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Background:

By the Riemann mapping theorem, for any non-empty, simply connected open subset $U \subset \mathbb{C}$, $U \neq \mathbb{C}$ there exists a biholomorphic map (in particular a homeomorphism) $U \rightarrow \mathbb{D}$ where $\mathbb{D}$ is the unit disk. Since $\mathbb{D}$ is homeomorphic to $\mathbb{C}$, any two simply connected subsets of $\mathbb{C}$ are homeomorphic. Trivially the same result holds for $\mathbb{R}^2$. I'm wondering if a similar result can be found for subsets of $\mathbb{R}^3$. Specifically, I want to know which open subsets of $\mathbb{R}^3$ are homeomorphic to $\mathbb{R}^3$ (or the unit ball).

The answer to this question (and this one) seems to claim that an open subset of $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^n$ if and only if it is contractible and "simply connected at infinity" (please correct me on this if I'm mistaken). Thus as far as I can tell, the problem reduces to describing contractible open subsets of $\mathbb{R}^3$ which are simply connected at infinity. Wikipedia leads me to believe that here, contractible simply means "no holes". I have not been able to find a definition for "simply connected at infinity" that I could understand.

Question: My question is threefold:

  1. Is it true that a domain in $\mathbb{R}^3$ is homeomorphic to $\mathbb{R}^3$ iff it is contractible and simply connected at infinity? Is there a simpler sufficient and necessary condition that works in $\mathbb{R}^3$?
  2. Is there a simple sufficient condition that guarantees that a contractible, open subset of $\mathbb{R}^3$ is homeomorphic to $\mathbb{R}^3$? For example, by this question and answer, it seems that convexity is a sufficient condition. Is boundedness a sufficient condition?
  3. Can you give an example of a contractible open subset of $\mathbb{R}^3$ that is not homeomorphic to the ball? The simpler, the better.

Side note: It appears to me that the usual counterexample to the question "are contractible $n$-manifolds homeomorphic to the $n$-ball" is the Whitehead manifold. However, this appears to me to be a 3-manifold embedded in 4-dimensional Euclidean space and thus not a subspace of $\mathbb{R}^3$. Is this a meaningful distinction?

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Peter
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  • I do not understand your side note. The Whitehead manifold is constructed as an open subset of $S^3$; because it's a proper subset, and $S^3$ minus a point is homeomorphic to $\Bbb R^3$, there is a Whitehead manifold in $\Bbb R^3$. –  Mar 06 '16 at 17:41
  • @MikeMiller I didn't know that $S^3$ minus a point is homeomorphic to $\mathbb{R}^3$. If that is the case, then I guess this answers point 3 of my question (thought a simpler example would be nice). – Peter Mar 06 '16 at 17:46
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    If it helps, the fundamental group at infinity $\pi_1^\infty(X) = \varprojlim \pi_1(X\setminus K)$, where $K$ runs over compact subspaces of $X$; then $X$ is simply-connected at infinity if $\pi_1^\infty(X) = 1$. (The ends of a manifold are similarly defined to be the group $\varprojlim \pi_0(X\setminus K)$.) Boundedness is not sufficient, since $\mathbb{R}^n$ is homeomorphic to $(0, 1)^n$. – anomaly Mar 06 '16 at 18:12
  • @anomaly: What are $\pi_1$ and $\pi_0$ here? – Peter Mar 06 '16 at 18:16
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    @Pétur: The set of path components and the fundamental group. It will probably be difficult to appreciate a chunk of ideas in geometric topology/low-dimensional topology without having some background in algebraic topology, where you'll learn about these objects. –  Mar 06 '16 at 18:18

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Pre-answer. Contractible does not mean "no holes". "No holes" is unfortunately completely meaningless. Contractible means that the identity map is null-homotopic (that is, there's a map $f: X \times I \to X$ such that $f(x,0) = x$ and $f(x,1) = c$ for some point $c \in X$.) If you want an intuitive statement of what this means, it's that you can slowly collapse the whole space at once down to a point.

1) Yes. I do not expect there to be a condition that's nicer/reasonably checkable. This condition is ultimately equivalent to saying that the 1-point compactification is still a manifold, which maybe sounds nicer, but is completely impossible to check.

2) Boundedness is not a sufficient condition - it's not a topological property in any way. Consider the map $\Bbb R^3 \to \Bbb R^3$ given by $x \mapsto x/(1+\|x\|)$. This is a homeomorphism onto its (bounded) image (it's contained in the open 3-ball). Restricting it to the Whitehead manifold, we have an open subset of the 3-ball that's homeomorphic to the Whitehead manifold. A mild generalization of convexity that still guarantees you're homeomorphic to $\Bbb R^3$ is being star-shaped. See here.

3) It's a bit much to expect a simple example, since it evaded Whitehead (indeed, he thought that contractible open 3-manifolds were all $\Bbb R^3$ until he later found his counterexample). Whitehead's example is as simple as you're going to get.

By variants on Whitehead's construction, by the way, you get uncountably many pairwise non-homeomorphic contractible open 3-manifolds. See this paper by McMillan.

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  • Thanks for clearing up my misconceptions. I guess what is a simple question in $R^2$ is already very complex in $R^3$. I will accept this answer, but any other answers (other than convexity) to point 2 would be welcome. – Peter Mar 06 '16 at 18:10
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    «It evaded Whitehead» is a great unit of measure for difficulty! – Mariano Suárez-Álvarez Mar 06 '16 at 18:11
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    @Pétur I added a mild generalization of convexity which is also sufficient. –  Mar 06 '16 at 18:12