the element $\left(a, \{ b_n\}_{n=1}^\infty\right)$ is in $\bigcap_{k=1}^\infty \mathfrak A(U_k)$ and since $a\neq 0$ this ideal is not contained in $\mathfrak A({\ensuremath{\mathbb{N}}\xspace})$. In particular, this implies that $\bigcap_{k=1}^\infty \mathfrak A(U_k) \neq \mathfrak A(U)$ and thus $\mathfrak A$ is not continuous.

For the other implication suppose that $\mathfrak A$ is not continuous and let $U_1 \supsetneq U_2 \supsetneq \dots$ be a (strictly decreasing) sequence of open subsets of $\widetilde {\ensuremath{\mathbb{N}}\xspace}$ such that $\bigcap_{k=1}^\infty \mathfrak A(U_k)  \neq \mathfrak{A} (U)$, where $U$ is the interior of $\bigcap_{ k = 1}^{\infty} U_{k}$. Note that $\bigcap_{k=1}^\infty \mathfrak A(U_k) \nsubseteq \mathfrak A({\ensuremath{\mathbb{N}}\xspace})$ otherwise $\bigcap_{k=1}^\infty \mathfrak A(U_k)  = \mathfrak{A} (U)$. Identifying $\mathfrak A$ with the pullback $\mathfrak A( \infty ) \oplus_{\overline \tau,\rho} \ell^\infty( \{ \mathfrak A( n ) \} )$, we may pick an element $\left(a, \{ a_n \}_{n=1}^\infty \right)$ in $\bigcap_{k=1}^\infty \mathfrak A(U_k)$ such that $a$ is non-zero. Now $\{ a_n \}_{n=1}^\infty$ is a lift of $\overline \tau(a)$ and $\| a_n\|=0$ for $n\in {\ensuremath{\mathbb{N}}\xspace}\setminus \bigcap_{k=1}^\infty U_k$ which is an infinite set.

\[c:continuous\] Let $\mathfrak A$ be a continuous $C^\ast$-algebra over $\widetilde {\ensuremath{\mathbb{N}}\xspace}$ and let ${  \overline \tau  \colon  \mathfrak A(\infty)  \rightarrow  \ell^\infty ( \{ \mathfrak A( n ) \} )  }$ denote the reduced Busby map. 