Is the cone generated by an open subset of $\mathbb S^n$ open in $\mathbb R^{n+1}$? I am trying to prove that the real projective space $\mathbb R\mathbb P^n$ is Hausdorff. This answer almost does what I want, except I am starting from the definition of $\mathbb R\mathbb P^n$ based on the quotient map $\pi:\mathbb R^{n+1}\setminus\{0\}\to\mathbb R\mathbb P^n$ sending each point to its span. Using this definition, the following theorem is necessary to make the argument in the cited answer work.
Let $U$ be an open subset of $\mathbb S^n$ and let
$$C(U)=\big(\mathbb R^*\big)U=\big\{x\in\big(\mathbb R^{n+1}\big)^*\,\big|\,x=\lambda t\text{ for some }\lambda\in\mathbb R^*,t\in U\big\}$$
denote the cone generated by $U$, where $A^*:=A\setminus\{0\}$. Prove: $C(U)$ is open in $\mathbb R^{n+1}$.
This is not as trivial as it might seem, because an open ball centered somewhere in $U$ whose intersection with $\mathbb S^n$ is contained in $U$ still might generate a cone that intersects $\mathbb S^n$ outside $U$. In fact, I have already proven that the cone generated by an open ball in $\mathbb R^{n+1}$ is open, so I only need to show that each point in $U$ is contained in an open ball in $\mathbb R^{n+1}$ whose associated cone does not intersect $\mathbb S^n\setminus U$. But I am stuck here.
 A: Since $\mathbb R^{n+1}\setminus\{0\}$ is open in $\mathbb R^{n+1}$ and $C(U) \subset \mathbb R^{n+1}\setminus\{0\}$, it suffices to show that $C(U)$ is open in $\mathbb R^{n+1}\setminus\{0\}$. We need two ingredients.

*

*The antipodal map $a : S^n \to S^n, a(x) = -x$, is a homeomorphism.


*The map
$$\phi : (0,\infty) \times S^n  \to R^{n+1}\setminus\{0\}, \phi(\lambda,x) = \lambda x$$
is a homeomorphism (its inverse is $\phi^{-1}(y) = (\lVert y \rVert,\frac{y}{\lVert y \rVert} )$).
Now let us prove that $C(U)$ is open. Clearly $a(U)$ is open in $S^n$, thus $U' = U \cup a(U)$ is open in $S^n$. Hence $(0,\infty) \times U'$ is open in  $(0,\infty) \times S^n$ and $\phi((0,\infty) \times U')$ is open in $\mathbb R^{n+1}\setminus\{0\}$. But we have
$$C(U) = \mathbb R^* \cdot U = \{\lambda x \mid \lambda \in \mathbb R^*, x \in U \} \\ = \{\lambda x \mid \lambda \in (0,\infty), x \in U \} \cup \{\lambda x \mid \lambda \in (-\infty,0), x \in U \} \\= \{\lambda x \mid \lambda \in (0,\infty), x \in U \} \cup \{\lambda (-x) \mid \lambda \in (0,\infty), x \in U \} \\= \{\lambda x \mid \lambda \in (0,\infty), x \in U \} \cup \{\lambda x' \mid \lambda \in (0,\infty), -x' = a(x') \in U \} \\= \{\lambda x \mid \lambda \in (0,\infty), x \in U \} \cup \{\lambda x' \mid \lambda \in (0,\infty), x' \in a(U) \} \\= \phi((0,\infty) \times U) \cup \phi((0,\infty) \times a(U)) \\= \phi((0,\infty) \times U \cup (0,\infty) \times a(U)) = \phi((0,\infty) \times U') .$$
