Metric equivalence of metrics in product metric spaces Assuming we have $\{(X_j,d_j)\}_{j=1}^n$ metric spaces and we have a metric $\rho$ which induces the same topology on $X:=\prod_{j=1}^nX_j$ from the product topology. Is it always the case for any $1\geq \alpha_1,...,\alpha_n>0$, that the metric $\hat{d}(x,y)=\sum_{j=1}^n \big( d_j(x_j,y_j)\big)^{\alpha_j}$ for $x=(x_1,...,x_n ) \in X$ and $y=(y_1,...,y_n)\in X$, is equivalent to $\rho$?[changed from d here] i.e., they induce the same topology?
I think that the answer is yes, but I wanted to verify. My logic goes as follows:

*

*For any $1\geq \alpha>0$ and $a_1,a_2>0$ we have that $(a_1+a_2)^\alpha\leq a_1^\alpha+ a_2^\alpha$.


*Given a metric $d$ on a metric space $Z$, $\tilde{d}(z_1,z_2)=\big( d(z_1,z_2) \big)^\alpha$ is a metric on $Z$. Furthermore, balls w.r.t $d$ contain balls w.r.t $\tilde{d}$. So $d$ and $\tilde{d}$ generate the same topology.


*A base for $X=\prod_{j=1}^n$ is given by the product of open sets in each coordinate. Therefore, since the metrics $\hat{d_j}= d^{\alpha_j}$ have the same open sets as $d_j$, a base consisting of product of balls w.r.t $\hat{d_j}$ is also a base for $X$.


*Since the topology induced by $\rho$ equals to the product topology, $\rho$ and $\hat{d}$ are equivalent.
I think that this is true, but I have a feeling I might have overlooked something. My original motivation is to argue by some general abstract arguments that $\sum_{j=1}^d \vert x_j-y_j\vert^{\alpha_j}$ is always a metric equivalent to the Euclidean one.
 A: This answer summarizes my comments.

*

*You need to assume $0<\alpha_i\leq 1$ to ensure the new function $\hat{d}$ is a metric, since in that case we have for all $i$:
$$ (d_i(x,y)+d_i(y,z))^{\alpha_i} \leq d_i(x,y)^{\alpha_i}+d_i(y,z)^{\alpha_i} \quad \forall x,y\in X_i$$
whereas if we use, for example $\alpha=2$, then we see $$(1+1)^2>1^2+1^2$$


*If you have two different metrics $\rho_1$ and $\rho_2$ on a nonempty set $\Omega$ and you show that for every $x \in \Omega$ and every $\epsilon>0$ there are positive constants $\delta_1$ and $\delta_2$ (which may depend on $x$ and $\epsilon$) such that
\begin{align}
&\rho_1(x,y) < \delta_1 \implies \rho_2(x,y)<\epsilon\\
&\rho_2(x,y)<\delta_2\implies \rho_1(x,y)<\epsilon
\end{align}
then the two metrics generate the same open sets, meaning that every union of open balls with respect to one metric is also a union of open balls with respect to the other metric.


*For your problem, if you define $\Omega = \times_{i=1}^n X_i$ and you use the fact that $\rho_1(x,y) = \sum_{i=1}^n d_i(x,y)$ is a metric that generates the desired product topology on $\Omega$, you can define $\rho_2(x,y)=\sum_{i=1}^n d_i(x,y)^{\alpha_i}$.  Then for each $\epsilon>0$ and each $x=(x_1, ..., x_n) \in \Omega$ you can find $\delta_1>0$ and $\delta_2>0$ that satisfy point 2 above.
