@Cla: That's not a bad idea. Every metrizable space embeds into a countable product of "hedgehogs" (where a "hedgehog" means the 1-point compactification of a bunch of copies of $[0,1)$). Maybe it's not too hard to find a spherically complete metric on these universal spaces.

Nevermind -- the formula in my previous comment need not be spherically complete. For example, if $d$ is a complete but not spherically complete metric on $X$, then the formula just gives $\delta = d$.

A metric space is completely metrizable if and only if it is a $G_\delta$ subspace of its Cauchy completion. The usual proof of the "if" direction gives you an actual formula for a complete metric $\delta$ on $X$ in terms of a complete metric $d$ on its completion: $$\delta(x,y) = d(x,y) + \sum_{n=1}^\infty 2^{-n} \frac{| d(x,U_n)^{-1} - d(y,U_n)^{-1}|}{1+| d(x,U_n)^{-1} - d(y,U_n)^{-1}|},$$ where $U_1,U_2,U_3,\dots$ are open and $X = \bigcap_{n=1}^\infty U_n$. I've stared at this formula for the last few minutes, and I can't decide whether it's spherically complete or not. But maybe?

Thanks very much for sharing this. I wasn't aware of this paper before. The remarks following Problem II in the introduction of Stern's paper, together with Theorem 4, seem to come very close to answering the other question I posted (the one linked to in this question) in the negative. If I'm reading things right, they say that a positive solution cannot come from an $\omega_1$-norm on a coanalytic set. Theorem 4 rules out a too-nice-looking solution to the problem in any form. Thanks again, and welcome to MathOverflow!

You're right. I meant something a little different from what I wrote: There is some $\alpha$ such that if $B$ is a $G_\delta$ set containing only well-founded relations, then $B$ cannot contain all relations of rank $<\alpha$ (and similarly for other classes beyond $G_\delta$). I don't suppose you know of any restrictions like this?

OK, thanks. That makes sense, although it's not what I was hoping you meant. (I was hoping you were going to tell me something along the lines of: there is some $\alpha$ such that if $B$ is a $G_\delta$ set containing only well-founded relations, then $B$ contains no relations of rank $\geq \alpha$ (and similarly for other classes beyond $G_\delta$).)