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@BillJohnson Thank you. To summarize, $X$ has the BAP iff there is a uniformly bounded net of finite-rank operators $X^*\to X^*$ that converges to the identity on $X^*$ in point-weak*-topology. However, considering the same for the point-weak-topology is equivalent to the BAP for $X*$ (which implies the BAP for $X$, but not conversely.)
I think the algebra $B_2$ contains every diagonal operator in $B(\ell_2)$ and also every permutation unitary. It would follow that $B_2$ also contains the uniform Roe algebra of every countable discrete group. Maybe it is known if uniform Roe algebras (of non-exact groups) contain $\prod_k M_k(\mathbb{C})$ ?
What I read in Corollary 4 is: Assume that $A$ is a projectionless C*-algebra. Let $a$ and $b$ be two orthogonal positive elements. Then $\| \|a\|b- \|a\|b \| = \|a\|\cdot\|b\|$.
Yes, a c.p.c. order-zero map $\varphi\colon A\to B$ between two C*-algebras corresponds to a *-homomorphism $\alpha\colon C_0((0,1])\otimes A \to B$ such that $\varphi(a)=\alpha(t\otimes a)$. Assume $B$ is $\sigma$-unital and $b$ is a strictly positive element in $B$. Given a set $F$ in $A$, consider $G=\{\alpha(t\otimes b)\}\cup \varphi(F)$. Then $\varphi(C^*(F))\subset \alpha(C_0((0,1])\otimes C^*(F)) = C^*(G)$.
Yes, this might seem counter-intuitive at first. But every element in $C(X)$ is contained in a finitely generated sub-C*-algebra of $C(X)$, and therefore $C(X)$ is the union (no closure needed) of all its finitely generated sub-C*-algebras.
I mean the AP as in Definition 1.1 of Haagerup, Kraus (1994) # Approximation properties for group C-algebras and group von Neumann algebras [Trans. AMS 344]