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I am trying to prove that $\mathcal{c}$ and $\mathcal{c_0}$ are isomorphic but not isometrically isomorphic.

I've read on this forum that isometric isomorphism preserves extreme points, but I don't see why. I know that I should use the fact that isometries preserve all metric properties and being an extreme point is a metric property but I can't come up with any proof.

If $K$ is a convex set and $a \in K$ is its extreme point, then whenever $a = \frac{x+y}{2}, \ x,y \in K, $ we have that $x=y=a$.

What should I do now?

Could you help me with that?

Thanks.

Bilbo
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    Let $f\colon X \to Y$ be the isometric isomorphism. Let $C = f(K)$. You want to show that $f(a)$ is an extreme point of $C$. So suppose $f(a) = \frac{u+v}{2}$. Then - what corresponds to $u$ and $v$ in $X$? – Daniel Fischer Oct 04 '14 at 13:23
  • $a$ is an extreme point of $K$, so $a = \frac{x+y}{2} \Rightarrow x=y=a$. $f(a)=\frac{u+v}{2} = \frac{f(w)+f(z)}{2}=f( \frac{w+z}{2})$, so $a=\frac{w+z}{2}$, so $a=w=z$, so $f(a)=f(w)=f(z)=u=v$. Is that correct? – Bilbo Oct 04 '14 at 13:44
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    Yes, that's correct. And that shows that the image of an extreme point is an extreme point. Actually, we didn't need that $f$ be an isometric isomorphism to deduce that. – Daniel Fischer Oct 04 '14 at 13:47
  • It seems that we only need isomorphism, but it can't be right, because I need to show that $c, c_0$ aren't isometrically isomorphic – Bilbo Oct 04 '14 at 13:50
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    It's right. But yes, you need more: The unit ball of $c$ has extreme points with norm $1$, the unit ball of $c_0$ doesn't. – David Mitra Oct 04 '14 at 14:02
  • Yes, I know that. In $c_0$ no closed unit ball has extreme points, if $x \in c_0, \ ||x|| \le 1, \ x \rightarrow 0$, then $\exists k \in \mathbb{N} : \forall n>k : |x_n| < \frac{1}{2}$, but then the segment $[x- \frac{1}{2}e_k, x+ \frac{1}{2}e_k]$ is contained in the unit ball – Bilbo Oct 04 '14 at 14:10
  • But in $c, \ (1,1,...)$ is an extreme point, because if it weren't, there would exist $0 \neq y \in c, \ \varepsilon >0$ s.t. $[x- \varepsilon y, x- \varepsilon y] \subset B(0,1)$, $y \neq 0$, so there exists $k$ s.t. $y_k \neq 0$, so either $x_k + \varepsilon y_k >1$ or $x_k - \varepsilon y_k >1$, so either $x + \varepsilon y $ or $x - \varepsilon y $ doesn't belong to $B(0,1)$ – Bilbo Oct 04 '14 at 14:14
  • Where exactly do we use the fact that $f$ is an isometry? – Bilbo Oct 04 '14 at 14:18
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    An isometry would map the unit ball of $c$ onto the unit ball of $c_0$. – David Mitra Oct 04 '14 at 14:19
  • Ok, thank you. Are the two proofs above correct, by the way? – Bilbo Oct 04 '14 at 14:20
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    The second has a small typo. (I would argue by observing the only way $1=(a+b)/2$ with both $|a|\le1$ and $|b|\le1$ is if $a=b=1$; but this is essentially what you did.) – David Mitra Oct 04 '14 at 14:30
  • All right. Thank you again for all your help! (The typo is the second minus sign in $[x- \varepsilon y, x- \varepsilon y]$ ) – Bilbo Oct 04 '14 at 14:38

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