$(\prod_{i\in I}X_i, \mathcal{T}_p)$ is metrizable iff $X_i$ is metrizable and all $X_i$ is singleton set except countable many indices.

Question

Note that $X_i$ is nonempty.

I guess that a singleton space as a factor of the product topology doesn't affect the whole product.

But I don't know how to show that in detial.

Just some hints are what I want.

Answer 1

As explained in the comments, it seems that you are asking why the product of unaccountably many spaces of cardinality at least then $2$ is not metrizable.

Let $X = \prod_iX_i$ a product of uncountably many metric spaces with the induced topology, all of size at least $2$. If $X$ is metrizable, then it is first countable.

Let $(x_i)_i \in X$ be any point, and $\{U^n\}_{n<\omega}$ a countable set of open neighborhoods of $(x_i)_i$. Then for every $n$ we have some basic open $\prod_i U_i^n \subset U^n$. In particular (by definition), $U_i^n = X_i$ for all but finitely many $i\in I$. Let $J = \{i\in I: \ \exists $n$ \, U_i^n\neq X_i \}$.

Can you prove $J$ is countable (and hence there is some $i_0\in I \smallsetminus J$)? Can you use that together with the fact that $X_{i_0}$ is metric with at least two pints, to find a neighborhood of $(x_i)_i$ that does not contain any of the $U^n$?