Let $(\mathfrak{A}_n)_{n \in \mathbb{N}}$ be a sequence of $C^*$-algebras. Let $\|\cdot\|_n$ denote the norm of $\mathfrak{A}_n$.Then its infinite direct sum is defined as the $c_0$-direct sum (cp. 1)
\begin{equation}
\bigoplus_{n \in \mathbb{N}} \mathfrak{A}_n = \left\{(A_n)_{n \in \mathbb{N}} \middle | A_n \in \mathfrak{A}_n \mathrm{~and~} \lim_{n \rightarrow \infty} \|A_n\|_n = 0\right\}
\end{equation}
together with the sup-norm $\| (A_n)_{n \in \mathbb{N}} \|_\infty = \sup_{n \in \mathbb{N}} \|A_n\|_n$ and componentwise defined multiplication/summation/involution?
My question is: Why isn't it defined as the $l^\infty$-direct sum given by
\begin{equation}
\infty-\bigoplus_{n \in \mathbb{N}} \mathfrak{A}_n = \left\{(A_n)_{n \in \mathbb{N}} \middle | A_n \in \mathfrak{A}_n \mathrm{~and~} \| (A_n)_{n \in \mathbb{N}}\|_\infty < \infty\right\}
\end{equation}
together with the sup-norm and multiplication/summation/involution defined componentwise.
Sources:
1: Example 1.2.5 of "Lecture notes on C^*-algebras" by Ian F. Putnam
Best Answer
Both sums are usually considered, but the second one is often called the product. The direct sum is more natural, because if you embed each $\mathfrak A_n$ in its component, the direct sum is precisely the C$^*$-algebra generated by these embeddings.
The product is "unnatural" in the sense that if you take the product of countably many separable C$^*$-algebras you get a non-separable C$^*$-algebra. So the product makes more sense for von Neumann algebras than it does for C$^*$-algebras.