Let $X=C[0,1]$ be the set of all continuous functions $f:[0,1] \longrightarrow \mathbb{R}$ together with the supremum norm $\lVert \cdot \rVert _{\infty}$. Let's consider the closed subespace $M$ of $X$ consisting of all the functions $f \in X $ such that $f(x_n)=0$ for all $n \in \mathbb{N}$, here $x_n=1/2+1/2^n$. I want to determine the space $X/M$.
Mi idea was to start with simpler cases in other to understand the general case. That is, if $M^{n}$ denote the closed subespace of $X$ consisting of all the functions $f \in X $ such that $f(x_k)=0$ for $k=1,\ldots,n$ then we can consider the mapping $g_n: X/M^{n} \longrightarrow (\mathbb{R}^n, \lVert \cdot \rVert _{\infty}) $ defined by $[f] \mapsto (f(x_1),\ldots,f(x_n))$ and this turns out to be a bijective isometry between these two normed spaces. Therefore, I thought that I could attack the general problem by defining a bijection analogously between $M$ and $l^{\infty}$ (in this case, that mapping would be $[f] \mapsto (f(x_n))_{n \in \mathbb{N}}$). However, even if the map appear to be injective it's not clear to me that it is a surjection. This tell me that maybe it's not $l^\infty$ the space that I should be considering for the bijection and therefore I am wrong.
Any help?
In advance thank you.
The quotient is $c$: Given any $(a_n) \in c$ there is a unique continuous fucntion $f$ such that $f(x_n)=a_n$ for all $n$ and $f(\frac 1 2)=\lim a_n$. I will let you check that $(a_n) \to f+M$ (where $f$ is chosen as above) is indeed an isometric isomorphism.