TITLE: Why is the complement of a discrete subspace of $\mathbb{R}^n$ ($n \ge 3$) simply-connected?
QUESTION [13 upvotes]: 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.

REPLY [4 votes]: 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.