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Quoted from the book Introductory functional Analysis by Erwin Kreyszig :

1.6-2 Theorem (Completion). For a metric space $X = (X, d)$ there exists a complete metric space $\bar{X}=(\bar{X}, \bar{d})$ which has a subspace $W$ that is isometric with $X$ and is dense in $\bar{X}$. This space $\bar{X}$ is unique except for isometrics, that is, if $\bar{X'}$ is any complete metric space having a dense subspace $W'$ isometric with $X$, then $\bar{X'}$ and $\bar{X}$ are isometric.

Why $W$ must be a dense subset of $\bar{X}$? Is it just because generalization is from completion of $\mathbb{Q}$ (to $\mathbb{R}$) or being dense is something to do with a super-space to be complete? And if so, why?

Edit - For example (0,1) is not a complete space but its closure [0,1] is. And still (0,1) is not dense in [0,1] because int((0,1)) is not empty set.

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    Concerning your edit: I believe you have the definitions mixed up. $(0,1)$ is dense in $[0,1]$ because $[0,1]$ is the closure of $(0,1)$. A set is said to be dense in a topological/metric space if its closure is the entire space. This has nothing to do with the interior! – haemi Aug 15 '17 at 12:58
  • @haemi, yes I had a problem with definition of closure and its consequences. Then I came to https://math.stackexchange.com/questions/949636/prove-that-inta-overlinex-a-where-x-a-is-the-closure-and-the-ba but got a bigger problem: Let $A={{0,1}}$ and $X=[0,1]$. Then how $\text{Int}(A) = \text{Cl(X - A)}$?! –  Aug 15 '17 at 13:02
  • if I understand correctly what's going on in the question you linked (as Daniel Fischer puts it quite nicely, the notation is evil), it should be $\text{Int}(A)=\text{Cl}(X-A)^\text{c}$ (you missed the complement, as I see it). In your example, $\text{Int}(A)=\emptyset$ and $\text{Cl}(X-A)=[0,1]=X$, so it all works out... – haemi Aug 15 '17 at 17:44
  • @haemi, the notations were confusing but I understand all now. Thanks –  Aug 15 '17 at 18:09
  • glad to help :) – haemi Aug 15 '17 at 19:27

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A completion of $(X,d)$ should be a complete metric space $(\hat{X},\hat{d})$ which contains (an isometric copy of) $X$. Without imposing additional requirements, a metric space will have various different "completions". For example, $\mathbb{Q}$ sits in the complete metric space $\mathbb{R}$ but $\mathbb{Q}$ also sits in the complete metric space $\mathbb{R}^2$. Let's assume that $(\hat{X},\hat{d})$ is a complete metric space such that $X \subseteq \hat{X}$. If $X$ is not dense in $\hat{X}$, we can consider the closure $\overline{X}$ of $X$ in $\hat{X}$. Since a closed subspace of a complete metric space is complete, we have found a smaller space $\overline{X}$ which is complete and contains a copy of $X$.

However, if in addition we require that $X$ should be dense in $\hat{X}$, it turns out that there exists essentially only one completion ("the smallest one") up to an isometry.

levap
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    @Edi: Let's say you have a metric space $(X,d)$ which is not complete and I give you a complete metric space $(\hat{X},\hat{d})$ and an isometric embedding of $(X,d)$ in $(\hat{X},\hat{d})$ so that we can think of $X$ as a subset of $\hat{X}$. How can we construct a completion of $X$ in your sense? If $X$ is dense in $\hat{X}$ we are done and $\hat{X}$ is a completion. If not, we can take instead the closure $\overline{X}$ of $X$ in $\hat{X}$. This will be a complete metric space (because it is a closed subspace of a complete metric space) and by definition, $X$ will be dense in $\hat{X}$. – levap Aug 15 '17 at 11:18
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    @Edi: However, if I start with an abstract metric space $(X,d)$, it does not sit in any other space so it doesn't make sense to "take the closure". The proof of the theorem you quote constructs a complete metric space $(\hat{X},\hat{d})$ such that $(X,d)$ embeds isometrically in $(\hat{X},\hat{d})$ and such that $\overline{X} = \hat{X}$. That's why it's even denoted in the text as $(\overline{X},\overline{d})$ - because the closure of the subset $X$ in $\overline{X}$ (denoted usually as $\overline{X}$) is the whole space. – levap Aug 15 '17 at 11:19
  • @Edi: I'm not sure what you mean. The interval $(-1,1)$ (open ball) is dense in $[-1,1]$. – levap Aug 15 '17 at 12:01
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I am not sure what is your question.

Why $W$ must be dense?

This is by the construction in the proof of the theorem.

Why is it important?

Because it basically tells you that in every metric space, only adding a 'few' limit points will make it complete. This is remarkable! Assume for a second you ignore the density part. Then a good question would be: given a metric space, what is the smallest complete space containing it? The answer we now know that it is it's 'closure'.

John D
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  • Can we construct another $W$ not being dense? Thanks. –  Aug 15 '17 at 11:25
  • For example an open ball is not dense in its closure! –  Aug 15 '17 at 11:35
  • Yes we can. Consider $X=(0,1)$ with the usual metric. Then $\hat{X}= [-1,1]$ with the usual metric is a complete space containing $X$ and $W=X$ it is not dense in $\hat{X}.$ What can be said from the result is that any complete space that contains $X$ and $X$ is dense in it is isometric to $[0,1]$ – John D Aug 15 '17 at 11:50