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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.

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

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  • $\begingroup$ 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/… $\endgroup$ – mm-crj Dec 21 '18 at 15:00
  • $\begingroup$ It can be closed when $S_n\subset B_r(p)$ $\endgroup$ – mm-crj Dec 21 '18 at 15:08
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    $\begingroup$ If the open ball $B_r(p)$ contains $S_n$, then $F$ is just $\mathbb{R}^{n+1}$ minus $0$. $\endgroup$ – W4cc0 Dec 21 '18 at 15:11
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    $\begingroup$ 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. $\endgroup$ – freakish Dec 21 '18 at 15:12
  • $\begingroup$ @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. $\endgroup$ – W4cc0 Dec 21 '18 at 15:17
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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 $R^{-1}(B_r(p)) = \{ te \mid t > 0, e \in E \}$.

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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 $V=F$. $\Box$

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