# An exercise about nuclear C*-algebra

Definition 2.3.1. A C*-algebra $A$ is nuclear if the identity map id$_{A}:A \rightarrow A$ is nuclear.

Exercise 2.3.7. If for each finite set $F\subset A$ and $\epsilon>0$ one can find a nuclear subalgebra $B\subset A$ such that $B$ almost contains $F$, within $\epsilon$ in norm, then $A$ is nuclear. In particular, the class of nuclear C*-algebras is closed under taking inductive limits with injective connecting maps.

My question are

1. I do not know the definition of "almost contain, within $\epsilon$ in norm".

2. Could someone give show me more details of this exercise or give me some hints?

Thanks

1. Likely means that for every $a\in F$ there exists $b\in B$ with $\|a-b\|<\varepsilon$.
2. You fix $F=\{a_1,\ldots,a_n\}\subset A$ finite and $\varepsilon >0$. By hypothesis there exists $B$ nuclear, and $\{b_1,\ldots,b_n\}\subset B$ with $\|a_j-b_j\|<\varepsilon/3$. As $B$ is nuclear there exist $k\in \mathbb N$, $\varphi:B\to M_k(\mathbb C)$, and $\psi:M_k(\mathbb C)\to A$, both completely contractive, such that $\|\psi\circ\varphi(b_j)-b_j\|<\varepsilon/3$ for all $j$. Then, extending $\varphi$ to $A$, $$\|\psi\circ\varphi(a_j)-a_j\|\leq\|\psi\circ\varphi(a_j-b_j)\|+\|\psi\circ\varphi(b_j)-b_j\|+\|b_j-a_j\|\\ \leq2\|a_j-b_j\|+\|\psi\circ\varphi(b_j)-b_j\|<3\varepsilon/3=\varepsilon.$$
• And I suppose you may use Arveson's extension Theorem on $\varphi$ to lift $B$ to $A$. – Yan kai Apr 1 '14 at 10:34