# How to prove that a particular cone is open [closed]

Let $$S_n$$ be the unit sphere in $$\mathbb{R}^{n+1}$$, and let $$p\in S_n$$. Moreover, call $$E=B_r(p)\cap S_n$$, with $$r$$ positive, and $$B_r(p)$$ the open ball or radius $$r$$ and center $$p$$. Let $$F=\{te\,:\, 0\neq t\in \mathbb{R},\, e\in E\}.$$ How do I formally prove that $$F$$ is open? I need this to have a formal prove that the projective space is homeomorphic to a quotient of the sphere, but I cannot see how to prove it formally.

## closed as off-topic by José Carlos Santos, Dando18, user10354138, Cesareo, KReiserDec 22 '18 at 1:37

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – José Carlos Santos, Dando18, user10354138, Cesareo, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.

• I am trying to look at it from a much simpler point of view. $B_r(p)$ is open and may or may not contain $S_n$ which is closed and cannot contain $B_r(p)$. Hence the $E$ must be closed no? math.stackexchange.com/questions/2243551/… – mm-crj Dec 21 '18 at 15:00
• It can be closed when $S_n\subset B_r(p)$ – mm-crj Dec 21 '18 at 15:08
• If the open ball $B_r(p)$ contains $S_n$, then $F$ is just $\mathbb{R}^{n+1}$ minus $0$. – W4cc0 Dec 21 '18 at 15:11
• So basically you take the quotient map $\pi:\mathbb{R}^{n+1}\backslash\{0\}\to\mathbb{P}\mathbb{R}^{n+1}$ with the real projective space on the right and you want to show that $\pi | {S^{n}}$ is open. Which follows if $\pi$ is because they are both onto. – freakish Dec 21 '18 at 15:12
• @freakish It depends on how you do define the real projective space. If you define it with a sphere, I would like to prove that $\pi$ is continuous. – W4cc0 Dec 21 '18 at 15:17

Define a retraction $$r : \mathbb{R}^{n+1} \setminus \{ 0 \} \to S^n, r(x) = x/\lVert x \rVert$$. This is a continuous function.

$$E$$ is open in $$S^n$$. Hence $$F_+ = r^{-1}(E) = \{ te \mid t > 0, e \in E \}$$, is open.

Similarly $$-r$$ is continuous and $$F_- = (-r)^{-1}(E) = \{ te \mid t < 0, e \in E \}$$ is open.

Now observe $$F = F_+ \cup F_-$$.

You can alternatively regard $$r$$ as a function $$R : \mathbb{R}^{n+1} \setminus \{ 0 \} \to \mathbb{R}^{n+1}$$. Then $$R^{-1}(B_r(p)) = \{ te \mid t > 0, e \in E \}$$.

Recall that if $$A\subseteq\mathbb{R}^{n+1}$$ and $$t\in\mathbb{R}$$ then $$tA=\{ta\ |\ a\in A\}$$.

Lemma 1. If $$C\subseteq S^n$$ is closed in $$S^n$$ then $$V=\bigcup_{t\neq 0}tC$$ is closed in $$\mathbb{R}^{n+1}\backslash\{0\}$$.

Proof. Assume that $$(a_n)$$ is a sequence in $$V$$ convergent to some $$a\in\mathbb{R}^{n+1}\backslash\{0\}$$. Then $$\lVert a_n\rVert$$ converges to $$\lVert a\rVert\neq 0$$. Now note that $$a_n/\lVert a_n\rVert\in C$$ and it is also convergent (as an image of $$(a_n)$$ via continuous function). Since $$C$$ is closed then it converges to some $$c\in C$$. We know what that $$c$$ is: $$c=a/\lVert a\rVert$$ again by continuity of $$v\mapsto v/\lVert v\rVert$$. It follows that $$a=\lVert a\rVert c$$ and so $$a\in V$$. $$\Box$$

Lemma 2. If $$U\subseteq S^n$$ is open then so is $$V=\bigcup_{t\neq 0}tU$$ in $$\mathbb{R}^{n+1}\backslash\{0\}$$.

Proof. By Lemma 1 $$C'=\bigcup_{t\neq 0}t(S^n\backslash U)$$ is closed in $$\mathbb{R}^{n+1}\backslash\{0\}$$. The statement follows because $$V$$ is the complement of $$C'$$ in $$\mathbb{R}^{n+1}\backslash\{0\}$$. $$\Box$$

Conclusion. Your $$F$$ set is open.

Proof. $$E$$ is an open subset of $$S^n$$ by definition. Apply Lemma 2 to $$V=F$$. $$\Box$$