Dimension of $c/c_0$

analysisfunctional-analysislp-spaces

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:

  1. Show $f(\lambda v)=\lambda f(v) \Rightarrow f(\lambda v) =\lim_{n\rightarrow \infty}\lambda v_n=\lambda \lim_{n\rightarrow \infty}v_n=\lambda f(v)$
  2. 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.

Best Answer

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.

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