# Is $\mathbb{R}^3 \setminus \mathbb{Q}^3$ simply connected?

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.

• There's a direct argument for this using the transversality-extension theorem. Commented Aug 30, 2015 at 6:22
• 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. Commented Aug 30, 2015 at 16:58
• @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. Commented Aug 30, 2015 at 21:14
• 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. Commented Aug 30, 2015 at 22:02
• 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. Commented Aug 31, 2015 at 10:19

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 point $$x_n$$ is open and dense. So the set of homotopies that miss all of $$X$$ is nonempty.

• 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.) Commented Aug 29, 2015 at 20:05
• 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 $\Bbb R$. Commented Aug 29, 2015 at 21:22
• @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 $\mathbb{R}^2$. Commented Aug 29, 2015 at 21:47
• @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. Commented Aug 30, 2015 at 0:28
• Great solution! I believe this also immediately generalizes to show $\pi_{n-2}(\mathbb{R}^n \setminus \mathbb{Q}^n) = 0$. Commented Aug 30, 2015 at 6:10

$$\newcommand\R{\mathbb{R}} \newcommand\cH{\mathcal{H}} \newcommand\bbD{\mathbb{D}}$$ I'm marking this community wiki since it's not so much an answer but an attempt to make it clear to the casual reader that despite Nick Mendler's answer it really is obvious that for a finite set $$S \subset \R^3$$, a continuous curve $$\gamma\colon S^1 \rightarrow \R^3 \setminus S$$, and a point $$y \in \R^3 \setminus S$$ the set of homotopies from $$\gamma$$ to the constant map $$y$$ that avoid $$S$$ is open and dense in the set of all homotopies. Here "obvious" means "immediate from standard tools/arguments that have little to do with $$\R^3$$". All I will use about $$\R^3$$ is that it is a smooth 1-connected manifold of dimension at least $$3$$.

The most convenient way to think of such homotopies is as the set $$\cH$$ of all continuous maps $$f\colon \bbD^2 \rightarrow \R^3$$ with $$f|_{S^1} = \gamma$$ and $$f(0) = y$$. Let $$\cH_S$$ be the subspace of $$\cH$$ consisting of such $$f$$ whose image is disjoint from $$S$$. Give $$\cH$$ the compact-open topology. We want to prove that $$\cH_S$$ is open and dense in $$\cH$$. That it is open is genuinely obvious, so the key thing to prove is that it is dense.

Since $$\mathbb{R}^3$$ is $$1$$-connected, the set $$\cH$$ is nonempty. Consider some $$f_0 \in \cH$$. Our goal is to perturb $$f_0$$ slightly (I'm not going to bother keeping track of $$\epsilon$$'s) to move it into $$\cH_S$$. Choose three nested proper open annuli $$A_1 \subset A_2 \subset A_3$$ in $$\bbD^2$$ with $$\overline{A}_1 \subset A_2$$ and $$\overline{A}_2 \subset A_3$$ such that $$f_0(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$. Here "proper open annuli" means that $$0 \notin A_1$$ and $$S^1 \cap \overline{A}_3 = \emptyset$$. That we can do this is clear: just make $$A_1$$ nearly all of $$\bbD^2$$.

The first standard fact we appeal to is that we can perturb $$f_0$$ to some $$f_1 \in \cH$$ with the following properties:

1. $$f_1$$ is smooth on $$\overline{A}_2$$; and
2. $$f_1 = f_0$$ on $$\bbD^2 \setminus A_3$$, which implies that $$f_1 \in \cH$$.

By making this perturbation small enough, we can assume that it is still the case that $$f_1(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$.

The second standard fact we appeal to is that we can perturb $$f_1$$ to some $$f_2 \in \cH$$ with the following properties:

1. $$f_2$$ is smooth on $$\overline{A}_2$$ and is transverse to $$S$$ on $$A_1$$; and
2. $$f_2 = f_1$$ on $$\bbD^2 \setminus A_2$$, which implies that $$f_2 \in \cH$$.

Again, by making this perturbation small enough, we can assume that it is still the case that $$f_2(\bbD^2 \setminus A_1)$$ is disjoint from $$S$$. Since on the $$2$$-manifold $$A_1$$ the map $$f$$ is transverse to the $$0$$-submanifold $$S$$ of the $$3$$-manifold $$\R^3$$, we actually have that $$f_2(A_1)$$ is disjoint from $$S$$. We conclude that $$f \in \cH_S$$, as desired.

