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I was thinking about a question on here earlier, and came up with this question.

[Added Hausdorff note, below.]

It is easy to see that the group of self-homeomorphisms of the real line acts $2$-transitively on the space, but not $3$-transitively.

Likewise, it is clear that, for example, $\mathbb R\setminus \{0\}$ is a space on which the group of self-homemorphisms is $1$-transitive but not $2$-transitive.

If you take the real line with the open sets of the form $(-\infty,a)$ for some $a$, then I guess this gives you an example, but that space is not Hausdorff.

I can't seem to think of an example of a connected Hausdorff space where the self-homeomorphisms are $1$-transitive but not $2$-transitive.

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    +1, nice question! This seems to be related to whether for any two points there are automorphisms that take $x$ to $y$ while leaving $z$ invariant, so one might look for an example of a connected space where this isn't the case.2012-07-20
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    I was able to think of an example that isn't Hausdorff, and so updated the question.2012-07-20
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    It might be useful to add or link to a definition of $k$-transitive. As terms go, it's a bit difficult to look up.2012-07-20
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    Weird, the Wolfram definition of $n$-transitivity seems wrong. http://mathworld.wolfram.com/TransitiveGroupAction.html2012-07-20
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    How is it wrong? It looks OK to me.2012-07-20
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    @joriki If the set $X$ is infinite, it's okay. But according to their definition, any action on a set of size $2n-1$ is $n$-transitive, since there are no set of distinct $\{x_1,..,x_n,y_1,...,y_n\}$2012-07-20
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    @joriki They also use the term "set" which is lazy language, since it is about ordered $n$-tuples if distinct elements, $(x_1,...,x_n)$ and $(y_1,...,y_n)$.2012-07-20
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    @Thomas: You're right; it's not just lazy language; it also makes it seem as if e.g. $y_2$ has to be distinct from $x_1$. Have you written feedback?2012-07-20
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    @joriki I can't find the link for feedback on the Wolfram site.2012-07-20
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    It's "Send a Message to the Team" in the sidebar on the left.2012-07-20
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    Very nice question.2012-07-20
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    Have you written feedback to MathWorld? If not, I'm happy to do it, I just want to avoid duplicating it.2012-07-22
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    @joriki feedback was sent to MathWorld2012-07-22

1 Answers 1

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Let $X=\big(\omega_1\times[0,1)\big)\setminus\{\langle 0,0\rangle\}$ ordered lexicographically. Clearly $X$ is connected and Hausdorff; indeed, $X$ is even hereditarily collectionwise normal. For each $x\in X$ the set $(\leftarrow,x)$ is homeomorphic to $\Bbb R$ with the Euclidean topology, so $X$ is transitive under autohomeomorphisms, but it’s clear that no autohomeomorphism of $X$ can reverse the order of a pair of points: that would require embedding $\omega_1^*$ in an initial segment of $X$.