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Two metric spaces are said to be homeomorphic if there is a bijection f between them such that $f$ and $f^{-1}$ are both continuous.

Consider $C[0,1]$ with metrics:

$d_\infty (f,g)=\max_{x\in [0,1]}|f(x)-g(x)|$

$d_1(f,g)=\int_0^1|f(x)-g(x)|dx$

We already know that the identity map $(C[0,1],d_1)→(C[0,1],d_∞)$ is not continuous (Prove that the identity map $(C[0,1],d_1) \rightarrow (C[0,1],d_\infty)$ is not continuous). Does this imply $(C[0,1],d_∞)$ and $(C[0,1],d_1)$ are not homeomorphic?

Or could you find a bijection which is continuous in both direction?

Any help is appreciated.

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  • $\begingroup$ Yes they are, as both are separable metric locally convex topological vector spaces. This is a classic theorem (but hard to prove in general). It might be easier in specific cases. E.g. the $\ell_p$ spaces have explicit homeomorphisms. $\endgroup$ – Henno Brandsma Apr 12 '16 at 4:56
  • $\begingroup$ No, the identity map is continuous in that direction. $\endgroup$ – zhw. Apr 12 '16 at 4:58
  • $\begingroup$ @zhw I'm sorry. I mistakenly typed the question and I have fixed the error. $\endgroup$ – user37299 Apr 12 '16 at 5:11
  • $\begingroup$ @HennoBrandsma: Does that theorem really hold without knowing that the spaces are complete? $\endgroup$ – Eric Wofsey Apr 12 '16 at 5:28
  • $\begingroup$ @EricWofsey you're right. We need completely metrisable instead of metrisable $\endgroup$ – Henno Brandsma Apr 12 '16 at 5:50
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No, they are not. Confirming a conjecture of Banach, Victor Klee proved that if there is a complete metric on a normed (or, more generally, metrizable topological vector) space inducing the norm (vector space) topology then the norm (uniformity induced by the vector space topology) is complete. This can be seen, e.g., in Köthe's book topological Vector Spaces I, page 165.

Clearly, $d_1$ is not complete.

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  • $\begingroup$ But we're talking about a homeomorphism, which doesn't have to preserve a uniformity. $\endgroup$ – Robert Israel Apr 12 '16 at 6:50
  • $\begingroup$ Sorry, a part of the argument was missing. Of course, it is the surprising point in Klee's theorem that the complete metric may induce a very different uniformity. $\endgroup$ – Jochen Apr 12 '16 at 6:56

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