# Are $(C[0,1],d_\infty)$ and $(C[0,1],d_1)$ homeomorphic?

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.

• 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. – Henno Brandsma Apr 12 '16 at 4:56
• No, the identity map is continuous in that direction. – zhw. Apr 12 '16 at 4:58
• @zhw I'm sorry. I mistakenly typed the question and I have fixed the error. – user37299 Apr 12 '16 at 5:11
• @HennoBrandsma: Does that theorem really hold without knowing that the spaces are complete? – Eric Wofsey Apr 12 '16 at 5:28
• @EricWofsey you're right. We need completely metrisable instead of metrisable – Henno Brandsma Apr 12 '16 at 5:50

Clearly, $d_1$ is not complete.