If $I_n = \{i \in \mathbb{N} : 1 \leq i \leq n\}$ and if $\mathcal{X}_n=\{(X_i,d_i) : i \in I_n\}$ is a finite family of metric spaces, we know that we can make their product $X = \prod_{i \in I_n}X_i$ a metric space by setting $d: X \times X \to \mathbb{R}$ to be simply $d((p_1,\dots,p_n),(q_1,\dots,q_n))=|(d_1(p_1,q_1),\dots,d_n(p_n,q_n))|$ where $|\cdot|$ is some norm in $\mathbb{R}^n$ (I've seem this done mostly with the $p$-norm).
Now, what if we do the following. Let $\Lambda$ be an arbitrary indexing set (can be even uncountable), then we have a family $\mathcal{X}_\Lambda = \{(X_\lambda, d_\lambda) : \lambda \in \Lambda\}$ of metric spaces. We define their product to be:
$$X = \prod_{\lambda \in \Lambda}X_\lambda = \left\{f : \Lambda \to \bigcup_{\lambda \in \Lambda}X_\lambda : f(\lambda) \in X_\lambda, \forall \lambda \in \Lambda\right\},$$
is it possible then to present a natural metric in $X$ in terms of the metrics in each $X_\lambda$? In other words: "is the arbitrary product of metric spaces a metric space again?"
I've seem a similar question here, but I think this isn't duplicate, since that question was regarding aspects about the category of metric spaces.
Thanks very much in advance!
Best Answer
No. Every metric space is first-countable, and e.g. $\mathbb{R}^\mathbb{R}$ is not first-countable.