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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. –  joriki Jul 20 '12 at 21:12
    
I was able to think of an example that isn't Hausdorff, and so updated the question. –  Thomas Andrews Jul 20 '12 at 21:18
    
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. –  MartianInvader Jul 20 '12 at 21:35
    
Weird, the Wolfram definition of $n$-transitivity seems wrong. mathworld.wolfram.com/TransitiveGroupAction.html –  Thomas Andrews Jul 20 '12 at 21:44
1  
@joriki feedback was sent to MathWorld –  Thomas Andrews Jul 22 '12 at 14:55

1 Answer 1

up vote 5 down vote accepted

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 $X$.

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