Now cross-posted to Mathoverflow.

Is there an infinite topological space $X$ with only countably many continuous functions to itself? Such a space would have only countably many points because the constant functions are continuous. A space with countably many points such that only the constant functions and the identity are continuous would work. I wasn't able to find something: all the topological spaces I can think of have $2^{\aleph_0}$ continuous maps to themselves.

We cannot find a counterexample among the metrizable spaces. Let $X$ be an infinite countable metric space with metric $d$.

If $X$ is discrete, it is clearly false, so suppose that there is some point $x$ which is not isolated. Let $(r_i)_{i \in \mathbb{N}}$ be a strictly decreasing sequence of real numbers converging to $0$ such that for any $n \in \mathbb{N}$, there is no point $y$ with $d(x,y) = r_n$, and such that there is a point $y$ with $r_n > d(x,y) > r_{n+1}$.

We define $$B_{n+1} = \{y \in X \mid r_n > d(x,y) > r_{n+1} \}\space \text{and} \space B_0 = \{y \in X \mid d(x,y) > r_0 \}$$ For any integer $n$, we choose such a point $y_n$ in $B_n$.

Then, we can define the continuous function $$f : X \to X \space \text{as} \space f(x) = x \space \text{and} \space f(y) = y_n \space \text{for}\space y \space \text{in}\space B_n$$ Then, for each $n \in \mathbb{N}$, you can choose to swap $y_{2n}$ and $y_{2n+1}$ or not, giving you $2^{\aleph_0}$ continuous maps.

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.

I looked for other examples in the Counterexamples in topology book, but nothing looked promising (not that I proved it for every countable space in the book.)

More generally, for any infinite cardinal $\kappa$, is there a topological space with $\kappa$ points and exactly $\kappa$-many continuous maps to themselves? An obvious example for $2^{\aleph_0}$ is $\mathbb{R}$ with the euclidean topology.

In fact, for any infinite cardinal $\kappa$, seeing $\kappa$ as a discrete space, we can define the space $X = \kappa^{\aleph_0}$ with the product topology, and this space has $2^\kappa$ continuous functions from itself to itself, because the continuous functions $f : X \to X$ are determined by their values on the sequences which are eventually constant. Hence, for any cardinal $\kappa$ such that $2^\kappa = \kappa^{\aleph_0}$, we know that the answer is positive for $2^\kappa$.

Edit: 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. Here's my last idea: if we take $F$ a filter on $\mathbb{N}$, adding the empty set, we obtain a topological space. Could some space obtained this way have only countably many continuous functions to itself? Could it be always true if the filter is an ultrafilter? I don't know how to answer this question.

  • 1,931
  • 5
  • 18
  • 4
    The Golomb space, that is $\Bbb N$ with the topology whose basis is given by arithmetic progression $\{a+bn\mid b\in\Bbb N\}$ with coprime $a,b$, could be a candidate, it is a countable connected Hausdorff space whose only continuous self homeomorphism is the identity – Alessandro Codenotti Aug 23 '21 at 15:23
  • 4
    The Golomb space has $2^{\aleph_0}$ continuous maps to itself, sadly. It is proven here, Theorem 4.9: https://dml.cz/handle/10338.dmlcz/147548 – QuinnLesquimau Aug 23 '21 at 19:09
  • 4
    Each such space $X$ (if it exists) must be $T_0$. If it is not, then it contains a two-point subspace $T$ with the trivial topology. Now there are uncountably many continuous maps $X \to T$. – Paul Frost Aug 24 '21 at 09:23
  • 4
    The filter space you mention at the end was one of the first ideas I tried when I saw the question earlier. Unfortunately, it does not work: if $A$ is any element of the filter $F$, then any function which is the identity on $A$ will be continuous. (So as long as the filter contains some coinfinite set, this gives uncountably many. If the filter contains only cofinite sets then either the topology fails to be $T_0$ or else it is either the cofinite topology or the cofinite topology with a generic point and any permutation of the non-generic points is continuous.) – Eric Wofsey Aug 25 '21 at 16:10
  • @Eric Wofsey Arg, it seems obvious now! Thank you! But now, I really have no idea left. – QuinnLesquimau Aug 25 '21 at 16:14
  • The best hope I can see to constructing a counterexample is some kind of transfinite recursion to construct the topology to one-by-one exclude every non-constant non-identity function from being continuous. I haven't been able to make it work, though. – Eric Wofsey Aug 25 '21 at 16:19
  • @EricWofsey Might a diagonalization argument show that no second-countable space has exactly $\aleph_0$ continuous self-maps (I'm basically thinking about forcing to construct a "generic" continuous self-map)? It's a bit late on my end but it seems plausible. – Noah Schweber Aug 26 '21 at 04:35
  • @NoahSchweber: Sounds plausible but second-countability is a huge restriction. – Eric Wofsey Aug 26 '21 at 04:37
  • @EricWofsey Oh I know, but already - unless I'm missing something - that isn't *quite* trivial. – Noah Schweber Aug 26 '21 at 04:52
  • @QuinnLesquimau Just a link to a related MO question: Taras Banakh (https://mathoverflow.net/users/61536/taras-banakh), Is the Golomb countable connected space topologically rigid?, URL (version: 2019-12-06): https://mathoverflow.net/q/285557 – Mirko Sep 12 '21 at 15:06
  • @QuinnLesquimau I think you should ask this at mathoverflow. – Noah Schweber Mar 21 '22 at 14:58
  • @NoahSchweber Good idea, I've just done it. – QuinnLesquimau Mar 21 '22 at 17:23

