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Let $((X_n, d_n))_{n = 1}^{\infty}$ be a sequence of metric spaces and let $X := \Pi_{n = 1}^{\infty} X_n$. Equipping the product space $X$ with the metric $$ d : X \times X \to [0,\infty), \quad (x,y) \mapsto \sum_{n = 1}^{\infty} \frac{1}{2^n} \frac{d_n (x_n, y_n)}{1 + d_n (x_n, y_n)} $$ makes the pair $(X, d)$ into a metric space. The following characterisation of convergence in $(X, d)$ is well-known.

A sequence $(x_k)_{k = 1}^{\infty} \subset X$ converges to an element $x \in X$ if, and only if, the sequence $(x_n^{(k)})_{k = 1}^{\infty} \subset X_n$ converges to $x_n \in X_n$ for each $n \in \mathbb{N}$.

It is quite surprisingly, but I am not able to find a direct proof of this fact in the literature. The forward implication is readily verified and the converse implication follows from Lebesgue's dominated convergence theorem. However, I am curious whether there is a more direct or standard proof of the converse direction. I do namely not think that the dominated convergence theorem is really needed for this.

Any comment or reference is highly appreciated.

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  • $\begingroup$ May be a corollary to another important theorem also be suitable or must it really be that direct? $\endgroup$ – TheGeekGreek Jan 29 '17 at 10:12
  • $\begingroup$ Yes, that would also be suitable. I am simply interested in what the most common or standard proof would be. The result follows of course also from net convergence in product spaces, but that argument seems also unnecessarily strong to me. $\endgroup$ – jvnv Jan 29 '17 at 10:15
  • $\begingroup$ It seems that you are a bit more advanced in that field than me. But nevertheless I try to help you. There is a theorem which says that the topology induced by the above metric is the same as the product topology on the product space. So you can simply use that convergence in the product is equivalent to convergence in the coordinates. Would that help? $\endgroup$ – TheGeekGreek Jan 29 '17 at 10:19
  • $\begingroup$ @TheGeekGreek. Thanks for pointing this out. However, your argument is exactly the one I have seen repeatedly in the literature, e.g., in Engelking's General Topology. I am therefore still curious whether there is a more direct approach. $\endgroup$ – jvnv Jan 29 '17 at 10:26
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    $\begingroup$ Ah ok (I still posted is as an answer, sorry). Then I may not help you, since the other reference for that corollary refers to nets. $\endgroup$ – TheGeekGreek Jan 29 '17 at 10:31
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The topology induced by your metric is the same as the usual product topology on the space $\prod_{i = 1}^\infty X_i$ where the topologies on $X_i$ are the ones induced by $d_i$. This fact can be found (with proof of course) on p. 259 in the book General Topology by Ryszard Engelking found here. Then you may use simply that a sequence converges in the product space if and only if every projected sequence converges. This is suggested as exercise 6 in Munkres Topology, p. 118 found here.

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