• Yes, such details are best to leave as an exercise to a reader, they are at a homework level in an algebraic topology class. Commented Jun 3 at 16:57
• @MoisheKohan: I agree. Commented Jun 3 at 16:59
• After all, if $h^s(t)$ is a free homotopy in $\mathbb R^3$ between loops $h^0(t)$ and $h^1(t)$ in $\mathbb R^3\setminus \{P\}$, and its image has no interior (e.g. it is Lipschitz) then a small translate $h^s(t)+v$ avoids $P$, and $h^0$ and $h^1$ are homotopic in $\mathbb R^3\setminus \{P\}$ resp. to $h^0+v$ and $h^1+v$ , therefore $h^0$ and $h^1$ are homotopic in $\mathbb R^3\setminus \{P\}$ too. Commented Jun 4 at 19:09

It's a good idea to use the Baire Category Theorem, but there is still work to be done to verify that the sets are open and dense.

Here's a summary of the proof for fast readers (for more clarification, I provide all the gory details afterward):

The set of homotopies avoiding $$x_n$$ is clearly open. Let $$h:\mathbb{D}\rightarrow \mathbb{R}^d$$ be a homotopy from $$\gamma$$ to $$y$$ and suppose $$x_n\not\in \gamma(S^1).$$ Approximate $$h$$ by a piecewise-linear (PWL) map $$h^*,$$ and let $$g^t$$ be the straight-line homotopy from $$h$$ to $$h^*.$$ Let $$\rho:\mathbb{D}\rightarrow[0,1]$$ be a continuous map with $$\rho|_{\partial\mathbb{D}}=0$$ and $$\rho|_U=1$$ where $$U\subset \textrm{int}(\mathbb{D})$$ is an open neighborhood of $$h^{-1}(x_n).$$ Defining $$H = g^{\rho(-)}(-),$$ we then have that $$H$$ is close to $$h,$$ $$H|_{\partial \mathbb{D}}=\gamma,$$ and $$H|_U$$ is PWL. Moreover, we can take $$U$$ sufficiently large so that $$H(\mathbb{D}\setminus U)$$ is sufficiently close to $$\gamma(S^1),$$ so that $$x_n\not\in H(\mathbb{D}\setminus U)$$. Because $$H|_U$$ is PWL, $$H(U)$$ has no interior, so it follows that $$x_n$$ is not in the interior of $$H(\mathbb{D}).$$ Hence we can find a point $$y\not\in H(\mathbb{D})$$ that is close to $$x_n.$$ In particular $$B_\varepsilon(y) \cap H(\mathbb{D})=\emptyset.$$ Then, apply a perturbation $$f_y:\mathbb{R}^d\rightarrow \mathbb{R}^d,$$ that enlarges the neighborhood $$B_\varepsilon(y)$$ so that $$x_n\in f_y(B_\varepsilon(y)).$$ We can certainly choose $$f_y$$ to be small and equal to identity on $$\gamma(S^1).$$ Hence the perturbed homotopy $$f_y\circ H$$ is close to $$h,$$ avoids $$x_n,$$ and $$(f_y\circ H)|_{\partial \mathbb{D}} = \gamma$$ as desired.

Formal proof:

Let $$\{x_1,x_2,\dotsc\}\subset \mathbb{R}^d$$ be given, and let $$\gamma:S^1\rightarrow \mathbb{R}^d\setminus \{x_1,x_2,\dotsc\}$$ be a loop.

Denote $$\mathbb{D} = \{z\in\mathbb{C}\mid \lvert z\rvert\leq 1\}$$, and identify $$\partial \mathbb{D} = S^1$$, and let $$C(\mathbb{D},\mathbb{R}^d)$$ be the space of continuous functions from $$\mathbb{D}$$ to $$\mathbb{R}^d$$ equipped with the uniform norm. Denote the subset $$\mathcal{H} = \{h\in C(\mathbb{D},\mathbb{R}^d)\mid h\rvert_{\partial \mathbb{D}} = \gamma\}.$$ Thus, $$\mathcal{H}$$ can be viewed as the space of all homotopies of $$\gamma$$ to a point. The uniform limit theorem ensures that $$\mathcal{H}$$ is complete. In particular, $$\mathcal{H}$$ is a Baire-space.

For each $$x_n$$, define $$S_n = \{h\in \mathcal{H}\mid x_n\notin h(\mathbb{D})\}.$$

If $$h_m$$ is a sequence in $$\mathcal{H}\setminus S_n$$, then there is a sequence $$z_m\in \mathbb{D}$$ for which $$h_m(z_m)=x_n$$. And, if $$h_m\rightarrow h\in \mathcal{H}$$, then—because convergence is uniform—we have that for every $$\varepsilon$$ there is an $$m$$ large enough so that $$d(h(z_m),x_n) = d(h(z_m),h_{m}(z_{m}))<\varepsilon.$$ So for any convergent subsequence $$z_{m_j}\rightarrow z$$ we have that $$h(z) = h(\lim_{j\rightarrow\infty}z_{m_j})= \lim_{j\rightarrow\infty}h(z_{m_j}) = x_n.$$ In particular, $$h\notin S_n$$. Thus $$H\setminus S_n$$ is closed, and so $$S_n$$ is open.

