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!
Best Answer
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 $(c_0, \lVert \cdot \rVert)$.