Cross-posted from MSE.

Is there an infinite countable topological space $X$ with only countably many continuous functions to itself?

It cannot be a metrizable space. Another large class of examples that I know of are Alexandrov topologies, however each Alexandrov topology corresponds to a preorder, and the continuous maps between two Alexandrov topologies correspond to the morphisms between the preorders. An infinite countable preorder has always $2^{\aleph_0}$ endomorphisms, hence I cannot find a counterexample there either. It also cannot be a filter (+ the empty set), because any function which restricts to the identity on a set in the filter is continuous (thanks to Eric Wofsey for this last fact).

Using the $\pi$-Base, an online database of topological spaces inspired by the book Counterexamples in topology and expanding it, I obtained this list of possible spaces. I proved for every one of these spaces that there were too many continuous maps, except for the Relatively prime integer topology (also known as the Golomb space) and the Prime integer topology. The first one was proved to have too many continuous maps, and the second one is very similar to the first one, so I don't place much hope on it. We need to look somewhere else.

On MSE, Mirko indicated the existence of the following paper:
Constructions and Applications of Rigid Spaces, I
V. Kannan, M. Rajagopalan

In it, it is proven (Theorem 2.5.6) that, for any cardinal $\kappa$, if $(2^\kappa)^+ < 2^{2^\kappa}$, then there is a Hausdorff topological space of cardinality $\kappa$ which is strongly rigid, i.e. such that any continuous endofunction is either constant or the identity, which is a lot stronger than what we are trying to prove.

  • 1
    $\begingroup$ Related: mathoverflow.net/questions/286785: for the Golomb space there's no non-identity self-homeomorphism but OP mentions that there are continuum many continuous self-maps (T. Banakh, J. Mioduszewski, S.Turek, On continuous self-maps and homeomorphisms of the Golomb space, Comment. Math. Univ. Carolin. 59:4 (2018) 423–442. arxiv.org/abs/1711.06749). For the Kirch space, see arxiv.org/abs/2006.12357 doi.org/10.1016/j.topol.2021.107782; I'm not sure what's known about self-maps. $\endgroup$
    – YCor
    Mar 21, 2022 at 17:40
  • $\begingroup$ @YCor the Golomb space has continuum many continuous self-maps, see dml.cz/handle/10338.dmlcz/147548 $\endgroup$ Mar 21, 2022 at 17:48
  • $\begingroup$ I know (I gave the reference!). I asked about the Kirch space. $\endgroup$
    – YCor
    Mar 21, 2022 at 17:50
  • 8
    $\begingroup$ There is no infinite regular (Hausdorff) space with only countably many continuous maps to itself. Any space with only countably many such maps is countable (because constant maps are continuous) and, if Hausdorff, it cannot be compact (because it would then be metrizable). Therefore, if regular, it has a partition into infinitely many open-and-closed subsets. If D is an infinite discrete subset of the space, any function from the space to D which is constant on each element of the partition is continuous, and there are continuum many such functions. $\endgroup$
    – Anonymous
    Mar 26, 2022 at 11:45
  • 2
    $\begingroup$ @Anonymous Would you agree that the result is valid also for infinite regular non-Hausdorff spaces? Starting with such a space with countably many continuous self-maps, if one of the equivalence classes under topological indistinguishability is infinite, any self-maps sending that equiv class to itself and fixing the other elements is continuous and there are continuum many such maps. So we can assume all the equiv classes are finite. Then the Kolmogorov quotient is infinite, regular and $T_0$, hence $T_3$, and by the Hausdorff case, there are continuum many continuous self-maps ... (ct'd) $\endgroup$
    – PatrickR
    Nov 15, 2022 at 5:56