To see that $$S_n$$ is dense in $$\mathcal{H}$$, let $$h\in \mathcal{H}\setminus S_n$$ and $$\varepsilon>0$$ be given, and suppose that $$\varepsilon$$ is sufficiently small so that the open ball $$B_{3\varepsilon}(x_n)$$ is disjoint from $$\gamma(S^1)$$ (this can be done because $$\gamma(S^1)$$ is closed and disjoint from every $$x_n$$).

For every $$m\in\mathbb{N}$$, let $$T_m$$ be a triangulation of $$\mathbb{D}$$ so that every triangle $$\tau\in T_m$$ has diameter $$\lvert\tau\rvert<2^{-m}$$. Let $$h_m$$ be the piecewise linear (PWL) approximation of $$h$$; that is, $$h_m$$ is linear on every triangle $$\tau\in T_m$$, and $$h_m$$ agrees with $$h$$ on every vertex of every triangle.

Then, because $$h$$ is uniformly continuous, and the vertices of the triangles in $$T_m$$ are $$2^{-m}$$ dense in $$\mathbb{D}$$, we can certainly take an $$m\in \mathbb{N}$$ sufficiently large so that

$$\lVert h_m - h\rVert_{\infty}\leq\varepsilon.$$

If $$h_m\in S_n$$, we are done. So suppose otherwise that $$x_n\in h_m(\mathbb{D})$$.

As $$h$$ is uniformly continuous, we can take $$0 sufficiently large so that $$h(\mathbb{D}\setminus B_R(0))\subset B_\varepsilon(\gamma(S^1)).$$ Also, let $$R$$ be large enough so that $$h^{-1}(x_n)\subset B_R(0).$$

Let $$g^t$$ be the straight-line homotopy from $$h$$ to $$h_m,$$ and let $$\rho:\mathbb{D}\rightarrow[0,1]$$ be a continuous map with $$\rho|_{\partial\mathbb{D}}=0$$ and $$\rho|_{B_R(0)}=1.$$ Defining $$H = g^{\rho(-)}(-),$$ we then have that

\begin{aligned}||H-h_m||_\infty &= ||(1-\rho)h + \rho h_m - h_m||_\infty \\ &\leq ||1-\rho||_\infty\cdot||h-h_m||_\infty \\&\leq \varepsilon. \end{aligned}

Also, for every $$z\in \mathbb{D}\setminus B_R(0),$$ we have \begin{aligned}\inf_{s\in[0,1]}|H(z)-\gamma(s)| &= |H(z) - h_m(z)+h_m(z)-h(z)+h(z)-\gamma(s)| \\ &\leq \inf_{s\in[0,1]}|h(z)-\gamma(s)| + 2\varepsilon \\ &< 3\varepsilon,\end{aligned} and because $$B_{3\varepsilon}(x_n)\cap \gamma(S^1)=\emptyset,$$ this shows that $$x_n\not \in H(\mathbb{D}\setminus B_R(0)).$$

Now, consider that $$H(B_R(0)) = \bigcup_{\tau\in T_m}h_m(\tau\cap B_R(0))$$, and each $$h_m(\tau\cap B_R(0))$$ has no interior in $$\mathbb{R}^d$$ (this is where we need $$d\geq 3$$). Also, $$T_m$$ contains finitely many triangles, and so $$H(B_R(0))$$ is a finite union of sets with no interior, and hence has no interior.

Because $$x_n\not\in H(\mathbb{D}\setminus B_R(0)),$$ and $$H(B_R(0))$$ has no interior, we can find a $$y\not\in H(\mathbb{D})$$ for which $$\lvert x_n-y\rvert<\varepsilon$$.

Next, define

$$f_{y}:\mathbb{R}^d\setminus\{y\}\rightarrow\mathbb{R}^d: f_{y}(v) = (\lvert v-y\rvert+\varepsilon^*)\frac{(v-y)}{\lvert v-y\rvert} + y,$$ where $$\varepsilon^* = \begin{cases}\varepsilon &: \lvert v-y\rvert<\varepsilon\\ 2\varepsilon - \lvert v-y\rvert &: \varepsilon\leq \lvert v-y\rvert<2\varepsilon \\ 0&:2\varepsilon\leq \lvert v-y\rvert.\end{cases}$$

Consider that $$f_y$$ is continuous on $$\mathbb{R}^d\setminus\{y\}$$. And since $$y\notin H(\mathbb{D})$$, it follows that $$f_y\circ H$$ is continuous. Also $$f_y$$ is the identity on $$\mathbb{R}^d\setminus B_{2\varepsilon}(x_n)$$; so the fact that $$B_{2\varepsilon}(x_n)$$ is disjoint from $$\gamma(S^1)$$ implies that $$(f_y\circ H)\rvert_{\partial \mathbb{D}} = f_y\circ \gamma=\gamma$$, and so we have that $$f_y\circ H \in \mathcal{H}$$.

