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Is uncountable product of metrisable spaces with product-topology metrisable?

The book only mentions that $I=[0,1]$，then $I^I$ is not metrisable, for it doesn't fit the 1st-countability axiom.

I have two questions here:

  1. What does $I^I$ mean? Why doesn't it fit the axiom?

  2. Is this case a special one? Or,is there any example that makes the uncountable product space metrisable?

José Carlos Santos
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yLccc
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2 Answers2

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An uncountable product of non trivial first countable spaces is never first countable. In particular, the uncountable product of non trivial metrizable spaces is never metrizable.

To see this, let $ X_i , \ i \in I$ be an uncountable family of non trivial first countable spaces. Let $X= \prod X_i$. Let us assume that $X$ is first countable. For $ i \in I, $ pick a non empty open $ U_i \subsetneqq X_i $ (such a set exists, since $X_i$ is not trivial). Denote $\pi_i \colon X \to X_i $ the natural projection. Let $ x = (x_i)_{i \in I} $ such that $$ x \in \prod_i U_i = \bigcap_i π_i ^{-1} (U_i).$$

Then $ x \in π_i^{-1} (U_i) $ (which is open), for every $ i \in I $. Let $ (V_n)_{n \in \mathbb N} $ be a countable basis of $ x $. For every $ i \in I $, there exists $ n_i \in \mathbb N$ such that $ x \in V_{n_i } \subset π_i^{-1} (U_i) $. Consider the (well defined) map $ \phi \colon I \to \mathbb N $, by $ \phi (i) = n_i $. Since $ I $ is uncountable, there exists $ n \in \mathbb N $ such that the set $ \phi ^{-1} ( \{n \}) $ is uncountable. Let $ J = \phi ^{-1} ( \{ n\} ) $. Then for $ j \in J $, we have that $ \phi (j) = n $ and $V_n \subset π_j^{-1} (U_j)$. Hence, $$ π_j (V_n) \subset U_j \subsetneqq X_j , \ \ \ j \in J.$$

So there are uncountably many $ j \in J$ such that the $j-$th coordinate of $ V_n $ is different than $ X_j $. This is impossible, since the $i$-th coordinate of $ V_n $ is $X_i$ for all but finitely many $i \in I$.

Evangelopoulos Foivos
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    Your first statement isn’t quite true: the product is metrizable iff at most countably many of the factors are non-trivial. – Brian M. Scott Apr 04 '21 at 20:25
  • Yeah, you are right. I always have in mind that the spaces are non trivial – Evangelopoulos Foivos Apr 04 '21 at 20:49
  • Does your 'V' mean $V_n$ here? I am confused by this V. By the way, you take the proper open subset $U_i$ of $X_i$, how to make sure this kind of proper open subset can be find? Is this the 'at most countable many non-trivial space issue? – yLccc Apr 05 '21 at 13:56
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    @yLccc yes, $V$ was supposed to be $V_n$, thanks for pointing that out. Such a set exists if we assume that each $X_i$ is non trivial. – Evangelopoulos Foivos Apr 05 '21 at 18:43
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The space $I^I$ is $\prod_{x\in I}I_x$, where, for each $x\in I$, $I_x=I$. Indeed, the first countability axiom does not hold in this space. In fact, take $p\in I^I$. Let$$\{V_n\mid n\in\Bbb N\}\tag1$$be a countable set of neighborhoods of $p$. For each $N\in\Bbb N$, let $A_n$ be a open subset of $I^I$ such that $p\in A_n$ and that $A_n$ can be written as $\prod_{x\in I}J_{n,x}$, with each $J_{n,x}$ an open subset of $I_x(=I)$ and such that $J_{n,x}\ne I_x$ only for finitely many $x$'s. Since $I$ is uncountable, there will be $x$'s such that $J_{n,x}=I$, for every $n\in\Bbb N$. Therefore, if $x_0$ is one such $x$, if $J$ is an open interval of $I$ to which $p_x$ (the canonical projection of $p$ onto $I_x$) belongs, and if $A=\prod_{x\in I}J_x$, with$$J_x=\begin{cases}J&\text{ if }x=x_0\\I&\text{ otherwise,}\end{cases}$$then, $A$ is an open set, $p\in A$, but $A$ contains no $V_n$, in spite of the fact that $V$ is a neighborhood of $p$. This proves that $(1)$ is not a fundamental system of neighborhoods of $p$.

The same argument shows that if $M$ is an uncountable metric space, then $M^M$ is not metrizable.

José Carlos Santos
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