Let $M$ be a finite-dimensional subspace of $c\_{0}$, and let $\varepsilon>0$.



> 
> **Question.** Does there exist a finite rank projection from $c\_{0}$, of norm $\leq 1+\varepsilon$, onto a subspace $N$ of $c\_{0}$ with
> $M\subseteq N$ and Banach-Mazur distance
> $\textrm{d}(N,l\_{\infty}^{n})\leq 1+\varepsilon$, where
> $n=\textrm{dim} N$ ?
> 
> 
> 



Finite-dimensional subspaces of $c_{0}$
For non-zero sequence $x=(x\_1, x\_2, \ldots)$ denote $L(x)=\min\{i:x\_i\ne 0\}$ (leader of $x$). By Gauss elimination, $M$ contains a basis $(p\_1, \ldots, p\_d) $ with distinct leaders $m\_1<m\_2<\ldots<m\_d$ respectively. Let $f\_1, f\_2, \ldots$ denote consecutive standard basic vectors $e\_k$ with $k\notin \{m\_1,\ldots,m\_d\}$. Consider the span $N$ of $M$ and $f\_1, \ldots, f\_s$ with very large $s$. On this $(d+s)$-dimensional space $N$ with the natural basis $\{p\_1,\ldots,p\_d,f\_1,\ldots,f\_s\}$, the coordinate functionals corresponding to $p\_i$'s are uniformly (by $s$) bounded: actually their expressions via the usual coordinate functionals do not depend on $s$. Now you cut all $p\_j$'s on the corresponding level $d+s$: denote $n\_i=P\_{d+s} p\_i$ where $P\_{d+s} x=(x\_1,\ldots,x\_{d+s},0,0,\ldots)$. The span $N$ of $n\_1,\ldots,n\_d,f\_1,\ldots,f\_s$ is the coordinate subspace $l\_\infty^{d+s}\subset c\_0$. The map $\Phi\colon M\to N$ which acts as $$\Phi\left( \alpha\_1p\_1+\ldots+\alpha\_sp\_s+\beta\_1f\_1+\ldots+\beta\_s f\_s\right)=
\alpha\_1n\_1+\ldots+\alpha\_sn\_s+\beta\_1f\_1+\ldots+\beta\_s f\_s$$
satisfies $\|x-\Phi x\|<\varepsilon\_s$ for all $x$ on the unit sphere of $M$ and $\varepsilon\_s$ tending to 0 for large $s$ (since $\alpha\_i's$ are uniformly bounded as observed above, and $n\_i$'s become close to $p\_i$'s when $s$ is large). This proves that $N$ and $M$ are Banach -- Mazur close.


The existence of a norm-1 projection to $l\_\infty^n$ is a general fact, as observed by Dirk Werner in the comments.


