Skip to main content
1 of 5
John Baez
  • 22.3k
  • 3
  • 85
  • 170

Is there a "universal" connected compact metric space?

Fact 1. The Cantor set $K$ is "universal" among nonempty compact metric spaces in the following sense: given any nonempty compact metric space $X$, there exists a continuous surjection $f\colon K \to X$.

Fact 2. The closed interval $I$ has a similar "universal" property among nonempty compact connected and locally connected metric spaces: given any such space $X$, there exists a continuous surjection $f \colon K \to X$.

This makes me wonder: is there a compact connected metric space $J$ such that for any nonempty compact connected metric space $X$, there exists a continuous surjection $f \colon J \to X$?

Such a space $J$, if it exists, would be 'intermediate' between $I$ and $K$: there would need to be continuous surjections

$$ K \to J \to I $$

Fact 1 is sometimes called the Alexandroff–Hausdorff theorem, since appeared in the second edition of Felix Hausdorff’s Mengenlehre in 1927 and also in an article by Pavel Alexandroff published in Mathematischen Annalen in the same year. Fact 2 was proved by Hans Hahn in 1914 and reproved by him more nicely in 1928. For a nice history of these results, see:

One may rightly complain that "universal" is the wrong word above, since we're not claiming there exists a unique continuous surjection, and indeed there's usually not. A better term is versal. There can be two non-homeomorphic spaces having the same versal property. For example, $I^2$ would work just as well as $I$ in Fact 2, thanks to the existence of space-filling curves.

Nonetheless we can create a category in which these versal properties become universal, by a cheap trick. Let $\mathrm{CompMet}$ be the collection of all homeomorphism classes of nonempty compact metric spaces, and put a partial order on this where $[X] \ge [Y]$ iff there exists a continuous surjection $f \colon X \to Y$. Then homeomorphism class of the Cantor set is initial in the $\mathrm{CompMet}$.

I'd appreciated any interesting information on this poset $\mathrm{CompMet}$.

John Baez
  • 22.3k
  • 3
  • 85
  • 170