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 $f \colon I \to X$.

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 $f \colon X \to Y$. The homeomorphism class of the Cantor set is the top element of the poset $\mathrm{CompMet}$. My question asks if the subset of $\mathrm{CompMet}$ coming from connected compact metric spaces has a top element.

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

For example, I think that there's a map sending each element of $\mathrm{CompNet}$ to its number of connected components, and I think that this is an order-preserving map from $\mathrm{CompMet}$ to the cardinals less than or equal to the continuum. But there also seems to be an order-preserving map sending each element of $\mathrm{CompNet}$ to its number of path-connected components. Are there other interesting maps like this?

  • 3
    $\begingroup$ This is nit-picky, but wouldn't it be more appropriate to call these universal metrizable spaces, rather than metric spaces? Since we are asking for existence of continuous maps, not isometries, and as a result there is no preferred metric to choose on $K$ or $I$, just any that is compatible with the topology $\endgroup$ Commented Nov 12, 2017 at 17:53
  • $\begingroup$ Yes, we can say "metrizable". $\endgroup$
    – John Baez
    Commented Nov 20, 2017 at 8:54
  • 1
    $\begingroup$ Another versal property: the pseudo-arc maps onto every chainable continuum, meaning one for which every open cover refines to a cover $U_1, \dots , U_n$ where $U_i \cap U_j \ne \emptyset$ iff $|i - j| \le 1$. This was proved by Mioduszewski in 1962. $\endgroup$
    – John Baez
    Commented Nov 20, 2017 at 17:16
  • $\begingroup$ At least, so say Irwin and Solewecki. But looking at Mioduszewski's paper linked to above, I see he claims that pseudo-arc maps onto every 'snake-like' continuum. He doesn't define this concept, but elsewhere I read that a continuum is snake-like if it's chainable and has an extra property too. So, I don't get why Irwin and Solewecki claim what they do. $\endgroup$
    – John Baez
    Commented Nov 20, 2017 at 17:45
  • 2
    $\begingroup$ "chainable," "linearly chainable," and "snake-like" are often used to mean the same thing. $\endgroup$ Commented Nov 24, 2017 at 18:43

2 Answers 2


There is no such continuum. See

Z. Waraszkiewicz, Sur un problème de M.H. Hahn, Fund. Math. 22 (1934) 180–205.

Waraszkiewicz constructed an uncountable family $W$ of continua in the plane called Waraszkiewicz spirals so that no continuum can be mapped continuously onto every continuum in $W$. Start with the space $X=S^1\cup [1,2]$ and consider the maps $f,g:X\to X$ where

  • $f(x)=x$ if $x\in S^1$, $f(x)=\exp(4\pi ix)$ for $1\leq x\leq \frac{3}{2}$, and $f(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$
  • $g(x)=x$ if $x\in S^1$, $g(x)=\exp(-4\pi ix)$ for $1\leq x\leq \frac{3}{2}$, and $g(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$

Now $W$ is the collection of all inverse limits of all inverse sequences $(X_i,h_i)$ where $X_i=X$ for all $i$ and $h_i\in\{f,g\}$.

For a short and readable description (where I found the construction) see:

W.T. Ingram, Concerning images of continua, Topology Proceedings 16 (1991) 89-93.

This non-existence results has been improved upon a few times, for instance in the following:

S.B. Nadler, The nonexistence of almost continuous surjections between certain continua, Topology and its Applications, 154 (2007) 1008-1014.


Here's a related positive result:

Alan Dow and KP Hart proved in 1998 that the (non-metrizable) continuum $\beta[0,1)\setminus [0,1)$ maps onto every continuum of weight $\leq \omega_1$, and therefore onto every metrizable continuum. See https://arxiv.org/pdf/math/9805008.pdf.

  • $\begingroup$ What's the "weight" of a continuum? $\endgroup$
    – John Baez
    Commented Nov 14, 2017 at 15:55
  • $\begingroup$ minimum cardinality of a basis $\endgroup$ Commented Nov 14, 2017 at 16:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.