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This question is motivated by this problem of Dominic van der Zypen.

Problem. Let $X=\prod_{i=1}^nX_i$ be the Tychonoff product of linearly ordered compact Hausdorff spaces $X_1,\dots,X_n$. Is it true that each subspace $Y\subset X$ has large inductive dimension $\mathrm{Ind}(Y)\le n$ and the covering dimension $\mathrm{dim}(Y)\le n$?

Remark 1. By induction it can be shown that each subspace $Y$ of $X$ has small inductive dimension $\mathrm{ind}(Y)\le n$.

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  • $\begingroup$ That's a really natural follow-up to the question you link to, I think an answer to this will be really helpful. $\endgroup$
    – Dominic van der Zypen
    Commented Jul 17, 2018 at 14:41
  • $\begingroup$ @DominicvanderZypen The problem with this question is with some possible pathologies (like the Tychonoff plank). For non-normal spaces, the large Inductive and covering dimensions have two version: for open covers and functionally open covers and those are different. So those question are for deep specialists that feel all such sublteties. By the way, Dominic: what about my counterexample to your question (mathoverflow.net/questions/305000/…). Is everything understanable, or there are some doubts? $\endgroup$
    – Taras Banakh
    Commented Jul 17, 2018 at 14:52
  • $\begingroup$ Thanks @tarasbanakh the counterexample you provided is very well understandable and useful! $\endgroup$
    – Dominic van der Zypen
    Commented Jul 18, 2018 at 6:37
  • $\begingroup$ How exactly are topology and order related? You assume the order relation is a closed subset of the cartesian square? Or is the topology the interval topology? Maybe there is some standard terminology that I don't know... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 18, 2018 at 14:42
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    $\begingroup$ @მამუკაჯიბლაძე I have in mind that linearly ordered spaces carry the order topology (i.e., the topology generated by the linear order). The product of linearly ordered spaces carries the topology of Tychonoff product. $\endgroup$
    – Taras Banakh
    Commented Jul 18, 2018 at 16:17

2 Answers 2

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In an old PhD-thesis "Finite products of locally compact ordered spaces" by J. van Dalen (Vrije Universiteit, Amsterdam) from 1972, I found (I could not find a paper with the result, so far, as I have no access to a university library, but I have the thesis on my shelves) the following Corollary 12.1 (page 45):

Let $m$ be a natural number and let $X_1, \ldots, X_m$ be non-degenerate connected ordered spaces. Then $\operatorname{ind}(\prod_{i=1}^m X_i)= m$.

And corollary 12.2 , which has the same data plus the extra assumption that $\prod_{i=1}^m X_i$ is normal (to make $\operatorname{Ind}$ sensibly defined, I suppose) and concludes that $\dim(\prod_{i=1}^m X_i) = \operatorname{Ind}(\prod_{i=1}^m X_i) = m$ as well.

In the references I also found

I.K. Lifanov, "Dimensionality of the product of ordered continua", Dokl. Akad. Nauk SSSR 177 (1967), 778-781 (Sov. Math. Dokl. 8 (1967), 1500-1503)

and

I.K. Lifanov, "The dimension of a product of unidimensional bicompacta", Dokl. Akad. Nauk SSSR 180 (1968), 534-537 (Sov. Math. Dokl. 9 (1968), 648-651)

which look relevant too.

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    $\begingroup$ The fact that the product $\prod_{i=1}^m X_i$ has dimension $m$ was more-or-less clear (see Theorem 3.2.13 and Problem 1.8.K in Engelking's book "Theory of Dimensions, Finite and Infinite"). The problem was about subspaces of products. Since the dimension functions $\mathrm{Ind}$ and $\mathrm{dim}$ are not monotone, this question is not trivial. $\endgroup$
    – Taras Banakh
    Commented Jul 19, 2018 at 4:42
  • $\begingroup$ @TarasBanakh that the product has dimension $\le m$, is not that hard, as a LOTS has dimension $\le 1$. The thesis does not mention subspaces. $\endgroup$
    – Henno Brandsma
    Commented Jul 19, 2018 at 4:58
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(All spaces are assumed to be Hausdorff). For every closed subspace $A$ of an arbitrary compact space $X$, we have:

$$ \dim(A)\ \le\ \dim(X) $$

This more than answers one of the questions. (The other q. would make me think, ouch!).

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    $\begingroup$ Thank you for the answer. I expected that you will be interested. But the question was about arbitrary (not necessarily closed) subspaces of products. For such subspaces the monotonicity law does not hold: for example, the Tychonoff plank has covering dimension 1 but is a subspace of the product $[0,\omega_1]\times [0,\omega]$ of two zero-dimensional compact spaces, which has dimension zero. $\endgroup$
    – Taras Banakh
    Commented Jul 18, 2018 at 7:45
  • $\begingroup$ I deleted the second question about the connected Tychonoff plank as this space has dimension 2. $\endgroup$
    – Taras Banakh
    Commented Jul 18, 2018 at 7:47
  • $\begingroup$ Taras, I am sorry for my lack of concentration. It was a wishful reading on my part, I guess. $\endgroup$
    – Wlod AA
    Commented Jul 18, 2018 at 18:35

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