Why is the complement of a discrete subspace of $\mathbb{R}^n$ ($n \ge 3$) simply-connected?

Question

I'm stuck with an Exercise in Hatcher's Algebraic Topology. (Exercise 1.2.6)

This problem asks me to show that the complement of a discrete subspace of $\mathbb{R}^n$ is simply-connected if $n\ge 3$, using the fact that if $Y$ is obtained by attaching $3$-cells to a path-connected space $X$, then the inclusion $X \hookrightarrow Y$ induces an isomorphism between fundamental groups. The following is a "proof" given by several people.

Let $A$ be a discrete subspace of $\mathbb{R}^n$. ($n \ge 3$) For each $a \in A$, take an open ball $B_a$ centered at $a$, so that $B_a$'s are all disjoint. Then $\mathbb{R}^n-A$ deformation retracts to $\mathbb{R}^n-\bigcup_{a\in A} B_a$. By attaching $n$-cells to $\mathbb{R}^n-\bigcup_{a\in A} B_a$, one for each $a \in A$, we can obtain $\mathbb{R}^n$. Therefore, we have $$\pi_1(\mathbb{R}^n-A) \approx \pi_1(\mathbb{R}^n-\bigcup_{a\in A} B_a) \approx \pi_1(\mathbb{R}^n) = 0.$$

However, this "proof" seems incorrect, since the quotient topology on $\mathbb{R}^n$ given by attaching $n$-cells to $\mathbb{R}^n-\bigcup_{a\in A} B_a$ may not agree with the usual topology on $\mathbb{R}^n$. To see this, consider $n = 3$ and $A = \{(1/n,0,0)\}_{n=1}^{\infty}$. Then $\mathbb{R}^3-A$ is open in the quotient topology, although it is not open in the usual topology. (Actually, I'm also not sure why $\mathbb{R}^n-A$ deformation retracts to $\mathbb{R}^n-\bigcup_{a\in A} B_a$)

Does anyone see why the complement of a discrete subspace of $\mathbb{R}^n$ is simply-connected if $n\ge 3$? I've been trying to construct an explicit homotopy, but it seems not so easy since the subset $A$ and the loop may look very wild. Thank you for your help.

Answer 1

Professor Allen Hatcher replied that he suspects that he was implicitly assuming that the subset is closed. So the exercise may be revised to include the hypothesis that the subset is closed. He also kindly sent me a sketch of an argument which applies even if the subset is not closed. Here is the sketch by Professor Hatcher:

Let $X$ be the discrete subset of $\mathbb{R}^n$ and let $L$ be the set of limit points of $X$. Then $L$ is closed, as is the union of $X$ and $L$. Let $\gamma$ be a loop in $\mathbb{R}^n - X$. This is nullhomotopic in $\mathbb{R}^n$, so there is a map $f : D^2 \to \mathbb{R}^n$ whose restriction to $S^1$ (the boundary of $D^2$) is the given loop $\gamma$. Let $V$ be the intersection of $f^{-1}(\mathbb{R}^n - L)$ with the interior of $D^2$, so $V$ is an open set in the interior of $D^2$. We can triangulate $V$ into simplices whose diameters approach $0$ on $S^1$ and on $f^{-1}(L)$.

The idea now is to perturb $f$ on each $2$-dimensional simplex $\sigma$ of the triangulation of $V$ so that the new $f$ has $f(\sigma)$ disjoint from $X$. Since $f(\sigma)$ is compact and disjoint from $L$, it has a positive distance $d(\sigma)$ from $L$. We perturb $f$ on $\sigma$ to a map which is linear on each simplex of some subdivision of $\sigma$ into smaller simplices, then we further perturb the new $f$ so that the image of each simplex in the subdivision of $\sigma$ is disjoint from $X$. This is possible since there are just finitely many points of $X$ within distance $d(\sigma)/2$ from $f(\sigma)$. The assumption $n \ge 3$ is used here. The perturbation of f on $\sigma$ can be chosen as small as we like, so we make it small compared to $d(\sigma)$ and to the diameter of $\sigma$.

This process is repeated inductively for each $2$-dimensional simplex $\sigma$ of $V$. For the induction step, we may have already perturbed $f$ on some vertices or edges of the boundary of $\sigma$ to have image disjoint from $X$, and in this case we do not need to deform $f$ again on this part of the boundary of $\sigma$.

