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Similarly is the complement of any countable set in $\mathbb R^3$ simply connected?

Reading around I found plenty of articles discussing the path connectedness $\mathbb R^2 \setminus \mathbb Q^2$ and even an approach using cofiltered limits to approach that problem, but I am not read enough in that literature to see if this could be applied here.

Stefan Hamcke
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Nick R
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    There's a direct argument for this using the transversality-extension theorem. – Ryan Budney Aug 30 '15 at 06:22
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    While I harbor some doubts that one could do better than Martin's answer below, why not explain this alternative argument? It could be educational, even for professionals. – Todd Trimble Aug 30 '15 at 16:58
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    It's essentially the same transversality argument one uses to show $\pi_k S^n$ is trivial for $k < n$. Only you notice you can make the map simultaneously transverse to a countable collection of manifolds. – Ryan Budney Aug 30 '15 at 18:53
  • For the title question, maybe some examples will help: Consider the circle $x^2+y^2=\sqrt[3]{4}$ or the square $\max(|x|,|y|)=\sqrt{2}$ in $R^2-Q^2$. Every contraction of them to a point in $R^2$ will go through all of their interior points in $R^2$, and in particular will not avoid $Q^2$. But the corresponding curves on the $z=0$ plane of $R^3-Q^3$ are both homotopic to curves on the $z=\sqrt{2}$ plane of $R^3-Q^3$, where they can be contracted. –  Aug 30 '15 at 20:28
  • @ToddTrimble: This is more general than just a countable collection of points; there is a version of the transversality theorem for maps $g: M \to N$ transverse to $f(A) \subset N$ noncompact (and not necessarily closed): transverse maps are still dense, just not necessarily open, in the space of all maps - even when you specify $\partial g$ ahead of time. As usual, our manifolds here are second countable, or this is completely false. This is somewhere in Hirsch. – mme Aug 30 '15 at 21:03
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    @MikeMiller Why not write up the argument carefully as an answer before the question is closed? The comment as written is a little too telegraphic for me to follow easily, and I don't know which thing of Hirsch you mean. A careful, detailed, self-contained answer would be peachy. – Todd Trimble Aug 30 '15 at 21:14
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    Is it obvious that transversality arguments work for weird non-differentiable curves? In a manifold you know any curve is homotopic to a differentiable one, but showing that fact in $\mathbb{R}^3 \setminus \mathbb{Q}^3$ doesn't seem any easier than the original question. – Martin M. W. Aug 30 '15 at 22:02
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    Adams mentions in "Lectures on Lie Groups" that there is a good theory of transversality and homotopy formulated using Hausdorff dimension, but he does not give a reference and I have never seen one. Assuming that that is correct, it would immediately answer the question asked here. – Neil Strickland Aug 31 '15 at 10:19

1 Answers1

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Yes, the complement of any countable set in $\mathbb{R}^3$ is simply connected, by the Baire category theorem.

Say your set is $X = \{x_1, x_2, ... \}$, and let $y$ be any point in $\mathbb{R}^3 \setminus X.$

Let $f:S^1 \rightarrow \mathbb{R}^3 \setminus X$, and consider the space of homotopies $h:S^1 \times [0,1] \rightarrow \mathbb{R}^3$, where $h(x, 0) = f(x)$ and $h(x, 1) = y$. With the natural topology the space of homotopies is a Baire space, and for each $n$, the set of homotopies that avoid the points ${x_1, ... , x_n}$ is open and dense. So the set of homotopies that miss all of $X$ is nonempty.

Martin M. W.
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    Does this argument generalize beyond countable to show also that $\mathbb{R}^3\setminus X$ is simply connected whenever $X$ has size less than $\text{cov}(\cal M)$? (The cardinal characteristic $\text{cov}(\cal M)$ is the covering number of the meager ideal, or in other words, the smallest number of meager sets that union to the whole space; it is more interesting when the continuum hypothesis fails.) – Joel David Hamkins Aug 29 '15 at 20:05
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    I think that some essential details are lacking. Where does this argument use that the ambient space is $\Bbb R^3$? Because the statement would be false for $\Bbb R$. – Victor Protsak Aug 29 '15 at 21:22
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    @Victor The argument takes for granted that the ambient space stays simply connected (even connected) when you remove just a finite set of points--fine in $\mathbb{R}^3$ but not $\mathbb{R}$ or $\mathbb{R}^2$. – Martin M. W. Aug 29 '15 at 21:47
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    @JoelDavidHamkins I think so. It might be interesting to understand this number more fully. One should be able to replace the space of homotopies with a finite-dimensional space and understand exactly what sort of spaces are being removed to get a better lower bound. – Will Sawin Aug 30 '15 at 00:28
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    Great solution! I believe this also immediately generalizes to show $\pi_{n-2}(\mathbb{R}^n \setminus \mathbb{Q}^n) = 0$. – Nick R Aug 30 '15 at 06:10