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I don't know if the term isotropic is correct in this context, but I was wondering if there exists a non trivial metric $\rho $ in $X=\mathbb{R}^{\mathbb{N}}$ such that $$\forall i \in \mathbb{N} \quad d(e^i,0) \text{ doesn't depend on }i$$

I tried with the product metric but with that I got $d(e^i,0) = 2^{-i}$. The $\ell^1$ metric is isotropic but this is a metric in $\ell^1 \subsetneq X$

Any help will be appreciated

user90803
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    What do you mean by "the product metric"? In any case, I would guess that you actually want something much stronger than what you're asking for, such as for the metric to also be translation-invariant and satisfy $d(tx,0)=td(x,0)$ for $t\geq 0$ or something like that. – Eric Wofsey Jan 19 '16 at 04:45
  • Also, do you care what the induced topology of the metric is? Do you want it to be compatible with the product topology? – Eric Wofsey Jan 19 '16 at 04:51

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If you want the topology induced by the metric to be the product topology, there is no such metric, because $e^i$ converges to $0$ as $i\to\infty$ in the product topology so $d(e^i,0)$ must go to $0$. If you don't care what the topology is, there are many reasonable metrics you could define. For instance, you could define $d((x_n),(y_n))=\min(1,\sup_n |x_n-y_n|)$. (Note that the topology induced by this metric has some properties that might be surprising; for instance, it is disconnected.)

Eric Wofsey
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  • Thanks for your answer. Do you have a reference for this metric? may be applications or a generalization that may be interesting? – user90803 Jan 19 '16 at 05:01