2 Answers 2


A partial answer: the only place where Kannan and Rajagopalan use the inequality $(2^\kappa)^+<2^{2^\kappa}$ is in the application of the Theorem on page 121. That theorem is a consequence of Corollary 10.15 in Comfort and Negrepontis' The Theory of Ultrafilters. However the particular case that they use can be proven without an appeal to that book. They show that for their set $F$ one can find a partition $\mathcal{A}$ of $\kappa$ into $\kappa$ many sets of cardinality $\kappa$ such that $\bar A\cap F\neq\emptyset$ for all $A\in\mathcal{A}$. Using that the space $\{0,1\}^{2^\kappa}$ has a dense subset of cardinality $\kappa$ one can map $\kappa$ onto a dense subset of that cube such that $\mathcal{A}$ is the set of point-inverses of that map, call it $f$. Then $\beta f$ not only maps $\beta\kappa$ onto that cube it also maps $F$ onto it. Take a closed subset $K$ of $F$ such that $f$ is surjective on $K$ and irreducible. For every $\alpha<2^\kappa$ let $I_\alpha=\{\beta\in\kappa:\pi_\alpha(f(\beta))=1\}$ and $J_\alpha=\kappa\setminus I_\alpha$. Then $\bigl\{(I_\alpha,J_\alpha):\alpha<2^\kappa\bigr\}$ is an independent family; even independent modulo the filter $\mathcal{F}=\{X\subseteq\kappa:K\subseteq\bar X\}$. The proofs of Theorems 2.2 and 2.7 in K. Kunen, Ultrafilters and independent sets, Trans. Amer. Math. Soc. 172 (1972), 299–306. go through with $\mathcal{F}$ as its starting point, so that $K$ contains a set of $2^\kappa$ many Rudin-Keisler incomparable ultrafilters.

Now specialize this to $\kappa=\omega_0$ and you have a ZFC-construction of the space in Kannan and Rajagopalan's Theorem 2.5.6.

  • 2
    $\begingroup$ Wasn't the question completely answered in @Anonymous's comment of March 26? $\endgroup$
    – LSpice
    Nov 9, 2022 at 23:01
  • 2
    $\begingroup$ @LSpice Isn't that only for regular spaces? $\endgroup$ Nov 10, 2022 at 1:33
  • 1
    $\begingroup$ @NoahSchweber, ah, yes, I missed that that wasn't part of the standing hypothesis. $\endgroup$
    – LSpice
    Nov 10, 2022 at 3:06
  • 1
    $\begingroup$ So to be clear, you are saying that we can prove directly in ZFC that there is a countable space which is strongly rigid? If so, it seems to be a complete answer. $\endgroup$ Nov 10, 2022 at 14:55
  • 2
    $\begingroup$ Yes,I say my answer is partial because I did not fully check Kannan and Rajagopalan's construction. They use rather much notation and I didn't have the time to go through it line by line. But their cardinality assumption is not needed for their special closed subset of $\beta D$. $\endgroup$
    – KP Hart
    Nov 10, 2022 at 21:01

The answer to this question is affirmative: there exists a countable Hausdorff space $X$ such that every continuous map $X\to X$ is either constant or the identity.

Many such spaces are constructed in this preprint of Banakh and Stelmakh, using the existence of continuum many Rudin-Keisler incomparable ultrafilters, like in the answer of K.P. Hart. Unfortunately, preparing the preprint (with my coauthor) took too much time (almost 3 months), so I has been late with my answer :(

  • 1
    $\begingroup$ How much choice is needed in your proof? $\endgroup$
    – Joel Adler
    Nov 24, 2022 at 9:11
  • 1
    $\begingroup$ To be more precise: To have ultrafilters we need AC. What about an example in ZF? $\endgroup$
    – Joel Adler
    Nov 24, 2022 at 12:44
  • $\begingroup$ @JoelAdler Yes, I certainly use the Axiom of dependent choice and also the existence of countably many pairwise Rudin-Keisler incomparable ultrafilters. Probably, this is a bit weaker than the full AC, but only just a bit weaker. I do not know whether such an example can be constructed in ZF. $\endgroup$ Nov 25, 2022 at 7:19
  • $\begingroup$ Very nice construction! Do you know whether a countable strongly rigid Hausdorff space can admit a nonconstant continuous function to a space with good separation properties? $\endgroup$ Nov 26, 2022 at 13:13
  • $\begingroup$ @AlessandroCodenotti Thank you! The strongly rigid countable Hausdorff spaces from that paper are Brown and hence admits no non-constant continuous maps to Urysohn spaces (i.e., spaces in which distinct points have disjoint closed neighborhoods). $\endgroup$ Nov 26, 2022 at 19:31

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.