Uncountable product in the category of metric spaces.

Question

I need to prove that category $\mathrm{Met}$ of metric spaces and continuous maps doesn't possess uncountable product of non-one point spaces.

Definition. A pair $(X,\{\pi_\nu:\nu\in\Lambda\})$ where $X\in\mathrm{Ob(Met)}$, $\pi_\nu\in \mathrm{Hom_{Met}}(X,X_\nu)$ is called a product of family of metric spaces $\{X_\nu:\nu\in\Lambda\}$ if for each $Y\in\mathrm{Ob(Met)}$ and $\{\varphi_\nu:\nu\in\Lambda\}$ where $\varphi_\nu\in \mathrm{Hom_{Met}}(Y,X_\nu)$ there exist unique $\varphi\in\mathrm{Hom_{Met}}(Y,X)$ such that $\varphi_\nu=\pi_\nu\varphi$.

My question. Assume that for all $\nu\in\Lambda$ the space $X_\nu$ contains at least two points. How one can prove that for such family of metric spaces their product doesn't exist in $\mathrm{Met}$?

My attempt We can consider arbitrary set $Y$ with discrete metric then every set theoreitc map $\varphi_\nu$, $\nu\in\Lambda$ will be continuous. Hence we see that if product $X$ exist, then it has to be set theoretic product of sets $\{X_\nu:\nu\in\Lambda\}$. On the other hand, we can consider this imaginary product as object of category topological spaces $\mathrm{Top}$. If we could extend universal property of $X$ to the cases when $Y$ is just topological space, then we would have that $X$ is a Tychonoff product. Since $\Lambda$ is uncountable and spaces $\{X_\nu:\nu\in\Lambda\}$ are not singletons, then topology of $X$ is not first countable and hence not metriazible. But this not a proof, since we can not extend universal property.

Could you give me a hint, or give me some related references for my question?

Answer 1

This was a cool question man. I think the following works:

1. The category of metric spaces admits countable products, namely the product $\prod (X_i, d_i)$ is given by $$(\prod X_i, d) \quad \quad d(x,y) = \sum_i \frac{d'_i(x_i, y_i)}{2^i}$$where $d'_i = \min\{d_i, 1\}$. This induces the standard product topology. In particular, any open set contains one of the form $$O \times \prod_{j \geqslant i} X_j\tag{*}$$where $O$ is open in the product $\prod_{j < i} X_j$.

2. As noted in the comments/question, as a set $\prod_\nu X_\nu$ would have to be the set product with the usual projection maps, in particular the $\pi_\nu$ are cts, hence the metric topology would contain the product topology.

3. For a point $x = (x_v)_{v \in \nu}$, we can construct nbhs $$O_v = U_v \times \prod_{v' \neq v} X_v$$where $U_v \subsetneq X_v$.

4. Consider the basis for nbhs of $x$ given by open balls of radius $\frac{1}{n}$. By pigeonhole, for some $n$ the ball of radius $\frac{1}{n}$ would be contained in uncountably many $O_v$, i.e. $$B_{1/n}(x) \subseteq \prod_I U_i \times \prod_{v \notin I} X_v$$where $I$ is uncountable. Thus we have $$B_{1/n}(x) \subseteq \prod_J U_j \times \prod_{v \notin J} X_v$$where $J$ is just a countable subset of $I$.

5. Consider the subset $$\prod_J X_J \times \prod_{v \notin J} x_v$$equipped with the induced metric/topology; by the univ. property of products one checks this is isomorphic to $\prod_J X_J$. At the end of point 4, we have found an open set of $\prod_J X_J$ which cannot contain any open set of the form $(*)$.

Lemme know if you find errors/questions etc.