About the Banach algebra $\ell^{\infty} (K( \ell^{1}))$ If $P_{n}: \ell^{1} \rightarrow \ell^{1}$ is the projection onto the first n coordinates, then it's well known that $ P_{n} K(\ell^{1}) P_{n}$ is isomorphic to $B(\ell^{1}_{n})$, the Banach space of all bounded linear maps on the finite dimensional space $\ell^{1}_{n}$, and $K(\ell^{1})$ is the Banach space of compact operators on $\ell^{1}$.
I do have few questions here:
1- Is it true that $\cup_{n=1}^{\infty} B(\ell^{1}_{n})$ is dense in $K(\ell^{1})$? 
2- Can we embed the space $ c_{0}-\oplus_{n=1}^{\infty} B(\ell^{1}_{n})$ into $c_{0}- \oplus_{n=1}^{\infty} K(\ell^{1})$, 
where for a family of Banach space $(X_{i})$, $c_{0}-\oplus_{n=1}^{\infty} X_{i}$ = { $(x_{i}) \in \prod X_{i}: x_{i} = 0$ for all but finitely many i}?
3- What happens when we replce $\ell^{1}$ by any other sequence space like $\ell^{p}$ for $p \in (1, \infty)$? and 
4-  What happens if we consider the space $\ell^{\infty}-\oplus_{n=1}^{\infty} B(\ell^{p}_{n})$ and $\ell^{\infty}-\oplus_{n=1}^{\infty}(K(\ell^{p})$?
Thanks in advance.
 A: *

*Yes, as every compact operator on $\ell_1$ is approximable by finite-rank operators.

*The $c_0$-sum is not what you say; you have defined the algebraic direct sum. In any case, this is possible. Where exactly are you stuck?

*
*

*and 2. will work as these spaces have a Schauder basis. You simply re-do the same constructions.


*Suppose that we have two sequences of Banach spaces $(X_n)_{n=1}^\infty$, $(Y_n)_{n=1}^\infty$ with $X_n\subset Y_n$ and such that $X_n$ is reflexive ($n\in \mathbb{N}$). As $(\bigoplus_{n=1}^\infty X_n)_{c_0}^{**}  = (\bigoplus_{n=1}^\infty X_n^{**})_{\ell_\infty}$ (similarily for $Y_n$), we have
$(\bigoplus_{n=1}^\infty X_n)_{c_0}^{**}  \subseteq (\bigoplus_{n=1}^\infty Y_n^{**})_{\ell_\infty}$. However, by reflexivity,
$$(\bigoplus_{n=1}^\infty X_n)_{\ell_\infty}=(\bigoplus_{n=1}^\infty X_n^{**})_{\ell_\infty}  \subseteq (\bigoplus_{n=1}^\infty Y_n^{**})_{\ell_\infty}.$$
Consequently, $(\bigoplus_{n=1}^\infty X_n)_{\ell_\infty}\subseteq (\bigoplus_{n=1}^\infty Y_n)_{\ell_\infty}.$ Apply this now to $X_n=B(\ell_1^n)$, which are finite-dimensional and $Y_n=\mathcal{K}(\ell_1)$ ($n\in \mathbb{N}$).
