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Let $d$ and $d^\prime$ be metrics on a non-empty set. Then $d$ is said to be equivalent to $d^\prime$ if there exist positive real numbers $\lambda$ and $\mu$ such that $$ \lambda d^\prime (x,y) \leq d(x,y) \leq \mu d^\prime(x,y) \ \mbox{ for all } x, y \in X.$$ In this case, the open sets in $\left( X, d \right)$ are precisely the same as the open sets in $\left( X, d^\prime \right)$, and the convergent, Cauchy, or bounded sequences in $\left( X, d \right)$ are convergent, Cauchy, or bounded, respectively, in $\left( X, d^\prime \right)$.

Now my question is, if $\left( X_1, d_1 \right), \ldots, \left( X_n, d_n \right)$ are any metric spaces, if $X = \prod_{i=1}^n X_i$, and if $d$ and $d^\prime$ are any two metrics on $X$ which involve the metrics $d_1, \ldots, d_n$ in their formulas and such that with respect to each of $d$ and $d^\prime$, each of the projection maps onto the factors $X_i$ is continuous, then are $d$ and $d^\prime$ always equivalent?

If not, then what is the actual situation that holds in this case?

Actually, this question occurs to me from Prob. 6, Chap. 4 in the book principles of Mathematical Analysis by Walter Rudin, 3rd edition. Rudin has not mentioned which metric to consider for the product of which the graph of $f$ is a subset.

I know that on a finite-dimensional vector space $X$, any two norms are equivalent and hence any two metrics induced by some norms on $X$ are equivalent.

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This is false for rather trivial reasons. For instance, you could take $d$ to be the sup metric, and $d'(x,y)=\min(d(x,y),1)$. Then $d'$ is not equivalent to $d$ (as long as $d$ is unbounded), but they both satisfy your hypotheses. For another example, let $d'$ be defined by $d'(x,y)=1$ whenever $x\neq y$. (Yes, perhaps this definition does not "involve" $d_1,\dots,d_n$, but if you want to object to example you really need to define what you mean by that more precisely.)

I don't know what Rudin's conventions are, but usually you only care about topological properties of the product, in which case you just use the product topology. This topology can be induced by any number of metrics, such as the sup metric.

Eric Wofsey
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  • Interesting examples! Probably worth mentioning that equivalence of metrics is stronger then generating the same topology (if it is true, of course; it is, if I remember correctly). – lisyarus Mar 24 '17 at 05:40
  • @lisyarus I believe that "equivalence of metrics" is ambiguous, and one possible meaning ("topologically equivalent") is "inducing the same topology." – bof Mar 24 '17 at 06:20
  • @bof Thank you for clarifying. I suppose the equivalence I am talking about is called strong equivalence. – lisyarus Mar 24 '17 at 07:30