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A topological space $(X,\tau)$ is said to be homogeneous if for all $x,y$ there is a homeomorphism $\varphi:X\to X$ such that $\varphi(x) = y$.

Is there an infinite homogeneous Hausdorff space $(X,\tau)$ such that every continous map $f: X\to X$ is either a homeomorphism, or constant?

EDIT. I forgot to add "infinite" in the original question.

Dominic van der Zypen
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  • you missed 'homogeneos' before 'Hausdorff', right? – Fedor Petrov Mar 06 '18 at 08:20
  • Right - and I also forgot "infinite"... – Dominic van der Zypen Mar 06 '18 at 10:18
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    well, you should not change the question once there is already an answer. In this way a perfectly correct answer becomes a wrong one. Please write an edit, explaining the modifications you made. – Francesco Polizzi Mar 06 '18 at 10:25
  • Yes, sorry for that. Maybe I should have written an entirely new question? – Dominic van der Zypen Mar 06 '18 at 10:34
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    @FrancescoPolizzi I disagree with you. Even if this corresponds to the original formulation, this does not deserve an answer, but rather an comment saying that the question should be reformulated. If this were the intended question, it would be closed as off-topic. – YCor Mar 06 '18 at 13:06
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    @YCor I agree with you. Franceso's answer is - although correct - trivial and unfortunately can not be generalised to larger spaces. But of course it was my mistake not to exclude this example – Dominic van der Zypen Mar 06 '18 at 13:19
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    @YCor: of course you can disagree. At any rate, this was the question, it was unanswered, open for more than one hour and it also had un upvote. Sometimes it happens in life that people miss trivial counterexamples, this is not my fault. – Francesco Polizzi Mar 06 '18 at 13:21
  • @FrancescoPolizzi that's right, I missed that example – Dominic van der Zypen Mar 06 '18 at 13:22
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    That said, I still think that modifying a question with an answer, making a right answer a wrong one is a bad practice, regardless if someone finds such an answer interesting or not. – Francesco Polizzi Mar 06 '18 at 13:22
  • @FrancescoPolizzi Would you suggest making an entirely new question? (I'm sure this is not the first question missing trivial counterexamples by not excluding the empty space, or spaces with more general low cardinalities.) – Dominic van der Zypen Mar 06 '18 at 13:24
  • @DominicvanderZypen: I guess it would be useless, by now – Francesco Polizzi Mar 06 '18 at 13:24
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    I guess a better place to discuss this topic is meta.MO: https://meta.mathoverflow.net/questions/3637/editing-a-question-admitting-a-trivial-example – Dominic van der Zypen Mar 06 '18 at 13:30

2 Answers2

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Topological groups are homogeneous. In

J. van Mill, "A topological group having no homeomorphisms other than translations," Transactions of the AMS 280 (1983), pp. 491-498 (link),

Jan van Mill constructed an infinite topological group whose only self-homeomorphisms are group translations. Such a space is called "uniquely homogeneous" -- it is homogeneous, but for any pair of points there is exactly one self-homeomorphisms of the space witnessing homogeneity. Jan's group also has the amazing property that removing any point results in a rigid space.

In the same paper (section 4), van Mill shows that, assuming the Continuum Hypothesis, there is a topological group whose only continuous self-maps are either group translations or constant functions.

Thus the answer to your question is "consistently yes, and you can come close in ZFC." I do not know whether anyone else has come along and improved Jan's CH result to a ZFC result (but a quick glance through the papers citing Jan's seems to indicate that no one has).

Will Brian
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  • Amazing - thanks for thinking of this group example @WillBrian, and for the link to this paper, which is completely new to me – Dominic van der Zypen Mar 06 '18 at 13:23
  • @DominicvanderZypen Very strange that this paper is completely new for you as it has been already mentioned in the (accepted) answer to another your question (https://mathoverflow.net/questions/285777/anti-fixed-point-property/285779#285779) – Taras Banakh Mar 10 '18 at 17:10
  • @Will Brian Concerning your question on weakening of CH, yes, it can be weakened to $\acute{n}=\mathfrak c$, see (https://mathoverflow.net/questions/285777/anti-fixed-point-property). Concerning the cardinal $\acute{n}$, look at (https://mathoverflow.net/questions/285780/are-the-sierpi%C5%84ski-cardinal-acute-mathfrak-n-and-its-measure-modification) where you actively participated in the discussion. Probably you just forgot. – Taras Banakh Mar 10 '18 at 17:14
  • @Will Brian This means that the results obtained during mathoverflow dscussions should be published in ordinary journals? Otherwise they are not searchable and can be forgotten even by participants of these discussions? Very surprising. Because at some moment I started to think that mathoverflow can substitute ordinary way of publishing (at least short) results in journals. – Taras Banakh Mar 10 '18 at 17:17
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    @DominicvanderZypen The same paper of van Mill was also mentioned in another (accepted) answer to your question (https://mathoverflow.net/questions/209595/t-2-space-x-with-x-cong-textautx/). So, the same result of van Mill answers three your questions and you still write that it is completely new for you. What does it mean? That you do not read and analyse answers to your questions? Or have a short memory? Sorry for this comment. But I am just curious. – Taras Banakh Mar 10 '18 at 17:35
  • @TarasBanakh: You should read my last paragraph more closely. I do not ask what is the optimal set-theoretic hypothesis under which Jan's argument can be carried out. I suppose it's nice that you've come along and identified this hypothesis as something much weaker than CH, but I was not asking about that -- I was asking whether Jan's result can be proved in ZFC. This has little to do with the questions you link to. (And no, for what it's worth, I have not forgotten your question about Sierpinski's cardinals. It was/is a very good question, and I truly enjoyed thinking about it.) – Will Brian Mar 12 '18 at 10:55
  • @WillBrian Thanks for the explanations. Indeed, it is a good question if a rigid Boolean group constructed by van Mill under CH exists in ZFC. – Taras Banakh Mar 12 '18 at 21:24
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Remark. This answer was written before the "infinite" assumption was added.

Take the discrete space with two points.

It is clearly homogeneous and Hausdorff, and its only self-maps are the two constant maps, the identity and the involution exchanging the two points.

Francesco Polizzi
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