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

Question

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.

Answer 1

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.

Answer 2

$\newcommand\R{\mathbb{R}} \newcommand\cH{\mathcal{H}} \newcommand\bbD{\mathbb{D}}$ 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.

Answer 3

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 $0<R<1$ 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 $\mathbb{R}^d.$

Answer 4

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.