Dimension of $c/c_0$ Be defined...
\begin{align*}
&c_0:=\{x:=(x_n)_{n\in \mathbb{N}}, x_k \in \mathbb{R}: \lim_{n\rightarrow \infty} x_n = 0\}\\
&c:=\{x:=(x_n)_{n\in \mathbb{N}}, x_k \in \mathbb{R}: \lim_{n\rightarrow \infty} x_n \quad \text{exists}\}
\end{align*}
in $l^{\infty}$-norm
Prove that $\dim(c/c_0)=1$

My Idea:
I want to find out what's $\dim(c/c_0)$ whereas $c/c_0$ is the quotient space. Meaning: $c/c_0:=\{[v]: v\in c\}$ and $[v]:=\{w\in c: \underbrace{v\sim w}_{(v-w) \in c_0}\}$
Since I'm presumably in subspaces $c_0,c$ of $l^{\infty}$, I'm not sure if I can even find a basis of $c$ or $c_0$ since $l^{\infty}$ doesn't really have a base in classical sense (I think?)

So I'd like to prove it with Rank-nullity-theorem: Let $f:V\rightarrow W$ linear map. Then: $\dim(V)=\dim\ker(f)+\dim \text{im}(f)$.
Let $f:c/c_0\rightarrow \mathbb{R}$ be defined as $v\mapsto\lim_{n\rightarrow \infty} v_n$.
Proof that it's linear:

*

*Show $f(\lambda v)=\lambda f(v) \Rightarrow f(\lambda v) =\lim_{n\rightarrow \infty}\lambda v_n=\lambda \lim_{n\rightarrow \infty}v_n=\lambda f(v)$

*Show $f(v+w)=f(v)+f(w)\Rightarrow f(v+w) = \lim_{n\rightarrow \infty} u_n+w_n = \lim_{n\rightarrow \infty}u_n + \lim_{n\rightarrow \infty}w_n=f(v)+f(u)$
Now obviously, the kernel of this map is the zero-class $[0]$ since $f([0]) = 0$, which means that $\dim\ker(f)=0$.
The image is then all real numbers i.e. $f(c/c_0)=\mathbb{R}$ and $\dim \text{im}(f) = \dim(\mathbb{R}) = 1$.
Rank-nullity-theorem gives us $\dim(c/c_0)=1+0 = 1$
Is that proof correct? Thanks in advance for any useful input.
 A: Using the Rank-nullity Theorem is technically correct, but it is kind of bad form. You are using it to hide information you already have.
The function $f$ you constructed is already a vector space isomorphism between $c/c_0$ and $\mathbb R$. Since the rank is finite, it is automatically bounded, so you get the isomorphism as Banach spaces (which RN does not give you). And with very little more effort you can show that $f$ is isometric, so you get that $c/c_0$ and $\mathbb R$ are isometrically isomorphic. And you get for free that they are isometrically isomorphic as Banach algebras, since your $f$ is also multiplicative.
A: You need to be more careful with the details:

*

*$f$ maps the equivalence class of $v = (v_n)_{n \in \mathbb N} \in c$ to the real number $f([v]) := \lim_n v_n$. In particular, one needs to ask itself why $f$ is well-defined, that is, if $[v] = [w]$ (i.e. $v \sim w$) then $f([v]) = f([w])$?

*Sure, linearity of $f$ follows from the limit rules, but also follows from the definition of the operations in $c/c_0$: $[v]+[w] := [v+w]$ and $\lambda[v] := [\lambda v]$.

*The kernel is $\{[0]\}$, but not for the reason that you gave. If $f([v]) = 0$, then $v \in c_0$ and so $v \sim 0$, meaning $[v]=[0]$.

*Surjectivity of $f$ is easy to prove, so you can avoid the rank-nullity theorem by concluding that $f$ is an isomorphism, and then $c/c_0 \cong \mathbb R$ as $\mathbb R$-vector spaces.

A more clean approach is by using the first isomorphism theorem:

Consider $\varphi : c \to \mathbb R$ given by $\varphi(v) = \lim_n v_n$. Then by the limit rules, $\varphi$ is linear, and since $\varphi$ is surjective, it follows that $c/(\ker \varphi) \cong \mathbb R$, but $\ker \varphi = c_0$.

A: Hint:
Any $\boldsymbol{x}\in\mathcal{c}$ can be expressed as
$$\mathbf{x}=\big(\boldsymbol{x}-\lim_n\boldsymbol{x}(n)\,\cdot\boldsymbol{u}\big) +\lim_n\boldsymbol{x}(n)\,\cdot\boldsymbol{u}
$$
where $\boldsymbol{u}(n)=1$ for all $n$. The term
$$
\boldsymbol{y}=\boldsymbol{x}-\lim_n\boldsymbol{x}(n)\,\cdot\boldsymbol{u}\in\mathcal{c}_0
$$
This shows that $\mathcal{c}/\mathcal{c}_0$ is generated by the coset $\boldsymbol{u}+\mathcal{c}_0$; that is, $\operatorname{dim}\big(\mathcal{c}/\mathcal{c}_0\big)=1$.