Also note that $$\lvert f_{y}(v)-v\rvert \leq \varepsilon$$ for every $$v\in \mathbb{R}^d\setminus\{y\}$$, and so $$\lVert(f_y\circ H) - H\rVert_\infty \leq \varepsilon.$$ Thus the triangle inequality yields $$\lVert(f_y\circ H) - h\rVert_{\infty} \leq 3\varepsilon.$$

Finally, it is clear that the open ball $$B_\varepsilon(y)$$ is disjoint from the image of $$f_y$$, and so $$B_\varepsilon(y)$$ must be disjoint from $$(f_y\circ H)(\mathbb{D})$$. But $$x_n\in B_\varepsilon(y)$$, so we have that $$x_n\notin (f_y\circ H)(\mathbb{D})$$, which shows that $$(f_y\circ H)\in S_n$$.

Thus it has been shown that each $$S_n$$ is open and dense in $$\mathcal{H}.$$ And because $$\mathcal{H}$$ is a Baire-space, the BCT implies that $$\bigcap_{n\in\mathbb{N}} S_n\neq \emptyset$$. Any $$h\in \bigcap_{n\in\mathbb{N}} S_n$$ is a homotopy in $$\mathbb{R}^n\setminus \{x_1,x_2,\dotsc\}$$ from $$\gamma$$ to a point.

NOTES: as to @JoelDavidHamkins's comment, this argument clearly generalizes for $$\{x_1,x_2,\dotsc\}$$ replaced by any set of cardinality less than $$\operatorname{cov}(\mathcal{M})$$.

This argument does not explicitly use the fact that $$\mathbb{R}^d$$ minus a finite set is simply connected, but it does use the fact that, the linear image of a $$(d-1)$$-dimensional polygon has no interior in $$\mathbb{R}^d.$$

• Is $f_y \circ h_m \in H$? It seems to me that it does not agree with $h$ on the boundary. Commented Jun 5 at 10:24
• The problem occurs already for $h_m$, unless $\gamma$ is already piecewise linear. Commented Jun 5 at 18:37
• Your right, it had to be fixed. I fixed it by blending $h_m$ with $h.$ Commented Jun 5 at 21:27
• A flag has been raised with respect to the large number of edits. The trouble is that each edit bumps the post to the front page, at the expense of other posts also vying for attention, and this is annoying to the community. Please try to condense as many edits as you can into a single revision. Commented Jun 9 at 22:59
• Got it, thanks for letting me know Commented Jun 11 at 23:48

Since the conversation seems to focus on the density issue in Martin M. W.'s proof: yes, density is really as obvious as the openness. If $$h^s(t)$$ is a free homotopy in $$\mathbb R^3$$ between loops $$h^0(t)$$ and $$h^1(t)$$ in $$\mathbb R^3\setminus \{P\}$$ and its image has no interior (e.g. $$h$$ it is Lipschitz) then a small translate $$h^s(t)+v$$ avoids $$P$$, and $$h^0+v$$ and $$h^1+v$$ are homotopic in $$\mathbb R^3 \setminus\{P\}$$ resp. to $$h^0$$ and $$h^1$$, therefore $$h^0$$ and $$h^1$$ are homotopic in $$\mathbb R^3\setminus \{P\}$$ too.

• I agree that if you assume any kind of regularity on your homotopies it becomes utterly trivial, but for density you need to handle homotopies that are arbitrary. This means you need some initial step regularizing your homotopy. And if you don’t assume some regularity on your loops (their images might have nonempty interior), then you are quickly led to something like what I did. Commented Jun 5 at 12:51
• Yes, but regularization of h is clear by Weierstrass, isn't it? since $h$ is inside an open set $H$ of a Banach space there is a smooth $\tilde h\in B(h,\epsilon)\subset H$, so one can go from $h^0$ to $\tilde h^0$ by an affine homotopy etc. Commented Jun 5 at 14:59
• Weierstrass is one way, I suppose. There are many ways to do this, and the disadvantage of Weierstrass is that it's more tied to open subsets of Euclidean space rather than general smooth manifolds. Commented Jun 5 at 15:05
• I think this method by translation is certainly the simplest, and therefore best, way of doing the perturbation. But I still think using piecewise linear approximations is the most elementary way of ensuring no interior. Commented Jun 5 at 18:25
• over here @TarasBanakh makes a brilliant argument without using Baire Category theorem. Also, he shows that $R^d \setminus C$ is simply connected as long as $C$ has cardinality less than the continuum. If the continuum hypothesis fails, this is a stronger result than can be gotten with the Baire Category theorem Commented Jun 5 at 22:23