y_{1} ) \| < \epsilon$. Thus, there exists $y_{2} \in \mathfrak{I}$ such that $\| x - y_{1} - y_{2} \| < \epsilon$. Since $y_{1} + y_{2} \in \bigcup_{ k = 1}^{\infty} ( \mathfrak{C}_{k} + \mathfrak{I} )$, we have just shown that $\bigcup_{ k = 1}^{\infty} ( \mathfrak{C}_{k} +\mathfrak{I} )$ is dense in $\mathfrak{A}$.

An extension $0 \to \mathfrak{B} \overset{\iota}{\to}  \mathfrak{E} \overset{\pi}{\to} \mathfrak{A} \to 0$ is said to be *quasi-diagonal* if $\mathfrak{B}$ has an approximate identity consisting of projections $\{ p_{n} \}_{n=1}^{\infty}$ such that $$\begin{aligned}
\lim_{ n \to \infty } \| \iota ( p_{n} ) x - x \iota( p_{n} ) \| = 0\end{aligned}$$ for all $x \in \mathfrak{E}$.

We end this section by showing that the extension $0 \to \mathfrak{A}({\ensuremath{\mathbb{N}}\xspace}) \to \mathfrak{A} \to \mathfrak{A}(\infty) \to 0$ is a quasi-diagonal extension under mild assumptions on the fibers. This fact will be used repeatedly throughout the paper.

\[p:quasidiag\] Let $\mathfrak A$ be a separable $C^\ast$-algebra over $\widetilde{\mathbb N}$ such that each $\mathfrak A(n)$ has real rank zero and $\mathfrak A(\infty)$ an AF-algebra. Then the extension $$0 \to \mathfrak A(\mathbb N) \to \mathfrak A \to \mathfrak A(\infty) \to 0$$ is a quasi-diagonal extension.

A functional calculus argument implies that every projection in $q_{\infty} 