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Let $\kappa>0$ be a cardinal and let $(X,\tau)$ be a topological space. We say that $X$ is $\kappa$-homogeneous if

  1. $|X| \geq \kappa$, and
  2. whenever $A,B\subseteq X$ are subsets with $|A|=|B|=\kappa$ and $\psi:A\to B$ is a bijective map, then there is a homeomorphism $\varphi: X\to X$ such that $\varphi|_A = \psi$.

Questions: Is it true that for $0<\alpha < \beta$ there is a space $X$ such that $|X|\geq \beta$, and $X$ is $\alpha$-homogeneous, but not $\beta$-homogeneous? Is there even such a space that is $T_2$? Also it would be nice to see an example for $\alpha=1, \beta=2$. And I was wondering whether there is a standard name for $\kappa$-homogeneous spaces. (Not all of these questions have to be answered for acceptance of answer.)

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    $\begingroup$ Shouldn't you be asking for a space that is $\alpha$-homogeneous, but not $\beta$-homogeneous? It seems that $\alpha$ and $\beta$ are mixed up in your question. $\endgroup$ – Joel David Hamkins Jul 28 '16 at 14:08
  • $\begingroup$ Set theorists would be inclined to make the definition for sets of size less than $\kappa$, rather than equal to $\kappa$, since this works better with limit cardinals. On that way of making the definition, your concept would be called $\kappa^+$-homogeneous. $\endgroup$ – Joel David Hamkins Jul 28 '16 at 14:10
  • $\begingroup$ Do you have any restrictions about the cardinality of $X$? If $\kappa$ is infinite and you allow $|X| = \kappa$ then the discrete topology on $\kappa$ is $\alpha$-homogeneous for all $\alpha < \kappa$ but not $\kappa$ homogeneous, as witnessed by $A = X$ and any $B \subset X$, $|B| = \kappa$, $B \neq X$. $\endgroup$ – Yair Hayut Jul 28 '16 at 21:13
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    $\begingroup$ For $n < \omega$ what you call $n$-homogeneity seems to be called strong $n$-homogeneity in the literature. (The "not strong" version just requires that there is a auto-homeomorphism $\varphi$ such that $\varphi [A] = B$, i.e., the two sets can be mapped onto each other by a homeomorphism, but you don't specify which point goes where.) $\endgroup$ – user642796 Jul 30 '16 at 12:14
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    $\begingroup$ It's a trivial modification of Joel and Yair's example, but finite discrete and indiscrete spaces of size $\alpha$ are $\alpha$-homogeneous but cannot be $\beta$-homogeneous for $\beta>\alpha$ because of your first condition. $\endgroup$ – Gabriel C. Drummond-Cole Sep 23 '16 at 19:13
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This is a great question!

The disjoint union of two circles is $1$-homogeneous, but not $2$-homogeneous. It is $1$-homogenous, since you can swap any two points and extend this to a homeomorphism (basically, "all points look alike"). But it is not $2$-homogeneous, since you can let $A$ be two points from one circle, and let $B$ be two points from different circles; there is no way to extend a bijection of $A$ with $B$ to a homeomorphism of $X$ (not all pairs look alike).

The real line $\mathbb{R}$ is $2$-homogeneous, but not $3$-homogeneous. For $2$-homogeneity, given any two pairs of reals $a,b$ and $x,y$, then no matter how you map these bijectively, you can extend to a homeomorphism of the line by affine translation (all pairs look alike). But the line is not $3$-homogeneous, since we can biject the triples $0,1,2$ mapping to $0,2,1$, respectively; this does not extend to a homeomorphism, since it doesn't respect between-ness (not all triples look alike).

The unit circle is $3$-homogeneous, but not $4$-homogeneous (thanks to Andreas Blass in the comments). For $3$-homogeneity, given two triples of points, we can match up the first in each by rotating the circle, and then match up the other two by stretching or by flipping and stretching, depending on whether the orientation was preserved or not (all triples look alike). It is not $4$ homogeneous, since we can have four points in clockwise rotation, and then try to fix the first two and swap the other two; this cannot extend to a homeomorphism, since fixing the first two fixes the orientation, which is not respected by swapping the other two (not all quadruples look alike).

I don't know examples yet that are $4$-homogeneous, but not $5$-homogeneous, or $n$-homogeneous but not $n+1$-homogeneous, for $n\geq 4$.

