TITLE: Example (?) of a Banach space containing an uncomplemented copy of itself
QUESTION [6 upvotes]: 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 $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$?

REPLY [2 votes]: 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.