A closed subspace of ell-infinity that is not complemented

$c_0$ is closed in ${\mathrm{\ell }}^{\mathrm{\infty }}$ and is a Banach space for the sup norm by [F1]; ${\mathrm{\ell }}^{\mathrm{\infty }}$ is not second countable and hence is not a Banach manifold in the library's sense, while the identity chart still makes the split-chart condition meaningful for the pair $({\mathrm{\ell }}^{\mathrm{\infty }},c_0)$ by [L1]. [F1, L1] 1.2 Suppose for contradiction that $c_0$ is complemented in ${\mathrm{\ell }}^{\mathrm{\infty }}$: by [L7] fix a bounded linear projection $P:{\mathrm{\ell }}^{\mathrm{\infty }}\to c_0$ of ${\mathrm{\ell }}^{\mathrm{\infty }}$ onto $c_0$, so that $Px=x$ for every $x\in c_0$. [assume-contra, L7] 1.3 Fix an enumeration $(q_n)$ of the rationals; for every irrational $x$ and every $k\ge 1$ let $n_k(x)$ be the least index $n$ not already used among $n_1(x),\dots ,n_{k-1}(x)$ with $q_n\in (x-\frac{1}{k},x+\frac{1}{k})$, and put $A_x:=\{n_k(x):k\ge 1\}$. [L3, L4, construct] 2.1 Each $A_x$ is infinite. If $x\mathrm{\ne }y$, choose $K$ so large that the intervals $(x-\frac{1}{k},x+\frac{1}{k})$ and $(y-\frac{1}{\mathrm{\ell }},y+\frac{1}{\mathrm{\ell }})$ are disjoint whenever $k,\mathrm{\ell }\ge K$. Any common index of $A_x$ and $A_y$ must therefore occur among the first $K-1$ choices for at least one of $x,y$, a finite set; hence $A_x\cap A_y$ is finite. It follows that $x\mapsto A_x$ is injective and $\{A_x\}$ is uncountable by [L3]. [step 1.3, L3, algebra] 3.1 Let $u_x$ be the indicator sequence of $A_x$ and let $Q:={\mathrm{\ell }}^{\mathrm{\infty }}\mathrm{/}{c}_0$ carry the quotient norm; then $u_x\in {\mathrm{\ell }}^{\mathrm{\infty }}$ and $[u_x]\mathrm{\ne }0$ because $u_x\mathrm{\notin }{c}_0$. For distinct $x_1,\dots ,x_m$, remove the finite union of all pairwise intersections from their supports. The resulting indicators have disjoint supports and differ from the $u_{x_j}$ by finitely supported, hence $c_0$, sequences. Therefore, for scalars $c_1,\dots ,c_m$, the quotient norm of ${\sum }_j{c}_j[u_{x_j}]$ equals ${\max }_j\mathrm{\mid }{c}_j\mathrm{\mid }$: the disjoint representative gives the upper bound, and each remaining infinite support attains the corresponding coefficient infinitely often, giving the lower bound against every $c_0$ perturbation. [step 1.3, step 2.1, F1, L6, algebra] 4.1 For every $g\in Q^{\ast }$ and every real $r>0$ the set $\{x:\mathrm{\mid }g([u_x])\mathrm{\mid }\ge r\}$ is finite: for distinct points $x_1,\dots ,x_m$ in it choose unimodular scalars $c_j$ with $c_{j}g([u_{x_j}])=\mathrm{\mid }g([u_{x_j}])\mathrm{\mid }$, so that by [step 3.1] and [L8] one has $mr\le {\sum }_j\mathrm{\mid }g([u_{x_j}])\mathrm{\mid }=\mid g({\sum }_j{c}_j[u_{x_j}])\mid \le \mathrm{\parallel }g\mathrm{\parallel }\cdot \parallel {\sum }_j{c}_j[u_{x_j}]{\parallel }_Q=\mathrm{\parallel }g\mathrm{\parallel }$ and hence $m\le \mathrm{\parallel }g\mathrm{\parallel }\mathrm{/}r$. [step 3.1, L8, choose, algebra] 5.1 No countable family in $Q^{\ast }$ separates the points of $Q$: given $(g_n)$, the set of $x$ with $g_n([u_x])\mathrm{\ne }0$ for some $n$ is the countable union over the pairs $(n,k)$ of the finite sets of [step 4.1] with $r=\frac{1}{k}$, hence countable by [L5]; since $\{A_x\}$ is uncountable by [step 2.1], some $x$ lies outside it, and then the nonzero vector $[u_x]$ is annihilated by every $g_n$. [step 4.1, step 2.1, L5] 6.1 Let $P$ be the projection assumed in step 1.2 and put ${\psi }_n(x+c_0):=x_n-(Px{)}_n$. By [L2] each ${\psi }_n$ is a well-defined bounded linear functional on $Q$ — well-defined because $P$ fixes every element of $c_0$ — with $\mathrm{\parallel }{\psi }_n\mathrm{\parallel }\le 1+\mathrm{\parallel }P\mathrm{\parallel }$ by [L8], and if ${\psi }_n(x+c_0)=0$ for all $n$ then $(x-Px{)}_n=0$ for all $n$, so $x=Px\in c_0$. The countably many functionals ${\psi }_n$ would therefore be a countable separating family in $Q^{\ast }$, contradicting [step 5.1]. [step 1.2, step 5.1, L2, L6, L8] 7.1 This contradiction with [step 5.1] shows that no bounded linear projection of ${\mathrm{\ell }}^{\mathrm{\infty }}$ onto $c_0$ exists; by [L7] $c_0$ is not complemented in ${\mathrm{\ell }}^{\mathrm{\infty }}$, although it is closed there by [step 1.1], and consequently the identity chart admits no split-chart decomposition of $c_0$, as claimed.