It is similar to "trees" of sets homeomorphic to star-shaped sets tangent to each other by a point (the edges correspond to tangency). Is that all, or are there contractible sets that don't look like this? If so, can we describe them all (up to homeomorphism)?
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asked Oct 14, 2022 at 18:34
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6$\begingroup$ I'm afraid no. There are contractibles that behave wildly: the Whitehead manifold is an open 3-manifold that is contractible, but not homeomorphic to $\Bbb{R}^3$. It is obviously smooth so can be realized as a smooth subset of some $\Bbb{R}^n$ and probably not a "tree" of starred sets - it is indeed a union of two $\Bbb{R}^3$ whose intersection is again $\Bbb{R}^3$, but they are twisted. $\endgroup$– ZeroxCommented Oct 14, 2022 at 18:56
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9$\begingroup$ @Zerox In fact, by construction the Whitehead manifold embeds as an open in $\mathbb R^3$. This suggests that even classifying open contractible subsets is going to be difficult in general $\mathbb R^n$ (though note that in $\mathbb R^2$ they are all homeomorphic to a disk, for instance by Riemann mapping theorem). $\endgroup$– WojowuCommented Oct 14, 2022 at 19:08
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1 Answer
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(Please, do not add any commas to mathematical formulas below).
Consider ALL compact subsets $\ X\subseteq\mathbb R^2\times\{1\}\ $ such that $\ |X|> 1.\ $ Consider the following topological cones:
$$ \mathcal C(X)\ :=\ \{\, (t\!-\!1)\cdot x+(0\,\ 0\,\ t): \ x\in X\ \ \text{and}\ \ 0\le t\le1\,\} $$
It's a simple exercise proving that any two of the above plain compacts $\ X\ $ and $\ Y\ $ are homeomorphic $\ \Leftarrow\Rightarrow\ $ cones $\ \mathcal C(X)\ $ and $\ \mathcal C(X)\ $ are homeomorphic.
No reasonable homeomorphic classification is known even in the limited case of plain compact spaces hence there is no known homeomorphic classification of contractible subspaces of $\ \mathbb R^3$.
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$\begingroup$ I've assumed $\ |X|>1\ $ in order to make life easier (smoother) for an eventual solver of the exercise mentioned in my answer above. $\endgroup$– Wlod AACommented Oct 14, 2022 at 20:28
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$\begingroup$ I didn't ask for a topological classification, I'm talking about a characterization (a more explicit equivalent condition). I gave an example of what the answer might look like. $\endgroup$ Commented Oct 15, 2022 at 14:30