\mathfrak{I} \cap \mathfrak{A}(n)  = 0  \right\}}}\end{aligned}$$ is finite. Set $U = \widetilde{ {\ensuremath{\mathbb{N}}\xspace}} \setminus F$. Then $U$ is an open subset of $\widetilde{{\ensuremath{\mathbb{N}}\xspace}}$ and $\mathfrak{I} \subseteq \mathfrak{A}(U)$. Let ${  \iota  \colon  \mathfrak{I}  \rightarrow  \mathfrak{A}(U)  }$ be the inclusion map. Then the diagram $$\begin{aligned}
\xymatrix{
0 \ar[r] & \mathfrak{A}(U \setminus \{ \infty \} ) \ar[r] \ar@{=}[d] & \mathfrak{I} \ar[r] \ar[d]^{\iota} & \mathfrak{A}(\infty) \ar[r] \ar@{=}[d] & 0 \\
0 \ar[r] & \mathfrak{A}(U \setminus \{ \infty \} ) \ar[r] & \mathfrak{A}(U) \ar[r] & \mathfrak{A}(\infty) \ar[r] & 0
}\end{aligned}$$ is commutative and the rows are exact. Thus, $\iota$ is surjective which implies that $\mathfrak{I} = \mathfrak{A}(U)$. We have just shown that the lattice map $\mathbb O(\widetilde{{\ensuremath{\mathbb{N}}\xspace}}) \to \mathbb I(\mathfrak A)$ is surjective, thus it remains to show that it is injective. Let $U,V\in \mathbb O(\widetilde{{\ensuremath{\mathbb{N}}\xspace}})$ such that $\mathfrak A (U) = \mathfrak A(V)$. If $U\subseteq {\ensuremath{\mathbb{N}}\xspace}$ then as $\mathfrak A(U)$ is tight over $U$ it follows that $V=U$. If $\infty \in U$ then $F= \widetilde{{\ensuremath{\mathbb{N}}\xspace}} \setminus U$ is a finite subset of ${\ensuremath{\mathbb{N}}\xspace}$ and $\mathfrak A \cong \mathfrak A(F) \oplus \mathfrak A(U)$ naturally. It follows that $F = \widetilde{{\ensuremath{\mathbb{N}}\xspace}} \setminus V$ and thus $U=V$.

A $C^{*}$-algebra $\mathfrak{A}$ has *weak cancellation* if any pair of 