Since the size of the perturbation of $f$ on $\sigma$ approaches zero as the size of $\sigma$ approaches zero, the limit of the infinite sequence of perturbations of $f$ exists and extends continuously over $D^2 - V$ to equal the original $f$ there. The new $f$ then gives a nullhomotopy of the loop $\gamma$ in $\mathbb{R}^n - X$.

Answer 2

Hatcher's solution is a little too advanced for that part of the book. I'm gonna post mine but it does use the fact that $A$ is closed.

Note that $A$ is either countable finite, the case finite it's even easier so let's focus on $A$ countable infinite. Let $\{a_n:n\in \mathbb{N}\}=A$ and for each $n$ let $R_n$ be a ball centered at $a_n$ so that all $R_n$'s are disjoint and let $B_n$ be a smaller ball inside $R_n$ such that $\overline{B_n}=D_n\subseteq R_n$ and $B_n$ has radius less than $1/n$. For the coproduct of the $D_ns$ and $X=\mathbb{R}^n-\cup_nB_n$, we need them to be disjoint but the boundaries of the $D_n$ are inside $X$ so I'm gonna be a little pedantic and disjoint them explicitly. Let $\alpha_n:D_n\to D_n'$ be homeomorphism such that $X$ and all the $D_n'$ and $X$ are disjoint. Let $\phi_n:\partial D_n'\to X,\phi_n(x)=\alpha^{-1}_n(x)$. be the attaching maps and $Y=X\coprod_{\phi_n}D_n'$ be the space obtained after attaching the $D_n'$s to $X$, so we have a quotient map $q:X\coprod_nD_n'\to Y$. Let $f:X\coprod_nD_n'\to \mathbb{R}^n$ defined as \begin{equation*} f(x)= \begin{cases} x & x\in X \\ \alpha^{-1}_n(x) & x\in D_n' \end{cases} \end{equation*} Note that $q(x)=q(y)\implies f(x)=f(y)$ so we may factor $f$ through $q$ obtaining a map $g:Y\to \mathbb{R}^n$ such that $f=gq$.

Sidenote: In the question the OP says that $g$ might not be an homeomorphism if $A$ is not closed and in fact it's not in the case he gives but there may be another weird homeomorphism between $Y$ and $\mathbb{R}^n$. I couldn't prove that the specific $Y$ given by the OP is not homeomorphic to $\mathbb{R}^n$. This doesn't affect this proof, however.

We now prove $\cup_nD_n$ is closed. Let $x\notin \cup_nD_n$ and $\epsilon=\inf_nd(x,D_n)$. We prove $\epsilon>0$. If $\epsilon=0$ then for each $\delta>0$ there is a $k\in \mathbb{N}$ such that $1/k<\delta/2$ and there is a $y\in D_k$ with $d(x,y)<\delta/2$. Since we also have $d(y,a_k)\le 1/k$ the triangle inequality implies that $d(x,a_k)<\delta$, since $\delta$ was arbitrary and $A$ is closed this implies $x\in A$, a contradiction, thus $\epsilon>0$. Now the ball of center $x$ and radius $\epsilon$ is disjoint from $\cup_nD_n$ and thus this set is closed.

Since each $D_n$ is inside an open set $R_n$ and all $R_n$ are disjoint one can prove that $\cup_nD_n$ has the coproduct topology $\coprod_nD_n$.

Let $h:\mathbb{R}^n\to Y$ defined as \begin{equation*} h(x)= \begin{cases} q(x) & x\in X \\ q(\alpha_n(x)) & x\in D_n \end{cases} \end{equation*} Since $\cup_nD_n$ and $X$ are both closed is enough to prove that $h|X$ and $X|\cup_nD_n$ are continous. But $h|X=q$ and since $\cup_nD_n$ has the coproduct topology $\coprod_nD_n$ is enough to prove that each $h|D_n$ is continuous but this is obvious since $h|D_n=q\alpha_n$.

One finally proves that $g$ and $h$ are inverses of each other and the proof is complete.

Sidenote 2: Note that we chose the balls $B_n$ as small as we wanted and that we also put the $D_n$s inside bigger disjoint balls $R_n$ to prove they have the coproduct topology . I'm not sure if one really has to do this, I think those hypothesis might be eliminated but the proof would be more difficult.