Let us consider the space of convergent sequences which is denoted by $c$. The space of all sequences $(x\_n)\in c$ with $\lim x\_n=0$ is also denoted by $c\_0$. Clearly $c\_0$ is a proper closed subspace of $c$ under the uniform topology. 


These two Banach spaces are the same if and only they are the same as two commutative C\*-algebras. But $c\_0$ is not unital, however $c$ is unital. 


Q1. How one may prove this fact just using functional analysis point of view (elementary tools)?


There exists a functional $\phi\in c^\*$ with $\phi(1)=1$ and vanishing on $c\_0$ where $1$ is the constant sequence $1,1,\cdots$. The functional $\phi$ corresponds a bounded sequence $(t\_n)$.


Q2. How could we determine (explicitly) all bounded sequences $(t\_n)$ vanishing on $c\_0$? 



$c_0$ is not isometrically isomorphic to $c$
The (multiplicative) Banach–Mazur distance between $c$ and $c\_0$ is exactly 3:



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> M. Cambern, [On mappings of sequence spaces](http://matwbn.icm.edu.pl/ksiazki/sm/sm30/sm3017.pdf), *Studia Math.* **30**. (1968), 73-77.
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Let me take this opportunity to advertise a rather crazy conjecture (due to Pełczyński, I think):



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> Suppose that two Banach spaces $C(K)$ and $C(L)$ are isomorphic. Is the Banach–Mazur distance between $C(K)$ and $C(L)$ an integer?
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