[proofplan] We argue by contradiction. If $h: \mathbb{R} \to \mathbb{R}^n$ were a homeomorphism, it would restrict to a homeomorphism $\mathbb{R} \setminus \{0\} \cong \mathbb{R}^n \setminus \{h(0)\}$. Homeomorphisms preserve connectedness of subspaces. We show $\mathbb{R} \setminus \{0\}$ is disconnected (it splits into two disjoint open intervals) while $\mathbb{R}^n \setminus \{p\}$ is path-connected for $n \geq 2$ (any two points can be joined by a path avoiding a single point). This contradicts the preservation of connectedness. [/proofplan] [step:Assume a homeomorphism exists and restrict to the complement of a point] Suppose for contradiction that $h: \mathbb{R} \to \mathbb{R}^n$ is a homeomorphism with $n \geq 2$. Let $p = h(0) \in \mathbb{R}^n$. Since $h$ is a bijection, the restriction \begin{align*} h|_{\mathbb{R} \setminus \{0\}}: \mathbb{R} \setminus \{0\} \to \mathbb{R}^n \setminus \{p\} \end{align*} is also a homeomorphism (a homeomorphism restricts to a homeomorphism between corresponding subspaces). [/step] [step:Show $\mathbb{R} \setminus \{0\}$ is disconnected] The space $\mathbb{R} \setminus \{0\} = (-\infty, 0) \cup (0, \infty)$ is the union of two disjoint, non-empty [open sets](/page/Open%20Set) in $\mathbb{R} \setminus \{0\}$. This is a separation, so $\mathbb{R} \setminus \{0\}$ is disconnected. [/step] [step:Show $\mathbb{R}^n \setminus \{p\}$ is path-connected for $n \geq 2$] Let $x, y \in \mathbb{R}^n \setminus \{p\}$. We construct a continuous path from $x$ to $y$ in $\mathbb{R}^n \setminus \{p\}$. **Case 1:** The line segment $\ell(t) = (1-t)x + ty$, $t \in [0,1]$, does not pass through $p$. Then $\ell$ is a path from $x$ to $y$ in $\mathbb{R}^n \setminus \{p\}$. **Case 2:** The line segment from $x$ to $y$ passes through $p$. Since $n \geq 2$, the affine line $L = \{(1-t)x + ty : t \in \mathbb{R}\}$ through $x$ and $y$ has dimension $1$, which is strictly less than $n$. Therefore $\mathbb{R}^n \setminus L \neq \varnothing$: choose $z \in \mathbb{R}^n \setminus L$. Since $z \notin L$ and $p \in L$, we have $z \neq p$. The line segment from $x$ to $z$ does not pass through $p$: if it did, then $p = (1-s)x + sz$ for some $s \in [0,1]$, but $p$ also lies on $L$, and since $z \notin L$, the point $(1-s)x + sz$ lies on $L$ only if $s = 0$ (giving $p = x$, contradicting $x \neq p$). By the same reasoning, the line segment from $z$ to $y$ does not pass through $p$. The concatenation of these two segments is a path from $x$ to $y$ in $\mathbb{R}^n \setminus \{p\}$. Therefore $\mathbb{R}^n \setminus \{p\}$ is path-connected, and by [connected open subsets of Euclidean space are path-connected](/theorems/301) (or directly: path-connected implies connected), $\mathbb{R}^n \setminus \{p\}$ is connected. [guided] The geometric idea is simple: in $\mathbb{R}^n$ with $n \geq 2$, a single point cannot block all paths between two other points, because there is enough room to "go around." Formally, we consider two cases. **Case 1 (direct segment avoids $p$):** If the line segment $\ell(t) = (1-t)x + ty$ does not pass through $p$, then $\ell$ is already a continuous path from $x$ to $y$ lying in $\mathbb{R}^n \setminus \{p\}$. **Case 2 (direct segment hits $p$):** If it passes through $p$, we need a detour. Since $n \geq 2$, the affine line $L = \{(1-t)x + ty : t \in \mathbb{R}\}$ is a $1$-dimensional subspace of $\mathbb{R}^n$, and $\mathbb{R}^n \setminus L$ is non-empty (a line in $\mathbb{R}^n$ with $n \geq 2$ does not exhaust the space). Pick any $z \in \mathbb{R}^n \setminus L$. We claim the segments $x \to z$ and $z \to y$ both avoid $p$. For the segment from $x$ to $z$: suppose $p = (1-s)x + sz$ for some $s \in [0,1]$. Since $p \in L$ and $x \in L$ but $z \notin L$, the convex combination $(1-s)x + sz$ lies in $L$ only when $s = 0$ (because for $s > 0$, the point is displaced from $L$ in the direction of $z \notin L$). But $s = 0$ gives $p = x$, contradicting $x \neq p$. The same argument applies symmetrically to the segment from $z$ to $y$. Concatenating the two segments gives a continuous path from $x$ to $y$ in $\mathbb{R}^n \setminus \{p\}$. Since $x, y$ were arbitrary points in $\mathbb{R}^n \setminus \{p\}$, the space is path-connected, hence connected. [/guided] [/step] [step:Derive a contradiction from the topological invariance of connectedness] The restriction $h|_{\mathbb{R} \setminus \{0\}}$ is a homeomorphism between $\mathbb{R} \setminus \{0\}$ and $\mathbb{R}^n \setminus \{p\}$. Homeomorphisms preserve connectedness: if two spaces are homeomorphic, one is connected if and only if the other is. But $\mathbb{R} \setminus \{0\}$ is disconnected while $\mathbb{R}^n \setminus \{p\}$ is connected. This is a contradiction, so no such homeomorphism $h$ exists. [/step]