[proofplan] We prove both directions by translating atomicity into a statement about projections. For $\ell^\infty(I)$, projections are characteristic functions of subsets of $I$, and the minimal nonzero projections are exactly the singleton characteristic functions. Conversely, starting from an atomic abelian von Neumann algebra, we choose a maximal orthogonal family of minimal projections, show its supremum is the identity, and then diagonalize every element over this [partition of unity](/page/Partition%20of%20Unity). The resulting scalar coordinate map is a unital normal $*$-isomorphism onto $\ell^\infty(I)$. [/proofplan] [step:Identify the minimal projections in $\ell^\infty(I)$] Let $I$ be a set, and let $\ell^\infty(I)$ denote the abelian von Neumann algebra of bounded functions $I\to \mathbb C$, with pointwise operations and the supremum norm. For a subset $S\subset I$, define the characteristic function $\mathbb 1_S:I\to \mathbb C$ by $\mathbb 1_S(i)=1$ for $i\in S$ and $\mathbb 1_S(i)=0$ for $i\notin S$. The projections of $\ell^\infty(I)$ are exactly the functions $e:I\to\{0,1\}$, hence exactly the functions $\mathbb 1_S$ with $S\subset I$. If $i\in I$, then $\mathbb 1_{\{i\}}$ is minimal: whenever $0\ne e\le \mathbb 1_{\{i\}}$ is a projection, we have $e=\mathbb 1_T$ for some nonempty subset $T\subset \{i\}$, so $T=\{i\}$ and $e=\mathbb 1_{\{i\}}$. Conversely, if $\mathbb 1_S$ is minimal and nonzero, then $S$ is nonempty; if $i\in S$, the nonzero projection $\mathbb 1_{\{i\}}$ satisfies $\mathbb 1_{\{i\}}\le \mathbb 1_S$, so minimality forces $S=\{i\}$. Thus the minimal nonzero projections in $\ell^\infty(I)$ are precisely the singleton projections $\mathbb 1_{\{i\}}$. [/step] [step:Show that $\ell^\infty(I)$ is atomic] Let $0\ne e\in \ell^\infty(I)$ be a projection. By the preceding step, there is a nonempty subset $S\subset I$ such that $e=\mathbb 1_S$. Choose $i\in S$. Then $\mathbb 1_{\{i\}}$ is a nonzero minimal projection and \begin{align*} \mathbb 1_{\{i\}}\le \mathbb 1_S=e. \end{align*} Therefore every nonzero projection in $\ell^\infty(I)$ dominates a nonzero minimal projection, so $\ell^\infty(I)$ is atomic. Since atomicity is preserved by unital normal $*$-isomorphism, every von Neumann algebra isomorphic to $\ell^\infty(I)$ is atomic. [guided] The point of this direction is that $\ell^\infty(I)$ has a completely visible projection lattice. A projection $e\in\ell^\infty(I)$ satisfies $e^2=e=e^*$, so for each $i\in I$ the scalar $e(i)$ satisfies $e(i)^2=e(i)$ and $\overline{e(i)}=e(i)$. Hence $e(i)\in\{0,1\}$. Therefore $e$ is the characteristic function of the subset \begin{align*} S:=\{i\in I:e(i)=1\}. \end{align*} Now let $0\ne e$ be a projection. Since $e=\mathbb 1_S$ and $e\ne0$, the set $S$ is nonempty. Choose $i\in S$. The singleton projection $\mathbb 1_{\{i\}}$ is nonzero and satisfies \begin{align*} \mathbb 1_{\{i\}}\le \mathbb 1_S=e. \end{align*} It is minimal because the only subsets of $\{i\}$ are $\varnothing$ and $\{i\}$. Thus every nonzero projection contains a minimal nonzero projection