.2012-07-02