Many admire the Euclidean space, and I am not an exception. I will try to catch the topological roundness of the $n$-ball in its greatest generality. I call the resulting axiomatized space to be a razor. If my definition turns out too general then (after getting interesting counter-examples) it will be a simple matter to narrow these definitions down (e.g. by finally agreeing--if necessary--to explicit homotopic constructions) till we are left with the most possible general characterizations (as opposed to simply general). Optimistically I hope for the minimal properties though (while among others the exotic cubes have to be considered).

THE GOAL (conjecture)   is to show that for every razor $\ (X\ T)\ $ and $S\subseteq X$ (see below) there exists a non-negative integer $\ n\ $ and subspace $\ Y\subseteq X\ $ such that $\ X\setminus S\subseteq Y\ $ and pairs $\ (Y\ \,Y\!\cap\! S)\ $ and $\ (\mathbb I^n\ \,\partial(\mathbb I^n))\ $ are homeomorphic.

All topological spaces are assumed here to be $T_1$-spaces (each single-point set should be closed). Also by definition:

$$X^{(2)}\ :=\ X\times X\setminus \Delta_X$$

where $\ \Delta_X\ :=\ \{(x\ x) : x\in X\}$,   for any set $\ X$.

At this time I am not making any assumptions about compactness, separability, finite dimension ... or even about the involved subsets being closed. But if you have to do them then do them, of course.


A razor is a topological space $\ X\ $ for which there exists a razor structure. And a razor structure in $\ X\ $ is defined as a triple $\ (X\ S\ f)\ $ where $\ S\ $ is a subset of $\ X\ $, function $\ f: X^{(2)}\rightarrow\ S\ $ is continuous, f.p.p. stands for the fixed point property, and the following conditions hold:

  1. $\ X\ $ has the f.p.p.
  2. $\ S\ $ does not have the f.p.p.
  3. $\ \forall_{x\in X}\forall_{y\in S\setminus\{x\}}\ \ f(x\ y)\ =\ y$

REMARK   Axiomat 1 is equivalent to property: $\ \ S\ $ is not a retract of $\ X\ \ $ (it's a standard observation in a situation like this).

  • $\begingroup$ Actually, it is about roundness+smallness. $\endgroup$ – Włodzimierz Holsztyński Sep 30 '13 at 6:37
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    $\begingroup$ Sorry, what's "the f.p.p."? $\endgroup$ – Todd Trimble Sep 30 '13 at 18:13
  • $\begingroup$ @Todd, sorry, it is "the fixed point property"--a standard abbreviation among the fpp folks :-) $\endgroup$ – Włodzimierz Holsztyński Sep 30 '13 at 18:22
  • $\begingroup$ My question got classified as "Community". May I know why? $\endgroup$ – Włodzimierz Holsztyński May 21 '14 at 5:33
  • $\begingroup$ @WlodzimierzHolsztynski Unanswered questions periodically get "bumped" to the front page by the "Community" user. See the description mathoverflow.net/users/-1/community $\endgroup$ – j.c. May 21 '14 at 6:50

A way to attack the above problem may be via non-$T_1$ spaces which otherwise satisfy the above three axioms--let's call a topological space like this, together with an appropriate function, to be a pre-razor. (In general, the finite topological spaces should play the role similar to the combinatorial simplicial complexes). Let me present the simplest pre-razor. It's defined as a 3-point space $\ (X\ T),\ $ where $\ X:= \{a\ b\ p\},\ $ such that $$ T := \{\emptyset\}\cup \{G\subseteq X: p\in G\}$$ Also $$ S:=\{a\ b\}$$

Thus $\ S\ $ is discrete, and it follows that $\ S\ $ does not have the fixed point property.

On the other hand $\ X\ $ is connected. Thus there does not exist any continuous $\ \phi:X\rightarrow X\ $ such that $\ \phi(X)=S.\ $ Thus every continuous $\ \phi:X\rightarrow X\ $ is either constant, and then it has a fixed point, or $\ p\in \phi(X).\ $ Then $\ \phi^{-1}(\{p\})\ $ is open, hence $\ \phi(p)=p.\ $ This proves that $\ X\ $ has the fixed point property.

Finally, let's define $\ f : X^{(2)}\rightarrow S\ $ as follows:

  • $f^{-1}(a) := \{ (b\ p)\ \,(b\ a)\ \,(p\ a)\}$
  • $f^{-1}(b) := \{ (a\ p)\ \,(a\ b)\ \,(p\ b)\}$

These sets are both open in $\ X^{(2)},\ $ say:

  • $f^{-1}(a) := \{b\ p\}\!\times\!\{a\ p\}\ \cap\ X^{(2)} $
  • $f^{-1}(b) := \{a\ p\}\!\times\!\{b\ p\}\ \cap\ X^{(2)} $

This shows that $\ f : X^{(2)}\rightarrow S\ $ is continuous, and it obviously satisfies the last axiom. Thus the system consisting of the topological space $\ (X\ T)\ $ and function $\ f:X^{(2)}\rightarrow S\ $ is a pre-razor.


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