Example (?) of a Banach space containing an uncomplemented copy of itself I was wondering the following:

Background Question: Does there exist a Banach space $X$ which contains a copy $X_0 \subset X$ of itself that is not complemented?

By "$X_0$ is a copy of $X$", I mean $X_0 \cong X$ via an invertible, bounded, linear map.  Some googling turned up that the answer to the above question is "yes". This thread points to a paper containing an example of an uncomplemented copy of $\ell^1$ in $\ell^1$.  Since the proof seems to depend on some pretty fiddly analysis, and since answering the above question is only incidental to the aims of that paper, I was wondering whether there might be a simpler example of this phenomenon. In particular, I thought that the following example might work:

Example(?): The most famous, and probably the most elementary, example of an uncomplemented subspace is the canonical copy of $c_0$ in $\ell_\infty$. With this in mind, I thought maybe one could use the following:
  \begin{align*}
X = c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots && X_0 = \{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots 
\end{align*}
  The hope is that, if $F \subset X$ was a complement for $X_0$, then this would imply $\pi_2(F) \subset \ell_\infty$ was a complement for $c_0 \subset \ell_\infty$ (a contradiction). Here, $\pi_2$ is projection onto the second factor $X \to \ell_\infty$. 

Towards completing the argument, I have already asked this question, but it hasn't attracted much attention. So, I am asking this (hopefully) better motivated question in addition. So, to reiterate, my question is:

Question: Is $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$?

 A: By the Banach-Mazur theorem, each separable Banach space embeds into $C[0,1]$–in particular $C[0,1]\oplus \ell_2$ does. However, no copy of this space in $C[0,1]$ can be complemented, because if it were, so would be $\ell_2$ (as the relation of being complemented is transitive) but this space is reflexive and $C[0,1]$ has the Dunford-Pettis property. Okay, embed now $C[0,1]\oplus \ell_2$ into $C[0,1]$ and form a direct sum with $\ell_2$ to have a concrete example.
Regarding your second question, these two spaces are isomorphic however if you treat the former as a subspace of the latter, then it is not complemented by the Phillips–Sobczyk theorem.
More generally, if $K$ is a compact metric space such that $C(K)$ is not isomorphic to $c_0$ then $C(K)$ contains an uncomplemented copy of itself (e.g., take $K=[0,1]$). By a result of Bourgain, $\ell_1$ (hence $L_1$ too) contain uncomplemented copies of themselves. The same is true for $\ell_p$ where $p\neq 2,\infty$ as well as for the Tsirelson space and its dual.
To the best of my knowledge the full list of so-far known spaces $X$ with the property that every isomorphic copy of $X$ in $X$ is complemented reads as follows:


*

*injective Banach spaces, e.g., $\ell_\infty(\Gamma)$ or $C(K)$ where $K$ is extremely disconnected,

*Hilbert spaces (separable or not),

*$c_0(\Gamma)$ for any set $\Gamma$,

*$c_0(\Gamma) \oplus H$ where $H$ is a Hilbert space,

*hereditarily indecomposable Banach spaces and indecomposable $C(K)$-spaces,

*certain finite sums of the above-mentioned spaces but not all sums: $c_0\oplus \ell_\infty$ contains an uncomplemented copy of itself.


Everybody is welcome to extend this list.
A: Tomek's answer is very comprehensive. However, I thought I would add a "lower level" explanation.

Proposition: $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ is not complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$.

The idea is to use the following simple lemma: 

Lemma: Let $X \subset Y \subset Z$ be nested Banach spaces.
  
  
*
  
*If $X$ is complemented in $Z$, then $X$ is complemented in $Y$.
  
*If $Y$ is complemented in $Z$, then $Y/X$ is complemented in $Z/X$.
  

Now we prove the proposition as follows. 


*

*Suppose that $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ is complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$. 

*Using (2) to quotient out $\{0\} \oplus \{0\}\oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$, we get that $\{0\} \oplus c_0$ is complemented in $c_0 \oplus  \ell_\infty$.

*Using (1), with the intermediate subspace being $\{0\} \oplus \ell_\infty$, we get that $\{0\} \oplus c_0$ is complemented in $\{0\} \oplus \ell_\infty$.

*The latter was equivalent to "the usual copy of $c_0$ is complemented in $\ell_\infty$", which contradicts Phillip's Lemma.

