Compatibility of topologies and metrics on the Hilbert cube Consider the Hilbert cube $Y = [0,1]^\mathbb{N}$.
It is easy to define four classes of metrics on $Y$ for $\gamma>0$ and $\omega>1$:
$$d^\gamma_{sup,pol}(x,y) = \sup_{k\geq 1} |x_k-y_k|/k^\gamma,$$
$$d^{\gamma}_{sum,pol}(x,y) = \sum_{k=1}^\infty  |x_k-y_k|/k^{1+\gamma}, $$
$$d^\omega_{sup,exp}(x,y) = \sup_{k\geq 1} |x_k-y_k|/\omega^k,$$
$$d^\omega_{sum,exp}(x,y) = \sum_{k=1}^\infty |x_k-y_k|/\omega^k  $$
Question, 


*

*do all these metrics define the same topology?

*does any of this metrics define the standard product topology?

*is there a standard reference for questions like the above?

 A: *

*Yes.

*Yes.

*I don't think there's a standard reference, but such metrics should be handled in many books on topology or analysis, though probably only special cases. One can then distil the essence oneself to get the general concept.


Proposition: Let $\bigl((X_n, d_n)\bigr)_{n\in \mathbb{N}}$ be a sequence of [non-empty, to avoid trivialities] metric spaces, with bounded diameters (there is a $K\in (0,+\infty)$ with $\operatorname{diam} X_n \leqslant K$ for all $n$). Let $(c_n)_{n\in \mathbb{N}}$ a sequence of strictly positive numbers converging to $0$. Then
$$\delta(x,y) = \sup \{ c_n\cdot d_n(x_n,y_n) : n \in \mathbb{N}\}\tag{1}$$
defines a metric on $\prod\limits_{n\in \mathbb{N}} X_n$ that induces the product topology. If $\sum\limits_{n=0}^\infty c_n < +\infty$, then
$$\sigma(x,y) = \sum_{n=0}^\infty c_n\cdot d_n(x_n,y_n)\tag{2}$$
also is a metric inducing the product topology.
Proof: The boundedness of the diameters gives the well-definedness of $\delta$ resp. $\sigma$ under the respective hypotheses. That $\delta$ and $\sigma$ are metrics is a straightforward verification. Since
$$d_k(x_k,y_k) \leqslant \frac{1}{c_k} \delta(x,y)\quad \text{resp.}\quad d_k(x_k,y_k) \leqslant \frac{1}{c_k}\sigma(x,y),$$
all the component projections are continuous (even Lipschitz-continuous) when the product is endowed with the topology induced by $\delta$ resp. $\sigma$, so the topology induced by either of these metrics is finer than the product topology.
It remains to be seen that the topology induced is coarser than the product topology. So let $x \in \prod X_n$, and $\varepsilon > 0$. We need to show that
$$B_\varepsilon^\tau(x) = \left\{ y \in \prod_{n\in\mathbb{N}} X_n : \tau(x,y) < \varepsilon \right\}$$
is a neighbourhood of $x$ in the product topology for $\tau \in \{\delta,\sigma\}$. Treating $\delta$ first, which is the easier case, we have
$$B_\varepsilon^\delta(x) = \prod_{n\in \mathbb{N}} B_{\varepsilon/c_n}^{d_n}(x_n),$$
where for all but finitely many $n$ we have $c_n\cdot K < \varepsilon$ and thus $B_{\varepsilon/c_n}^{d_n}(x_n) = X_n$, since $c_n \to 0$, so $B_\varepsilon^\delta(x)$ is an open neighbourhood of $x$ in the product topology.
For $\sigma$, we choose $n_0\in \mathbb{N}$ so that
$$\sum_{n=n_0}^\infty c_n\cdot K < \frac{\varepsilon}{2},$$
and let $$\eta = \frac{\varepsilon}{2\sum_{n=0}^{n_0-1} c_n}.$$
Then
$$U := \prod_{n=0}^{n_0-1} B_\eta^{d_n}(x_n) \times \prod_{n=n_0}^\infty X_n \subset B_\varepsilon^\sigma(x)$$
since for $y\in U$ we have
$$\sigma(x,y) = \sum_{n=0}^{n_0-1} c_n\cdot d_n(x_n,y_n) + \sum_{n=n_0}^\infty c_n\cdot d_n(x_n,y_n) < \sum_{n=0}^{n_0-1} c_n\cdot\eta + \sum_{n=n_0}^\infty c_n\cdot K < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}.$$
$U$ is an open neighbourhood of $x$ in the product topology, so $B_\varepsilon^\sigma(x)$ is a neighbourhood of $x$ in the product topology.
