Let $B$ be a $C^*$ algebra,$H_B$ be the completion of direct sum of countable number of copies of $B$.Why $\{\alpha_j\}$ is the standard orthonormal basis for $H_B$,where $\alpha_j$ has a $1_B$ in the j-th place and zeros elsewhere?
If $(x_1,\cdots,x_n,\cdots)$ is an element in $H_B$,how to show that it can be expressed as a linear combination of $\{\alpha_j\}$?What troubled me is that $x_1$ cannot be written as the form of $k1_B,k\in \Bbb C$
It is not true that every element of $H_B$ can be written as a linear combination of $\{a_j\}$. What is true is that the collection $$\left\{\sum_{n=0}^\infty \alpha_jx_j: x_j\in B,x_j=0\text{ for all but finitely many }j\right\}$$ is dense in $H_B$. Additionally, $\|\alpha_j\|_{H_B}=1$ for all $j$, and $\langle \alpha_j,\alpha_k\rangle=\delta_{jk}$ for all $j,k\in N$.
In short, $\{\alpha_j\}$ is orthonormal with respect to the $B$-valued inner product, and the $B$-linear span of $\{\alpha_j\}$ is dense in $H_B$.