Use Proposition 1.26 to show that the complement of a closed discrete subspace of \(\mathbb{R}^n\) is simply-connected if \(n \geq 3\).
Solution
Let \(D\subset\mathbb R^n\) be closed and discrete, with \(n\ge3\), and put \[ Y=\mathbb R^n\setminus D. \]
1. Choose pairwise disjoint closed balls\[ B_d=\overline{B}(d,r_d),\qquad d\in D, \]whose family is locally finite.
Since \(D\) is discrete, for each \(d\in D\) there is \(\epsilon_d>0\) such that \[ B(d,\epsilon_d)\cap D=\{d\}. \] Choose \[ 0<r_d<\min\{1,\epsilon_d/3\}. \] If \(d\ne e\), then \(\epsilon_d\le\|d-e\|\) and \(\epsilon_e\le\|d-e\|\), so \[ r_d+r_e<\frac23\|d-e\|<\|d-e\|. \] Hence the closed balls are pairwise disjoint. Moreover, if a compact set \(K\) meets \(B_d\), then \(d\) lies in the closed \(1\)-neighborhood of \(K\), which is compact. A closed discrete subset of Euclidean space meets a compact set in only finitely many points, so only finitely many of the balls meet \(K\). Thus the family is locally finite.
2. The complement \(Y\) deformation retracts onto\[ Z=\mathbb R^n\setminus\bigcup_{d\in D}\operatorname{int}(B_d). \]
Inside each punctured ball \(B_d\setminus\{d\}\), radially push every point away from \(d\) onto the boundary sphere \(\partial B_d\), fixing the boundary. Outside the interiors of the balls, fix every point. The balls are pairwise disjoint and locally finite, so these radial homotopies patch to a continuous global deformation retraction \[ Y\simeq Z. \]
3. The space \(\mathbb R^n\) is obtained from \(Z\) by attaching one \(n\)-cell for each \(d\in D\) along the sphere \(\partial B_d\).
Each omitted open ball is restored by adjoining the closed ball \(B_d\cong D^n\) along its boundary \[ \partial B_d\cong S^{n-1}\subset Z. \] Since the balls are disjoint, doing this for all \(d\in D\) recovers all of \(\mathbb R^n\).
4. The inclusion\[ Z\hookrightarrow\mathbb R^n \]induces an isomorphism on fundamental groups.
By 3, \(\mathbb R^n\) is obtained from \(Z\) by attaching cells all of the same dimension \(n>2\). Proposition 1.26(b) says that attaching cells of dimension greater than \(2\) does not change the fundamental group. Hence \[ \pi_1(Z)\cong\pi_1(\mathbb R^n)=0. \]
5. Therefore \(Y\) is simply connected.
By 2, \[ \pi_1(Y)\cong\pi_1(Z)=0. \] Also \(Y\) is path connected: any two points can be joined by a polygonal path, and since \(D\) is discrete one can perturb finitely many segments slightly to avoid the finitely many points of \(D\) met by a compact polygonal path. Thus \(Y\) is path connected with trivial fundamental group, hence simply connected.