1. **Problem Statement:** Prove that $\mathbb{R}^n \setminus \{0\}$ is simply connected for $n \geq 3$. This means every loop in $\mathbb{R}^n \setminus \{0\}$ can be continuously contracted to a point. 2. **Key Idea:** Use the hint that any loop is homotopic to a piecewise linear loop, and then show this piecewise linear loop can be contracted to a constant loop. 3. **Step 1: Homotopy to Piecewise Linear Loop** - Any continuous loop $\gamma : S^1 \to \mathbb{R}^n \setminus \{0\}$ can be approximated arbitrarily closely by a piecewise linear loop $\gamma_{PL}$. - This is because $\mathbb{R}^n \setminus \{0\}$ is an open subset of $\mathbb{R}^n$, and simplicial approximation applies. - Hence, there exists a homotopy $H_1 : S^1 \times [0,1] \to \mathbb{R}^n \setminus \{0\}$ with $H_1(\cdot,0) = \gamma$ and $H_1(\cdot,1) = \gamma_{PL}$. 4. **Step 2: Contracting the Piecewise Linear Loop** - The piecewise linear loop $\gamma_{PL}$ is a finite union of line segments in $\mathbb{R}^n \setminus \{0\}$. - Since $n \geq 3$, the space $\mathbb{R}^n \setminus \{0\}$ is path-connected and has trivial fundamental group. - We can contract each line segment to a point by a linear homotopy avoiding the origin. 5. **Step 3: Constructing the Contraction Homotopy** - For each vertex $v_i$ of the piecewise linear loop, define a homotopy $H_2$ that linearly contracts the loop towards a fixed base point $x_0 \neq 0$. - Explicitly, for $t \in [0,1]$, define: $$ H_2(s,t) = (1-t) \gamma_{PL}(s) + t x_0 $$ - Since $x_0 \neq 0$ and $\gamma_{PL}(s) \neq 0$ for all $s$, the entire homotopy avoids the origin. 6. **Step 4: Combine Homotopies** - The concatenation of $H_1$ and $H_2$ gives a homotopy from the original loop $\gamma$ to the constant loop at $x_0$. 7. **Conclusion:** - Every loop in $\mathbb{R}^n \setminus \{0\}$ for $n \geq 3$ is homotopic to a constant loop. - Therefore, $\pi_1(\mathbb{R}^n \setminus \{0\})$ is trivial, and the space is simply connected. **Final answer:** $\mathbb{R}^n \setminus \{0\}$ is simply connected for $n \geq 3$.