Suppose that $p\in \mathfrak A( \infty )$ is a non-zero projection and that $\{ q_n \}_{n=1}^\infty \in \ell^\infty( \{ \mathfrak A(n ) \} )$ is a projection which lifts $\overline \tau(a)$. Then $q_n=0$ for only finitely many $n\in {\ensuremath{\mathbb{N}}\xspace}$.

This follows from Lemma \[l:continuous\] since each $q_n$ is a projection for each $n\in {\ensuremath{\mathbb{N}}\xspace}$ and thus has norm $0$ or $1$.

\[c:tight\] Let $\mathfrak{A}$ be a $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$ such that $\mathfrak{A}(n)$ is non-zero and simple for all $n \in \widetilde{{\ensuremath{\mathbb{N}}\xspace}}$. Then $\mathfrak{A}$ is a tight $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$ if and only if $\mathfrak{A}$ is a continuous $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$.

Since $\mathfrak{A}$ is a tight $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$, we have that the map ${  \psi  \colon  \mathrm{Prim} (\mathfrak{A} )  \rightarrow  \widetilde{{\ensuremath{\mathbb{N}}\xspace}}  }$ is a homeomorphism. Hence, $\psi$ is an open map which implies that $\mathfrak{A}$ is a continuous $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$.

Suppose $\mathfrak{A}$ is a continuous $C^{*}$-algebra over $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$. Let $\mathfrak{I}$ be an ideal of $\mathfrak{A}$. Suppose $\mathfrak{I} \subseteq \mathfrak{A}({\ensuremath{\mathbb{N}}\xspace})$. Since $\mathfrak{A}(n)$ is simple for all $n \in {\ensuremath{\mathbb{N}}\xspace}$, we have that $\mathfrak{I} = \mathfrak{A}(U)$ for some $U \subseteq {\ensuremath{\mathbb{N}}\xspace}$. Suppose $\mathfrak{I}$ is not a subset of $\mathfrak{A}({\ensuremath{\mathbb{N}}\xspace})$. By Lemma \[l:continuous\], $$\begin{aligned}
F = {\ensuremath{\left\{  n \in {\ensuremath{\mathbb{N}}\xspace}\: : \:  