Codimension and sequences

Let $c$ be the set of all convergent sequences and $c_0$ the set of sequences tending to zero. How would one find the codimension of $c_0$? In particular, how would one find $\text{dim} \ c/c_0$?

So one needs to choose some arbitrary sequence in $c_0$ and represent it as a linear combination of elements in $c/c_0$?

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You don't want to think of $c$ and $c_0$ as "sets" if you are going to talk about codimension: they should be "introduced" as vector spaces. To paraphrase Dr Evil, they didn't go to vector space school to be called "Mr set". –  Arturo Magidin Jan 4 '11 at 6:40

Your final paragraph makes no sense. The elements of $c/c_0$ are equivalence classes of sequences in $c$: they are elements of the form $[a] = \{b\in c\mid a-b\in c_0\}$ where $a$ is some element of $c$. You cannot express an element of $c_0$ as a linear combination of elements of $c/c_0$, because they are not even the same kind of things: the elements of $c_0$ are sequences, the elements of $c/c_0$ are sets of sequences.
The simplest way to find the codimension is to find an onto linear transformation from $c$ to some vector space $\mathbf{V}$ with kernel exactly $c_0$; the codimension will be dimension of $\mathbf{V}$.
This is easy in this case: you can either work abstractly, directly with $c/c_0$, or try to find the map. Herer the former works: when will $a,b\in c$ be congruent modulo $c_0$? If and only if $a-b\in c_0$, if and only if $\lim(a-b)=0$, if and only if $\lim a = \lim b$. The equivalence classes correspond to the limits of the sequences. This suggests exactly which map to pick: let $T\colon c\to\mathbf{F}$ be the map to the underlying field given by $T(a) = \lim a$. It is straightforward to see that $T$ is linear, and that $\ker(T) = c_0$. Thus, $\mathbf{F} \cong c/c_0$. Since the dimension of $\mathbf{F}$ is $1$, the dimension of $c/c_0$ is $1$, so the codimension of $c_0$ in $c$ is $1$.
@Arturo: The dimension of $\textbf{F}$ is $1$ follows because for each $a \in c$ we can map it to $\lim a$ in $\textbf{F}$? –  PEV Jan 4 '11 at 17:05
@Trevor: No, the dimension of $\mathbf{F}$ is $1$ because it has a 1 element basis: $1$ is a basis for $\mathbf{F}$ as an $\mathbf{F}$-vector space. What you propose would make my argument above circular, since I use the fact that $\mathbf{F}$ is a 1-dimensional vector space over itself to conclude that the map $c\to\mathbf{F}$ given by $a\mapsto\lim a$ shows that $c/c_0$ is 1-dimensional. –  Arturo Magidin Jan 4 '11 at 17:11
@Arturo: In other words, $a = a_{1}x_{1}+ \lim a$ for $x \in c$ and $\lim a \in c_0$? –  PEV Jan 4 '11 at 17:14
@Arturo: If $\text{dim} \ E/E_1 = n$, then there exist $x_1, \dots, x_n$ such that for every $x \in E$ there are numbers $a_1, \dots, a_n$ and $y \in E_1$ such that $x = \sum a_{i}x_{i}+y$. So in this case $n=1$. So basically we can represent a convergent sequence $c$ as the sum of a sequence tending to $0$ and $\textbf{1}x_1$? –  PEV Jan 4 '11 at 17:20