1 Answers1


I have something that I know is not a complete answer, but I think good enough to put here as an extended comment.

First I thought the question might be related to rigid spaces, i.e. where the only autohomeomorphism is the identity. So I googled countable rigid topological space.

One if the results was the following paper:
Constructions and Applications of Rigid Spaces, I
V. Kannan, M. Rajagopalan

In section 2 (p.103 of their paper) they discuss strongly rigid spaces:
A Hausdorff space $X$ is said to be strongly rigid if every continuous self-map of $X$ is either the identity map or a constant map.

They discuss the difficulties in finding such spaces, but they have a construction which seems to answer the above question under the assumption that $\mathfrak c^+<2^{\mathfrak c}$ where $\mathfrak c=2^{\aleph_0}$. (Pretty long, I just glanced through the paper, they work with $\beta D$ where $D$ is discrete of infinite cardinality $m$, and their assumption is $(2^m)^+<2^{2^m}$; the main idea seems to be that $\beta D\setminus D$ contains many points of "different topological type", some results of Kunen. Or rather, I assumed it related to some results of Kunen, but looking again I see that the authors use the following result of Hajnal:
THEOREM. Let $D$ be an infinite discrete space of cardinality $m$. Assume $(2^m)^+<2^{2^m}$. Then every closed subset of $\beta D\setminus D$ having cardinality $2^{2^m}$ contains a subset of cardinality $2^m$ in which no two elements are comparable.
One may want to review Kunen's later results, to see if the condition $(2^m)^+<2^{2^m}$ could possibly be removed.)

See in particular Theorem 2.5.6 (p.124) and Remark 2.5.8 (p.127).

THEOREM 2.5.6. The space $S$ (of Construction 2.5.3) has the following properties:
(1) $S$ is a Hausdorff space.
(2) $S$ is connected. In fact, both $S\setminus\{\infty\}$ and $S\setminus\{-\infty\}$ are connected.
(3) $|S| = m.$
(4) $S$ is strongly rigid.

Of course, when $m=\aleph_0$ and $(2^{\aleph_0})^+<2^{2^{\aleph_0}}$ then we get a countable strongly rigid space $S$, and in particular only countably many continuous functions to itself, namely the identity map and all constant maps.

(Their Remark 2.5.8. We believe that we have given the first example of a countable strongly rigid space. Note that even countable connected Hausdorff spaces are rare ([22]).)

Some other papers that google showed me and might be relevant:

A Compact Space with a Measure That Knows Which Sets Are Homeomorphic

REMARK. It is a trivial remark, but just for the record, it is easy to see (and essentially already noted by the OP) that any example (of a space with only countably many continuous maps into itself) could have at most finitely many isolated points. Indeed, we could use the identity map on the subset of non-isolated points, and we could send each isolated point to any point we want, and if there were infinitely many isolated points then we get at least $\mathfrak c=2^{\aleph_0}$ many continuous maps. Well, I take this back, as it is not obvious why the resulting maps need to be continuous, it may not be that easy to see, though the idea seems plausible.
On the other hand, if we have a strongly rigid example $X$ then $X$ cannot have any isolated point (for if $p$ was isolated then we could take the identity on $X\setminus\{p\}$, and send $p$ to any $x\not=p$ to obtain a continuous map that is neither the identity, nor constant).
I am thinking along the lines of starting with a strongly rigid example $X$ and adding an isolated point $p$ to get a space $Y=X\cup\{p\}$ that is not strongly rigid yet has only countably many continuous maps into itself. This (or more generally, adding finitely many isolated points) was supposed to be easy, but I am all getting messed up with the details, and starting to doubt it (watching US Open tennis men's final at the same time, and that is actually over now, but let me leave this question alone for a moment).

  • 13,040
  • 16
  • 39