Under countable choice, a countable product of completely metrizable spaces is completely metrizable

Given: The objects, hypotheses, and choice principles stated above.

Let $((X_n,d_n){)}_{n\in \mathbb{N}}$ be complete metric spaces with $d_n\le 1$. On ${\prod }_n{X}_n$, the formula $D(x,y)={\sum }_{n=0}^{\mathrm{\infty }}{2}^{-(n+1)}{d}_n(x_n,y_n)$ defines a complete metric inducing the product topology. The empty product is the one-point space. (The standard weighted metric on a countable product of bounded complete metric spaces is complete).

Let $(X,d)$ be a metric space (def-metric-space) and define, for $x,y\in X$, $d^{\mathrm{\prime }}(x,y):=\min \{d(x,y),1\},d^{\mathrm{\prime }\mathrm{\prime }}(x,y):=\frac{d(x,y)}{1+d(x,y)}\mathrm{.}$ Both are well defined: $d(x,y)\ge 0$ (lem-metric-nonnegativity), so $1+d(x,y)>0$ and is invertible, and the minimum of a two-element set of reals exists (lem-finite-set-has-max, def-max-min). Then: 1. $d^{\mathrm{\prime }}$ and $d^{\mathrm{\prime }\mathrm{\prime }}$ are metrics on $X$. 2. $d^{\mathrm{\prime }}(x,y)\le 1$ and $d^{\mathrm{\prime }\mathrm{\prime }}(x,y)<1$ for all $x,y$; hence $(X,d^{\mathrm{\prime }})$ and $(X,d^{\mathrm{\prime }\mathrm{\prime }})$ are bounded metric spaces (def-metric-bounded-diameter), and if $X\mathrm{\ne }\mathrm{\varnothing }$ then $\mathrm{diam}(X)\le 1$ for both. 3. $d^{\mathrm{\prime }}$ and $d^{\mathrm{\prime }\mathrm{\prime }}$ are each uniformly equivalent to $d$, hence topologically equivalent to it (def-equivalent-metrics, thm-metric-equivalence-hierarchy). Consequently every metric space carries a bounded metric with exactly the same topology, so boundedness cannot be read off the topology alone. ($\min (d,1)$ and $d\mathrm{/}(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology).

The Axiom of Countable Choice, written ${\mathrm{A}\mathrm{C}}_{\omega }$, is the following statement. The statement is: for every family $(X_n{)}_{n\in \mathbb{N}}$ of nonempty sets indexed by $\mathbb{N}$ there is a function $f$ with domain $\mathbb{N}$ such that $f(n)\in X_n$ for every $n\in \mathbb{N}$. Equivalently, every at most countable family of nonempty sets has a choice function. (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)).

Let $(X,d)$ be a metric space (def-metric-space) and let ${\mathcal{T}}_d$ be its metric topology (def-metric-topology). Call ${\mathcal{T}}_d$ completely metrizable if some metric $\rho$ on $X$ is topologically equivalent to $d$, that is ${\mathcal{T}}_{\rho }={\mathcal{T}}_d$ (def-equivalent-metrics), and makes $(X,\rho )$ complete (def-complete-metric-space). Then: 1. Homeomorphism invariance. Let $(Y,e)$ be a metric space and let $h:X\to Y$ be a bijection (def-injection-surjection-bijection) such that $h$ and $h^{-1}$ are continuous (def-metric-continuity). If ${\mathcal{T}}_d$ is completely metrizable then so is ${\mathcal{T}}_e$. 2. Closed subspaces. If ${\mathcal{T}}_d$ is completely metrizable and $A\subseteq X$ is closed in $(X,d)$, then ${\mathcal{T}}_{d_A}$ is completely metrizable, $d_A$ being the subspace metric (def-isometry-and-metric-embedding). 3. The property is strictly weaker than completeness. Let $P:=(0,\mathrm{\infty })\subseteq \mathbb{R}$ (def-interval) carry $d(x,y):=\mathrm{\mid }x-y\mathrm{\mid }$ (lem-real-line-is-a-metric-space). Then $(P,d)$ is not complete, while ${\rho }_P(x,y):=\mathrm{\mid }x-y\mathrm{\mid }+\mid \frac{1}{x}-\frac{1}{y}\mid$ is a complete metric on $P$ with ${\mathcal{T}}_{{\rho }_P}={\mathcal{T}}_d$. So ${\mathcal{T}}_d$ is completely metrizable although no completeness assumption holds for $d$ itself. Complete metrizability is a condition on the collection of open sets alone: the metric is quantified over and does not survive into the statement. That is exactly what completeness fails to be, and claim 3 shows the two conditions are genuinely different rather than merely stated differently. (Complete metrizability: admitting a topologically equivalent complete metric is preserved by homeomorphism and by closed subspaces, and $(0,\mathrm{\infty })$ has it without being complete).