# Linear isometry between $c_0$ and $c$

The following question is an exercise and so I’m just looking for advices and not for answers if it’s possible.

I have the following sets in $l^\infty$
$$c_0 := \{x_n \in l^\infty: \lim x_n = 0\} \subseteq c := \{x_n \in l^\infty: \exists \lim x_n\}.$$
And I intend to prove that they are not isometrically isomorphic. I suppose that the problem is that for every $(x_n) \in c_0$ there exists a finite and non-empty set $\{x_{n_i}\}$ such that $|x_{n_i}| = \|x_n\|$ but this is false in $c$, but I don’t know how to continue with this idea. Can you help me?

An interesting fact about this two spaces is that although they aren’t linearly isometric they are linearly homeomorphic given by $T:c \to c_0, T(x_n) = (\lim x_n, x_n-\lim x_n)$. This is too strange to me.

#### Solutions Collecting From Web of "Linear isometry between $c_0$ and $c$"

Hints:

1) Prove that unit ball of $c$ have a lot of extreme points. In fact there are $\mathfrak{c}$ extreme points but this is not important for the solution.

2) Prove that unit ball of $c_0$ have no extreme points.

3) Prove that if $x$ is an extereme point of unit ball of some normed space $X$, and $i:X\to Y$ is an isometry, then $i(x)$ is an extreme point of unit ball of $Y$.

4) The rest is clear.