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But don't assume knowledge of any topological dimension theory. Here is a specific approach (an open problem):

Does there exist a separable metric space $X$ such that the following two conditions hold:

  • $|X|\ge 2$;
  • for every closed   $A\subseteq X$   such that   $X\setminus A$   is disconnected, there exists   $Y\subseteq A$   homeomorphic to   $X$;

?

(I formulated this question around 1960, perhaps without ever publishing it; I didn't work on it--too bad, since I like this question :-) )

BTW, you're welcome to use any means (including the dimension theory) to construct $X$. If you did it without using dimension theory or fpp for $I^n$ then you'd obtain a new proof of   $\dim I^n\ge n$   and of fpp for $I^n$, for every $n$ (it's well known and easy that if this fpp holds for every natural $n$, then for arbitrary cardinal $n$ too).

TERMINOLOGY:   fpp = Fixed Point Property.

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    $\begingroup$ What's fpp? ${}$ $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 16, 2013 at 6:30
  • $\begingroup$ fpp = the Fixed Point Property (I'll add this explanation to the formulation of the problem). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Feb 16, 2013 at 7:26
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    $\begingroup$ I find this question very confusingly written. In the title you say "$\infty$-dimensional" but not in the text. If infinite dimension is a consequence of your conditions, please indicate that. The text is fragmented (what is the first paragraph for?). You could just say "fixed-point property" in the single place where you use "fpp", so you don't have to explain what the acronym means. Also, it might help if you explain why $I^\omega$ isn't a solution, even if so for a trivial reason. $\endgroup$
    – Andrej Bauer
    Commented Feb 16, 2013 at 7:47
  • $\begingroup$ Oops, I was reading the second condition wrong, thinking you want a copy of $X$ inside $X \setminus A$. $I^\omega$ obviously is not a solution. $\endgroup$
    – Andrej Bauer
    Commented Feb 16, 2013 at 7:49
  • $\begingroup$ @A.B.--my bad. Any space $X$, which satisfies the above 2 conditions (from the question) must be $\infty$-dimensional--think of the Ind definition of the topological dimension. (The bad luck wants that my first attempt at posting my question ended in a failure; I've encountered a software system difficulty. My first edition was quite extensive. It's hard for me to concentrate on the second edition, when I am a kind of sick of writing about the same again). $\endgroup$
    – Włodzimierz Holsztyński
    Commented Feb 16, 2013 at 9:00

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