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I encountered a question about proving that $\mathbb{R}^\omega$, which is the countably infinite product of $\mathbb{R}$ under the product topology, is not metrizable. I have seen many solutions here by using the sequence lemma and the first countable argument, which is great. However, are there any ways to prove that by just using basic definitions of product topologies and metric spaces? I was trying to bring up some contradiction about violating the definition of being a metric but it seems nothing is wrong with that...

Edi
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This is false. A countable product of metrizable sets is metrizable (w.r.t. the product topology). In your case, $\Bbb{R}^\omega \cong \Bbb{R}^\Bbb{N}$ is for example metrizable with metric

$$d((x_n)_n, (y_n)_n):= \sup_{n \in \Bbb{N}}\frac{1}{n} (|x_n-y_n| \land 1)$$

J. De Ro
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  • so we are dividing the distance between $x_n$ and $y_n$ by $n$? – Edi Aug 25 '20 at 13:49
  • I first make sure that the usual metric on $\Bbb{R}$ becomes bounded: Indeed, replace $|x|$ by $|x| \land 1$. Check that the latter metric induces the usual topology on $\Bbb{R}$ as well. And then yes, it gets divided by $n$. – J. De Ro Aug 25 '20 at 13:55
  • Sorry, but I really need some definition clarification. In my knowledge, the basis of the product space is the set containing $\Pi_{i\in I} U_i = {(x_i)\in \Pi_{i\in I}X_i|x_i \in U_i}$ where each $U_i$ is open in $X_i$ and ${i \in I| U_i \neq X_i}$ is finite. And from wiki, I get something really similar: https://en.wikipedia.org/wiki/Box_topology

    What are the differences between them?

    – Edi Aug 25 '20 at 14:00
  • In one you don't allow infinitely many non-trivial factors in a basis and in the other you do. – J. De Ro Aug 25 '20 at 15:00
  • Oh wow, such a small difference but totally different structure – Edi Aug 25 '20 at 16:50
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    @Ming Here I show that a countable product of metric spaces (in the product topology) is again metrisable, giving another, similar (and equivalent) metric for that product. – Henno Brandsma Aug 25 '20 at 22:47