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Let $I = [0,1]$. For this post start the ordinals at $1$ and consider this a limit ordinal. Consider an ordinal $\lambda$ and let $\lambda_{*}$ be the greatest limit ordinal less than or equal to $\lambda$; so $\lambda = \lambda_{*} + n$ for some nonnegative integer $n$. Consider $I^{\lambda}$ as a topological space in the dictionary order. Let $\mathbb{0}$ denote a sequence of $\lambda_{*}$ zeroes and define $\mathbb{1}$ analagously with ones. (So $\mathbb{0} \times I^n$ denotes the subset of $I^{\lambda}$ whose elements begin with $\lambda_{*}$ zeroes and vary arbitrarily in the last $n$ coordinates.)

The title is actually incorrect when $\lambda$ is not a limit ordinal but I wasn't sure how to word it descriptively and succinctly (see first comment). I'm really asking the following. Is $I^{\lambda} - (\mathbb{0} \times I^n \cup \mathbb{1} \times I^n)$ homogeneous? Is it 2-transitive in the case of a limit ordinal?

I only have intuition for the finite cases and for the limit ordinal cases but from these this seems reasonable.

My idea is to argue something like the space $\mathbb{Z} \times (I^{\lambda} - \mathbb{1} \times I^n)$ under the dictionary order is homeomorphic to this space and has a compatible abelian group operation (the obvious one using base number arithmetic), but don't know how to make any of this precise.

Edit: Niels points out that successor ordinals and ordinals with uncountable cofinality will be counterexamples. I should have caught the first example but wouldn't have considered the second and so this question may have been overambitious/naive. In order to give some direction I'd like to narrow the question to the following. Is the space $I^{\omega}$ minus its end-points homogeneous? I suspect it is not because sequences converging to $(\frac{1}{2}, \frac{1}{2}, \ldots)$ from below appear to have a characteristic difference from those sequences converging to $(1,0,0,\ldots{})$ from below but I'm not sure how to make this precise and want to learn.

Geoffrey Sangston
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  • I wasn't sure how to phrase "essentially homogeneous" in a more descriptive but succinct manner. Is there a standard operation for getting $I^\lambda - (\mathbb{0} \times I^n \cup \mathbb{1} \times I^n)$ from $I^\lambda$? When $\lambda$ is a limit ordinal you just remove the boundary points and that operation definitely seems like it should have a name. In successor ordinals you're removing "boundary spaces". I guess the title could have been "Is $[0,1]^\lambda$ minus its boundary homogeneous?" and I could have clarified this in the post.

    Ah looks like you can edit titles.

     – Geoffrey Sangston Jun 11 '17 at 18:29
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    If $\lambda$ is a sucessor ordinal (other than 1), some points are boundary points of a path component while others are interior points. This rules out homogeneity. If $\lambda$ has uncountable cofinality, some points are limits of strictly monotone sequences while others are not. This also rules out homogeneity. That leaves only the case where $\lambda$ has cofinality $\aleph_0$. – Niels J. Diepeveen Jun 12 '17 at 10:52
  • @NielsDiepeveen I'm still confused about whether or not David Speyer's construction can work however (https://math.stackexchange.com/a/418753/444923). His space is an abelian topological group and so should be homogeneous but seems like it should fall prey to your idea here so should not be homogeneous. – Geoffrey Sangston Jun 12 '17 at 21:48
  • I worked through your example (and learned a lot, thank you). Suppose $\lambda$ has uncountable cofinality. Consider a point $p$ which is not eventually constant. Any sequence converging to $p$ must be indexed by an uncountable ordinal. A point which is eventually constant is a limit of a countable sequence. A homeomorphism preserves cardinality of sequences and limit points so no homeomorphism can take a point of this type to one of the previous. – Geoffrey Sangston Jun 12 '17 at 22:33
  • The fundamental limitation of David Speyer's approach is that no totally ordered abelian group other than the trivial group or $\mathbb{R}$ is connected in the order topology. However, that question only seems to call for a homogeneous, totally orderd space. (the term one-dimensional is used, but apparently not in the topological sense) – Niels J. Diepeveen Jun 12 '17 at 23:14
  • @NielsDiepeveen Is what I've written in the second comment correct? (I omitted details / kept rewriting it since it's hard to fit everything and I kept making mistakes).

    Wouldn't that proof apply to David Speyer's example as well though?

     – Geoffrey Sangston Jun 12 '17 at 23:32
  • I don't think it is quite correct. An example of a point that is the limit of a strictly increasing sequence (regardless of $\lambda$) is $\lim_{n\to\infty} ((n-1)/(2n), 0, 0, 0, ...) = 1/2, 0, 0, 0, ...$. – Niels J. Diepeveen Jun 13 '17 at 01:58
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    An example of a point that is not the limit of a strictly monotone sequence is $1/2, 1/2, 1/2, ...$. Suppose ${(x_n^k){k < \lambda}}{n=1}^\infty$ is such a sequence. For every $n$ there is $\alpha_n<\lambda$ s.t. $x_n^{\alpha_n} \ne 1/2$. If $\lambda$ has uncountable cofinalty, $\beta = \sup_n \alpha_n < \lambda$ and the points where the first $\beta+1$ terms are $1/2$ form a neighbourhood of $1/2, 1/2, 1/2, ...$ that does not contain any of the $x_n$. – Niels J. Diepeveen Jun 13 '17 at 02:00
  • Let us continue this discussion in chat. – Geoffrey Sangston Jun 13 '17 at 03:22
  • Actually now I doubt your idea. Consider some element $f : \lambda \to [0,1]$ for arbitrary $\lambda$. There are many countable sequences converging to $f$: suppose $f_{n}$ is a (countable) sequence whose elements equal $f$ everywhere except at say index $k$. Then we can write $f_{i}(k) = f(k) - \frac{1}{i + h}$ (where $h$ is a constant chosen to be large enough so that this is never equal to $0$). – Geoffrey Sangston Jun 13 '17 at 15:53
  • My previous example may have given you the wrong impression that convergence in the lexicographic order is somehow related to pointwise convergence, which it is not. The limit of $(n-1)/2n, 1, 1, ...$ is also $1/2, 0, 0, ...$. – Niels J. Diepeveen Jun 13 '17 at 20:09
  • If I understand correctly you're asserting that the sequence of points $(\frac{n-1}{2n},1,1,\ldots)$, indexed by $n$, converges both to $(\frac{1}{2},0,0,\ldots)$ and to $(\frac{1}{2},1,1,\ldots)$. This implies the space is not Hausdorff by theorem 17.10 of Munkres but theorem 17.11 of Munkres says every simply ordered set is Hausdorff in the order topology.

    I don't believe $\frac{1}{2},0,0,\ldots$ is a limit point of that sequence.

     – Geoffrey Sangston Jun 13 '17 at 20:34
  • Isn't $(\frac{n-1}{2n}, \frac{n-1}{2n}, \ldots)$ a strictly monotone sequence converging to $(\frac{1}{2}, \frac{1}{2}, \ldots)$? My post from two comments ago gives you a general way to get a countable monotone convergent sequence for any point. – Geoffrey Sangston Jun 13 '17 at 20:44
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    I never said it converges to $1/2, 1, 1, ...$. It does not, at least not in the relevant topology. – Niels J. Diepeveen Jun 13 '17 at 20:56
  • Ah I see your point. Because for any point in that sequence, 1/2,1/2,0,0,... is closer to 1/2,1,1,... – Geoffrey Sangston Jun 13 '17 at 21:05

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