Questions on Theorem 5.5.7. in Brown-Ozawa I am currently trying to digest the proof of Theorem $5.5.7.$ in the book "$C^*$-algebras and Finite-Dimensional Approxmiations" by N. Brown and N. Ozawa.
Background:
Let $(X,d)$ be a metric space (of bounded geometry/uniformly locally bounded) and denote by $B(x,S)\subseteq X$ the open ball centered at $x$ of radius $S$. Also, we denote by $\ell_n^2$ the $n$-dimensional Hilbert space with orthonormal basis $\{\eta_i\}$ and by $A(X),C_u^*(X)$ the translation algebra respectively the $\textit{uniform Roe algebra}$ associated to $X$ (i.e. the closure of $A(X)$ in $B(\ell^2(X))$). Finally, $(e_{ij})$ will denote the standard matrix units (in $\mathbb{M}_n(\mathbb{C})$).
I have the following three questions (and I would be very happy to receive even partial answers/ideas to any one of them):
$(1)$: How does one realize that $\prod_{n\in \mathbb{N}}\mathbb{M}_k(\mathbb{C})\cong \ell^{\infty}(\mathbb{N})\otimes \mathbb{M}_k(\mathbb{C})$? I suppose that the implication "$\supseteq$" is clear enough.
$(2)$: Assume that there exists $S>0$ such that for every $x\in X$ there is an element $\zeta_x\in \ell^2(X)$ with $||\zeta_x||=1$ and $\text{supp}\zeta_x\subseteq B(x,S)$. Now, the operator $V_x\colon \ell^2(X)\to \ell^2(B(x,S))$  given by $\delta_y\mapsto \zeta_y(x)\delta_y$ satisfies that $\sum_{x\in X}V_x^*V_x=1$ strongly ($\textbf{why?}$) and so defines a u.c.p. map ($\textbf{why?}$) $\psi\colon \prod_{x\in X}B(\ell^2(B(x,S))\to B(\ell^2(X))$ given by $$\psi((b_x)_{x\in X})=\sum_{x\in X}V_x^*b_xV_x$$
(here, complete positivity is true for every $x$-coordinate, but I fail to see how this gives me what I want).
$(3)$: Given $\mathcal{H}=\ell^2_n\otimes \ell^2_n$, we turn $\mathcal{H}\otimes C_u^*(X)$ into a Hilbert $C_u^*(X)$-module through the inner product given by $\langle \xi_1\otimes a_1, \xi_2\otimes a_2\rangle = \langle \xi_1,\xi_2\rangle_\mathcal{H} \cdot a_1^*a_2$. Now, let $\psi\colon \mathbb{M}_n(\mathbb{C})\to C_u^*(X)$ be completely positive and let $[b_{ij}]=[\psi(e_{ij})]^{1/2}\in \mathbb{M}_n(C_u^*(X))$. From this, we define the element
$$\xi_\psi=\sum_{j,k}\eta_j\otimes \eta_k\otimes b_{kj}\in \mathcal{H}\otimes C_u^*(X)$$
It is claimed that "Perturbing $\xi_\psi$, we may assume that $\xi_\psi\in \mathcal{H}\otimes A(X)$". I do not see which kind of "perturbation" that is needed to achieve this.
Thank you very much!
 A: I'll answer $(1)$. More generally, if $M$ is a von Neumann algebra, and $I$ is an index set, then
$$\prod_{i \in I} M \cong \ell^\infty(I)\overline{\otimes} M$$
where the tensor product is the von Neumann tensor product. In your case, $M$ is finite-dimensional, so the von Neumann tensor product is just the algebraic tensor product which leads to
$$\prod_{i \in I} M_k(\mathbb{C}) \cong \ell^\infty(I)\odot M_k(\mathbb{C}).$$
In an ad hoc way, we can proceed in multiple ways. For example, we have a canonical chain of $*$-isomorphisms
$$\prod_{i \in I} M_k(\mathbb{C})= \ell^\infty(I, M_k(\mathbb{C})) \cong M_k(\ell^\infty(I)) \cong  \ell^\infty(I) \odot M_k(\mathbb{C}).$$
A: *

*For C$^*$-algebras, the norm in the direct product is the supremum norm. This is natural if one thinks of the direct product as forming a "block diagonal". Thus
$$
\prod_n A=\{f:\mathbb N\to A,\ \text{ uniformly bounded}\}=\ell^\infty(\mathbb  N,A). 
$$
So one needs to check that $\ell^\infty(\mathbb N,A)\simeq \ell^\infty(\mathbb N)\otimes A$  when $\dim A<\infty$. If we fix a basis $a_1,\ldots,a_r$ of $A$ any $f:\mathbb N\to A$ is of the form $f(n)=\sum_{k=1}^r x_{n,k}\,a_k$, where $x_k=(x_{n,k})_n\in\ell^\infty(\mathbb N)$. Then we define the map
$$
\gamma:f\longmapsto \sum_{k=1}^r x_k\otimes a_k.
$$
Using that $a_1,\ldots,a_r$ form a basis, it is easy to see that $\gamma$ is bijective. Linearity and multiplicativity are also trivial, as is the fact that $\gamma$ preserves adjoints.


*In the canonical basis the operator $V_x$ is "diagonal", since its sends $\delta_y$ to a multiple. It follows that
$$
V_x^*\delta_y=\overline{\zeta_y(x)}\,\delta_y,\qquad\qquad\text{ and } \qquad\qquad V_x^*V_x\,\delta_y=|\zeta_y(x)|^2\,\delta_y. 
$$
Then, for any $F\subset X$ finite,
$$
\sum_{x\in F}V_x^*V_x\sum_y\alpha_y\delta_y=\sum_{x\in F}\sum_y\alpha_y|\zeta_y(x)|^2\,\delta_y=\sum_y\alpha_y\Big(\sum_{x\in F}|\zeta_y(x)|^2\Big)\,\delta_y\xrightarrow[F]{}\sum_y\alpha_y\delta_y
$$
since $\sum_{x}|\zeta_y(x)|^2=1$. That is, $\sum_xV_x^*V_x\to1$ sot.
When you apply $\psi^{(n)}$ to a positive $n\times n$ matrix $[(b^{(k,j)}_x)_x]$ (where positivivity means positivity for each $x$) you get
$$
\sum_x[V_x^*b^{(k,j)}_x V_x]_{k,j}=\sum_x\begin{bmatrix} V_x\\ &\ddots\\&&V_x\end{bmatrix}\big[b_x^{(k,j)}\big]\,\begin{bmatrix} V_x\\ &\ddots\\&&V_x\end{bmatrix}^*\geq0
$$


*Given $\varepsilon>0$, there exists elements $b_{kj}'\in A(X)$ such that $\|b_{kj}-b_{kj}\|<\varepsilon/n^2$. Then
$$
\xi_\psi'=\sum_{j,k}\eta_j\otimes \eta_k\otimes b_{kj}'\in \mathcal{H}\otimes A(X)$$ and $\|\xi_\psi'-\xi_\psi\|<\varepsilon$.