below it. This is exactly the projection-lattice definition of atomicity used in the statement. [/guided] [/step] [step:Choose a maximal orthogonal family of minimal projections] Assume now that $A$ is atomic. By [Zorn's lemma](/theorems/1226), choose a maximal family $(p_i)_{i\in I}$ of pairwise orthogonal nonzero minimal projections in $A$. Indeed, partially order the set of pairwise orthogonal families of nonzero minimal projections in $A$ by inclusion. The union of any chain is again a pairwise orthogonal family of nonzero minimal projections, because any two members of the union already occur together in one family in the chain. Hence Zorn's lemma gives a maximal such family. [/step] [step:Prove that the maximal family has supremum equal to the identity] By the completeness of the projection lattice of a von Neumann algebra, applied as in [citetheorem:9267], the supremum \begin{align*} p:=\bigvee_{i\in I}p_i \end{align*} exists in $A$. Suppose $p\ne 1$. Then $1-p$ is a nonzero projection in $A$. Since $A$ is atomic, there is a nonzero minimal projection $q\in A$ such that \begin{align*} q\le 1-p. \end{align*} For every $i\in I$, the inequality $p_i\le p$ gives $q p_i=0$. Thus the enlarged family consisting of $q$ together with all $p_i$ is still pairwise orthogonal, contradicting maximality. Therefore \begin{align*} \bigvee_{i\in I}p_i=1. \end{align*} [/step] [step:Diagonalize each element on the minimal projections] Fix $a\in A$ and $i\in I$. Since $A$ is abelian, $p_iap_i=ap_i$. By the [minimal projection corner criterion](/theorems/9290) [citetheorem:9290], applied to the minimal projection $p_i\in A$, we have \begin{align*} p_iAp_i=\mathbb C p_i. \end{align*} Therefore there exists a unique scalar $\lambda_i(a)\in\mathbb C$ such that \begin{align*} ap_i=\lambda_i(a)p_i. \end{align*} Uniqueness follows because $p_i\ne0$: if $\lambda p_i=\mu p_i$, then $(\lambda-\mu)p_i=0$, hence $\lambda=\mu$. Define \begin{align*} \Phi:A\to \ell^\infty(I),\qquad a\mapsto (\lambda_i(a))_{i\in I}. \end{align*} For each $i\in I$, \begin{align*} |\lambda_i(a)|\|p_i\|_{\mathrm{op}}=\|\lambda_i(a)p_i\|_{\mathrm{op}}=\|ap_i\|_{\mathrm{op}}\le \|a\|_{\mathrm{op}}\|p_i\|_{\mathrm{op}}. \end{align*} Since $p_i$ is a nonzero projection, $\|p_i\|_{\mathrm{op}}=1$, so $|\lambda_i(a)|\le \|a\|_{\mathrm{op}}$. Thus $\Phi(a)\in\ell^\infty(I)$. [/step] [step:Verify that the coordinate map is an injective unital $*$-homomorphism] Let $a,b\in A$, let $\alpha,\beta\in\mathbb C$, and write \begin{align*} ap_i=\lambda_i(a)p_i,\qquad bp_i=\lambda_i(b)p_i. \end{align*} Then \begin{align*} (\alpha a+\beta b)p_i=(\alpha\lambda_i(a)+\beta\lambda_i(b))p_i, \end{align*} so $\Phi$ is linear. Also, \begin{align*} (ab)p_i=a(\lambda_i(b)p_i)=\lambda_i(b)\lambda_i(a)p_i, \end{align*} so $\Phi(ab)=\Phi(a)\Phi(b)$ pointwise. Since $p_i=p_i^*$, taking adjoints in $ap_i=\lambda_i(a)p_i$ gives \begin{align*} a^*p_i=\overline{\lambda_i(a)}p_i, \end{align*} so $\Phi(a^*)=\Phi(a)^*$. Finally, $1p_i=p_i$, hence $\Phi(1)=\mathbb 1_I$. To prove injectivity, suppose $\Phi(a)=0$. Then $ap_i=0$ for every $i\in I$. Let $\mathcal F(I)$ denote the directed set of finite subsets of $I$, ordered by inclusion. For $F\in\mathcal F(I)$, define \begin{align*} e_F:=\sum_{i\in F}p_i. \end{align*} The net $(e_F)_{F\in\mathcal F(I)}$ increases to $\bigvee_{i\in I}p_i=1$, hence $e_F\to 1$ strongly. Since $ae_F=0$ for every $F$, strong convergence gives \begin{align*} a=\lim_F ae_F=0. \end{align*} Thus $\Phi$ is injective. [/step] [step:Construct every bounded diagonal element] Represent $A$ faithfully as a strongly closed unital $*$-subalgebra of $\mathcal L(H)$ for a [Hilbert space](/page/Hilbert%20Space) $H$; this is the operator-algebra representation in which all strong limits below are taken. Let $\lambda=(\lambda_i)_{i\in I}\in\ell^\infty(I)$, and define \begin{align*} M_\lambda:=\sup_{i\in I}|\lambda_i|. \end{align*} For each finite subset $F\in\mathcal F(I)$, define \begin{align*} a_F:=\sum_{i\in F}\lambda_i p_i\in A. \end{align*} The net $(a_F)_{F\in\mathcal F(I)}$ is uniformly bounded by $M_\lambda$, because the projections $p_i$ are pairwise orthogonal and hence \begin{align*} \|a_F\|_{\mathcal L(H)}\le M_\lambda \end{align*} for every finite $F\subset I$. For every vector $\xi\in H$, orthogonality of the projections gives \begin{align*} \|a_G\xi-a_F\xi\|_H^2\le M_\lambda^2\|(e_G-e_F)\xi\|_H^2 \end{align*} whenever $F\subset G$. For arbitrary finite $F,G\subset I$, apply this estimate to the comparable pairs $F\subset F\cup G$ and $G\subset F\cup G$; since $e_F\to1$ strongly, this shows that $(a_F\xi)_F$ is Cauchy in $H$ for every $\xi\in H$. Define $a:H\to H$ by $a\xi:=\lim_F a_F\xi$. The uniform bound gives $\|a\xi\|_H\le M_\lambda\|\xi\|_H$, so $a\in\mathcal L(H)$ and $a_F\to a$ strongly. Since each $a_F\in A$ and $A$ is strongly closed, $a\in A$. For each $j\in I$, if $F$ contains $j$, then \begin{align*} a_Fp_j=\lambda_jp_j. \end{align*} For every $\xi\in H$, strong convergence gives $a_Fp_j\xi\to ap_j\xi$, while the eventually constant identity $a_Fp_j\xi=\lambda_jp_j\xi$ gives the same limit. Hence \begin{align*} ap_j=\lambda_jp_j. \end{align*} Hence $\Phi(a)=\lambda$. Therefore $\Phi$ is surjective. [guided] The only subtle point is that $I$ may be uncountable, so we must not write an unqualified infinite norm-convergent series. Instead, we build the diagonal element as a strong operator limit of finite partial sums. We work in a fixed faithful representation of $A$ as a strongly closed unital $*$-subalgebra of $\mathcal L(H)$ for a Hilbert space $H$. Let $\mathcal F(I)$ be the directed set of finite subsets of $I$, ordered by inclusion. For $F\in\mathcal F(I)$, set \begin{align*} e_F:=\sum_{i\in F}p_i \end{align*} and \begin{align*} a_F:=\sum_{i\in F}\lambda_i p_i. \end{align*} The projections $p_i$ are pairwise orthogonal, so each $e_F$ is a projection and the net $(e_F)_F$ increases to $\bigvee_{i\in I}p_i=1$. Therefore $e_F\to1$ strongly. Now fix a vector $\xi\in H$ in a Hilbert space representation $A\subseteq\mathcal L(H)$. If $F\subset G$, then \begin{align*} a_G\xi-a_F\xi=\sum_{i\in