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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 fpp for $I^n$ for every $n$ (if this fpp holds for every natural $n$, then it's easy to show that for arbitrary cardinal $n$ too).

TERMINOLOGY:   fpp = Fixed Point Property.

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.

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 fpp for $I^n$ for every $n$ (if this fpp holds for every natural $n$, then it's easy to show that for arbitrary cardinal $n$ too).

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 fpp for $I^n$ for every $n$ (if this fpp holds for every natural $n$, then it's easy to show that for arbitrary cardinal $n$ too).

TERMINOLOGY:   fpp = Fixed Point Property.