The real plane $\mathbb{R}^2$ appears to be $n$-homogeneous for every finite $n$, but not $\omega$-homogeneous (thanks again to Andreas). It is $n$-homogeneous, because given any two sets of $n$ points, we can imagine the plane made of stretchable latex and simply pull the points each to their desired targets, with the rest of plane getting stretched as it will. One can see this inductively, handling one additional point at a time: having moved any finitely many points, nail them down through the latex; now any additional point can be stretched to any desired target, before also nailing it down, and so on (so all $n$-tuples look alike). It is not $\omega$-homogeneous, since a countable dense set can be bijected with a countable bounded set, and this will not extend to a homeomorphism.

Meanwhile, the infinite case is settled. For any infinite cardinal $\beta$, the discrete space of size $\beta$ is $\alpha$-homogeneous for every $\alpha<\beta$ but not $\beta$-homogeneous, in the OP's terminology. Any bijection of small subsets can be extended to a permutation, since there are $\beta$ many points left over, but if $X$ has size $\beta$, then we can take $A=X$ and $B=X-\{a\}$, which are bijective, but this bijection cannot be extended to a permutation of $X$.

In particular, the countable discrete space is $n$-homogeneous for every finite $n$, but not $\omega$-homogeneous, just like $\mathbb{R}^2$.

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  • $\begingroup$ The plane $\mathbb{R}^2$ is obviously $3$-homogeneous, but it is also $4$-homogeneous, and I think also $5$-homogeneous. For higher finite levels, I can't quite see what happens yet. $\endgroup$ – Joel David Hamkins Jul 28 '16 at 16:10
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    $\begingroup$ I think I can "see" that the plane is $n$-homogeneous for all finite $n$, but I don't see (yet) a clear proof. The picture I have is that the plane is made of very elastic material, and handles are attached to $n$ points. You can use the handles to drag those points wherever you want, and the rest of the plane "flows" along. (I vaguely recall a similar picture associated with an old theorem, perhaps of Alexander.) $\endgroup$ – Andreas Blass Jul 28 '16 at 18:03
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    $\begingroup$ Another example along the same lines as your answer: a circle is 3-homogeneous but not 4-homogeneous. $\endgroup$ – Andreas Blass Jul 28 '16 at 18:25
  • $\begingroup$ I had made a comment to that effect earlier---that the plane is $n$-homogeneous for every $n$---but then deleted it when I lost confidence in it. But now your handle manner of expressing it is very convincing, so I'm inclined to agree again. Meanwhile, of course it is not $\aleph_0$-homogeneous. And I like your circle example! Can you make such examples for every $n$? $\endgroup$ – Joel David Hamkins Jul 28 '16 at 18:25
  • $\begingroup$ Unfortunately, I don't see how to handle larger finite $n$. Model-theorists might be able to contribute something here, because the examples of the line and the circle correspond to a couple of standard examples of such homogeneity properties in model theory, namely dense linear orders and circular orders. (The latter is a ternary relation, $R(x,y,z)$ intended to mean that $x$, $y$, and $z$ occur in clockwise order around the circle.) I think the model theorists have examples of fancier sorts of homogeneity; I don't know whether they can be made into topological examples. $\endgroup$ – Andreas Blass Jul 28 '16 at 18:47
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The sort of space you describe is usually called strongly $\kappa$-homogeneous. If you google that phrase you will find some interesting results about these kinds of spaces (mostly concerning how this property relates to other homogeneity properties).

The earliest reference I could find to strongly $n$-homogeneous spaces (only finite values of $n$ are considered) is in a 1953 paper by C. E. Burgess (available here).

Despite the fact that these kinds of spaces have been in the literature for well over half a century, it is unknown whether there is a topological space that is strongly $4$-homogeneous but not strongly $5$-homogeneous. (This is stated explicitly in this paper by Ancel and Bellamy from last year -- see the second paragraph of the second page.) As far as I can tell (although I don't have an authoritative reference), it is also unknown whether there is a strongly $n$-homogeneous space that is not also strongly $(n+1)$-homogeneous for any finite $n \geq 4$.

Therefore Joel's answer (along with some of Andreas's comments below it) constitutes the state-of-the-art knowledge on this question.

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  • $\begingroup$ Of course, there are the trivial examples of finite discrete and indiscrete spaces, which give two strongly $n$-homogeneous but not $n+1$-homogeneous spaces for each $n$. $\endgroup$ – Gabriel C. Drummond-Cole Feb 27 '17 at 15:56

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