I'm trying to figure out if the space $c_{00}=\{(x_n)|\exists N\: \forall n>N \: x_n=0\}$ is a hyperplane and the maximum ideal in the space $c_0=\{(x_n)|\lim x_n=0\}$.
I understand that the closure $c_{00}$ will be $c_0$. But that doesn't help me solve the problem.
I tried to consider the factor $c_0 / c_{00}$ and it is the space of sequences converging to zero with different tails. But $c_0$ itself does not have a unital element in it either. Is this a sufficient reason to say that then $c_0 / c_{00}$ is not a field and $c_{00}$ is not the maximum ideal.
Perhaps I should consider some functionality with the $c_{00}$ kernel?
Help please with the idea of the solution.
I guess $c_{0 0}$ is not a hyperplane. If $X = \{\frac{1}{n}\}_{n \geq 1}\,\bigcup\,\{0\}$ is equipped with the normal metric in $\mathbb{R}$ then $c = \{\{x_n\}\,\vert\,\lim_{n \rightarrow \infty} x_n$ exits$\}$ is actually the entire $C(X)$. In this case $c_0$ is a hyperplane in $c$ but meanwhile $c_{0 0}$ is dense (in $\|\cdot\|_{\infty}$) in $c_0$. If there is a bounded linear functional defined on $c_0$ with kernel $c_{0 0}$, then we can extend that functional to $c$ but then that extension will have kernel equal to $c_0$. Alternatively you can use the fact that $c_0^{\ast} = \ell_1$ and hence no bounded linear functional will have its kernel equal to $c_{0 0}$