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Are there topological spaces $X,Y$, each having more than $2$ points, satisfying the following two properties?

  • $X\not\cong Y$, and
  • there is a bijection $\varphi: X\to Y$ such that for all $x\in X$ the spaces $X\setminus \{x\}$ and $Y\setminus \{\varphi(x)\}$ are homeomorphic.
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    $\begingroup$ I remember seeing a talk on something related to this, and I think a countably branching infinite tree T and a union of countably many copies of T have this property (mapping vertices to vertices). However I couldn't reprduce the exact details. $\endgroup$ – Joshua Erde Aug 4 '15 at 9:36
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    $\begingroup$ Homeomorphic via $\varphi$? $\endgroup$ – Alex Degtyarev Aug 4 '15 at 10:04
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    $\begingroup$ @AlexDegtyarev No, not necessarily via $\varphi$, just homeomorphic via some homeomorphism. - JoshuaErde, this sounds promising! $\endgroup$ – Dominic van der Zypen Aug 4 '15 at 10:06
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If $X$ and $Y$ are Hausdorff then the one easily sees that the homeomorphism can not be induced by $\varphi$. But you did not request that.

Let $C_0$ denote a disjoint union of countably many copies of the Cantor set. It is easily seen that the Cantor set $C$ is homeomorphic to the one point compactification of $C_0$. Since $C$ is homogeneous, by removing any point from $C$ we get $C_0$. Then $C_0$ less any point is again homeomorphic to $C_0$. Thus take $X\cong C$, $Y\cong C_0$ and $\varphi$ to be any bijection between them.

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    $\begingroup$ You actually don't need $X$ and $Y$ to be Hausdorff to show that the homeomorphism cannot be induced by $\phi$. You only need them to have at least three points (otherwise the various topology on a two point spaces are all counterexample). $\endgroup$ – Simon Henry Aug 4 '15 at 10:29
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It is known that the Cantor set minus a point is (up to homeomorphisms) the unique zero-dimensional separable metric space without isolated points that is locally compact and not compact. In particular, the Cantor set minus a point is homeomorphic to the Cantor set minus two points (or minus any finite number of points).

Now you can take as $X$ the Cantor set and as $Y$ the Cantor set minus a point.

See also this math.stackexchange question.

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For a minimal example consider the following two 3-element posets equipped with the Alexandroff topology (i.e. open sets are down-sets):

(a)   (b)                      (C)
  \   /                        / \
   \ /                        /   \
   (c)                      (A)   (B)

The bijection is given by: $a\mapsto A$, $b\mapsto B$, $c\mapsto C$.

The reconstruction conjecture for ordered sets (see e.g. Jean-Xavier Rampon, What is reconstruction for ordered sets?) - together with equivalence of finite posets and finite $T_0$ topological spaces (via the specialization order/Alexandroff topology functors) - says the are no such examples among finite $T_0$ spaces with at least four elements. However, the conjecture is open.

(The related reconstruction conjecture for graphs is better known and also open.)

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