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?

  • 2
    $\begingroup$ What do you require of homomorphisms in the category of metric spaces? Simple continuity with respect to the corresponding topology, or do you require that they respect the metric in some way? $\endgroup$ – Arturo Magidin May 1 '12 at 4:18
  • $\begingroup$ The topology on $X$ is obviously finer than the weak topology induced by the $\pi_\nu$, so maybe try to show they are equal. (The weak topology in this case is not first countable) I think this is done by showing $\pi_\nu$ are open maps. $\endgroup$ – Jose27 May 1 '12 at 5:34
  • $\begingroup$ @Norbert: When you change categories you must first show that the structure you are thinking about gets preserved. Here, you must show that the forgetful functor from metric spaces to topological spaces preserves infinite products. This isn't as obvious as the fact that the forgetful functor from metric spaces to sets preserves all limits (because it has a left adjoint). $\endgroup$ – Zhen Lin May 1 '12 at 8:57
  • $\begingroup$ @ArturoMagidin Morphism in $\mathrm{Met}$ are just continuous maps, whithout any additional restrictions. $\endgroup$ – Norbert May 1 '12 at 18:03

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.

  • $\begingroup$ Can you explain first inclusionin step 4? Where is a contradiction in step 5? $\endgroup$ – Norbert May 1 '12 at 11:04
  • $\begingroup$ Sure thing. First inclusion: informally, each $O_v$ is putting a ``constraint" in the $v^{th}$ coordinate. Since the $B_i$ are a basis, we have each $O_v$ contains some $B_i$. Since there are countably many $B_i$, and uncountably many $O_v$, some $B_i$ is contained in uncountably many $O_v$ (indexed by $I$), hence their intersection, which is the 1st line. Informally, we have found a ball with constraints in uncountably many coordinates. $\endgroup$ – uncookedfalcon May 1 '12 at 18:17
  • $\begingroup$ Contradiction: an open set in a countable product has "constraints" only in finitely many coordinates. The intersection of $B_{1/n}$ with the copy of $\prod_J X_J$ has constraints in every coordinate, a contradiction. $\endgroup$ – uncookedfalcon May 1 '12 at 18:25
  • $\begingroup$ Your solutions seems to be correct. Thank you. $\endgroup$ – Norbert May 1 '12 at 18:35
  • $\begingroup$ Ah! Tiny error, just corrected. But yeah cheers! $\endgroup$ – uncookedfalcon May 1 '12 at 19:24

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