G\setminus F}\lambda_i p_i\xi. \end{align*} The vectors $p_i\xi$ are mutually orthogonal, so the Hilbert space norm satisfies \begin{align*} \|a_G\xi-a_F\xi\|_H^2=\sum_{i\in G\setminus F}|\lambda_i|^2\|p_i\xi\|_H^2. \end{align*} Since $|\lambda_i|\le M_\lambda$ for every $i\in I$, we obtain \begin{align*} \|a_G\xi-a_F\xi\|_H^2\le M_\lambda^2\sum_{i\in G\setminus F}\|p_i\xi\|_H^2. \end{align*} The last sum is exactly $\|(e_G-e_F)\xi\|_H^2$. Because $e_F\to1$ strongly, the tails $(e_G-e_F)\xi$ become small. For arbitrary finite $F,G\subset I$, compare both $F$ and $G$ with the common upper bound $F\cup G$; the triangle inequality then shows that $(a_F\xi)_F$ is a Cauchy net in $H$. Define $a:H\to H$ by $a\xi:=\lim_F a_F\xi$. The uniform estimate $\|a_F\|_{\mathcal L(H)}\le M_\lambda$ gives $\|a\xi\|_H\le M_\lambda\|\xi\|_H$, so $a$ is a bounded operator and $a_F\to a$ strongly. Since each $a_F$ belongs to $A$ and $A$ is strongly closed, the limit $a$ also belongs to $A$. Finally, for a fixed $j\in I$, every finite set $F$ containing $j$ satisfies \begin{align*} a_Fp_j=\lambda_jp_j. \end{align*} For every $\xi\in H$, the strong convergence $a_F\to a$ gives $a_Fp_j\xi\to ap_j\xi$, and the identity $a_Fp_j\xi=\lambda_jp_j\xi$ is eventually constant in $F$. Therefore $ap_j=\lambda_jp_j$. Thus the coordinate of $\Phi(a)$ at $j$ is $\lambda_j$. Since $j$ was arbitrary, $\Phi(a)=\lambda$, proving surjectivity. [/guided] [/step] [step:Conclude the von Neumann algebra isomorphism] We have proved that \begin{align*} \Phi:A\to\ell^\infty(I) \end{align*} is a unital bijective $*$-homomorphism. The construction also shows normality. Let $(q_\alpha)_\alpha$ be an increasing net of projections in $A$ with supremum $q$. For each $i\in I$, the projections $q_\alpha p_i$ increase to $q p_i$, because multiplication by $p_i$ preserves order inside the abelian algebra $A$ and commutes with suprema below the minimal projection $p_i$. Since $q_\alpha p_i=\lambda_i(q_\alpha)p_i$ and $q p_i=\lambda_i(q)p_i$, the scalar coordinates satisfy \begin{align*} \lambda_i(q_\alpha)\uparrow \lambda_i(q) \end{align*} for every $i\in I$. Thus $\Phi$ preserves suprema of increasing nets of projections, which is the projection criterion for normality [citetheorem:9271]. For the inverse, a projection in $\ell^\infty(I)$ has the form $\mathbb 1_S$ for a subset $S\subset I$, and the construction gives \begin{align*} \Phi^{-1}(\mathbb 1_S)=\bigvee_{i\in S}p_i. \end{align*} If $(S_\alpha)_\alpha$ is an increasing net of subsets of $I$ with union $S$, then $(\mathbb 1_{S_\alpha})_\alpha$ increases to $\mathbb 1_S$ and \begin{align*} \bigvee_\alpha \Phi^{-1}(\mathbb 1_{S_\alpha})=\bigvee_\alpha\bigvee_{i\in S_\alpha}p_i=\bigvee_{i\in S}p_i=\Phi^{-1}(\mathbb 1_S). \end{align*} Hence $\Phi^{-1}$ also preserves suprema of increasing nets of projections, so $\Phi^{-1}$ is normal by [citetheorem:9271]. Therefore $\Phi$ is a unital normal $*$-isomorphism \begin{align*} A\cong \ell^\infty(I). \end{align*} Combining this with the first direction proves the equivalence. [/step]