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In the book I'm reading, the theorem is as follow:

If $\mathcal{X}$ is a normed space, then there is a Banach space $\hat{\mathcal{X}}$ and a linear isometry $U:\mathcal{X}\to \hat{\mathcal{X}}$ such that $U(\mathcal{X})$ is dense in $\hat{\mathcal{X}}$.

Then the author leaves an example：

As a practical point, it is easier to think of $\mathcal{X}$ as contained in $\hat{\mathcal{X}}$ rather than work through the isometry $U$. When we actually complete a specific normed space, this is what happens. For example, if $\mathcal{X} = c_{00}$ and it is given the supremum norm, then $\hat{c_{00}} = c_0$.

I'm wondering what does this example means. Why is the completion of finite sequences be infinite sequences? And moreover, how should I understand this completion procedure? Any help and hints will be appreciated!

Best regards!

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    $\begingroup$ Do you understand how the completion of the rationals is the reals? $\endgroup$
    – JonathanZ
    Commented Sep 12, 2022 at 3:26
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    $\begingroup$ Yes. And do you know that there are Cauchy sequences of rationals that don't have a limit in the rationals, but that every Cauchy sequence of reals has a limit in the reals? $\endgroup$
    – JonathanZ
    Commented Sep 12, 2022 at 3:31
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    $\begingroup$ Okay, good. That takes care of the motivation part, because an incomplete Banach space and its completion are like the incomplete rationals and the complete reals. Now, for that example, you should be able to find a sequence (of sequences) in $c_{00}$ that is Cauchy (wrt the given norm), that doesn't have a limit in $c_{00}$. $\endgroup$
    – JonathanZ
    Commented Sep 12, 2022 at 3:37
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    $\begingroup$ If it's the isometry that's giving you trouble, that's because your incomplete space might not be sitting nicely inside a complete space. E.g. if you only knew rationals as an integer over an integer, it might be hard to understand exactly what a sequence converging to $\pi$ is converging to. But embed those $a/b$'s in the space of decimals, and 3.1, 3.14, 3.141, ... has an obvious limit $\endgroup$
    – JonathanZ
    Commented Sep 12, 2022 at 3:45
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    $\begingroup$ Oh, and I left off the final bit from my comment two above: that Cauchy sequence that doesn't have a limit in $c_{00}$? It will have a limit in $c_0$. Also, anything in $c_{0}$ is the limit of some sequence in $c_{00}$. That's what it means for it to be the completion of $c_{00}$. $\endgroup$
    – JonathanZ
    Commented Sep 12, 2022 at 3:59

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First, note that $(c_{00}, \lVert \cdot \rVert)$, the normed space of all sequences $x = (x_1, x_2, \dots)$ with finitely many non-zero entries (i.e. $x_n = 0$ eventually) with $\lVert x \rVert_\infty = \sup_n |x_n|$ is not complete. The sequence (of sequences) $x^{(k)} = (1, 1/2, \dots, 1/k, 0, 0, \dots)$ for $k = 1, 2, \dots$ is a Cauchy sequence that is converging towards $x^{(\infty)} := (1, 1/2, 1/3, \dots)$. The problem is that $x^{(\infty)} \notin c_{00}$ since it has infinitely many non-zero entries.

What is a reasonable way to complete this space to obtain $\widehat{c}_{00}$? Naturally, we want to make sure that $x^{(\infty)}$ is in the space. The solution: the completion that your text is referring to is to consider the space of all Cauchy sequences in $c_{00}$ equipped with the "metric" $d(\widehat{x}, \widehat{y}) := \lim_{k \to \infty} \lVert x^{(k)} - y^{(k)} \rVert$ for any two Cauchy sequences $\widehat{x} = (x^{(1)}, x^{(2)}, \dots)$ and $\widehat{y} = (y^{(1)}, y^{(2)}, \dots)$ of elements $x^{(k)}, y^{(k)} \in (c_{00}, \lVert \cdot \rVert)$. This is not quite a metric, but can be fixed by identifying elements up to equivalence (i.e. $\widehat{x} \sim \widehat{y} \iff d(\widehat{x}, \widehat{y}) = 0$). The metric space $(\widehat{c}_{00}, d)$ now consists of all the equivalence classes of Cauchy sequences in $(c_{00}, \lVert \cdot \rVert)$, and it is an exercise to check this construction carefully.

Thus, back to our motivation, the sequence $x^{(\infty)} = (1, 1/2, 1/3, \dots)$ can now be identified (up to equivalence) with the Cauchy sequence $(x^{(1)}, x^{(2)}, \dots)$ . The isometry associated with this construction naturally changes the space. The practical point simply means that it is more fruitful to think of the completed space as still living in the sequence space, and this naturally leads to thinking of the completion of $(c_{00}, \lVert \cdot \rVert)$ as $(c_0, \lVert \cdot \rVert)$.

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  • $\begingroup$ Then how should I identify the original sequences in $c_{00}$ in the new completed $\hat{c_{00}}$? For example I choose a sequence $x \in c_{00}$, maybe I can correspondingly choose the sequence as $\{x,x,x ……\}$ , am I right? $\endgroup$
    – Re-ocean
    Commented Sep 12, 2022 at 7:40
  • $\begingroup$ Exactly: technically, the equivalence class of sequences that are equivalent to $(x, x, x, \dots)$. (Part of the technical points to check is that the choice of representative doesn't matter.) You can see how this is a bit hard to work with intuitively, and hence why we can simply think of $c_{00}$ as living inside $c_{0}$ instead. $\endgroup$
    – JKL
    Commented Sep 12, 2022 at 14:49

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