How to prove that a particular cone is open [closed]
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Let $S_n$ be the unit sphere in $mathbb{R}^{n+1}$, and let $pin S_n$. 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 $$F={te,:, 0neq tin mathbb{R},, ein E}.$$ 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.
real-analysis general-topology analysis
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
$endgroup$
closed as off-topic by José Carlos Santos, Dando18, user10354138, Cesareo, KReiser Dec 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.
|
show 1 more comment
$begingroup$
Let $S_n$ be the unit sphere in $mathbb{R}^{n+1}$, and let $pin S_n$. 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 $$F={te,:, 0neq tin mathbb{R},, ein E}.$$ 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.
real-analysis general-topology analysis
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
$endgroup$
closed as off-topic by José Carlos Santos, Dando18, user10354138, Cesareo, KReiser Dec 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.
$begingroup$
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
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It can be closed when $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
1
$begingroup$
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
1
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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
|
show 1 more comment
$begingroup$
Let $S_n$ be the unit sphere in $mathbb{R}^{n+1}$, and let $pin S_n$. 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 $$F={te,:, 0neq tin mathbb{R},, ein E}.$$ 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.
real-analysis general-topology analysis
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
$endgroup$
Let $S_n$ be the unit sphere in $mathbb{R}^{n+1}$, and let $pin S_n$. 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 $$F={te,:, 0neq tin mathbb{R},, ein E}.$$ 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.
real-analysis general-topology analysis
real-analysis general-topology analysis
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
edited Dec 21 '18 at 15:01
W4cc0
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
asked Dec 21 '18 at 14:51
W4cc0W4cc0
1,89621227
1,89621227
closed as off-topic by José Carlos Santos, Dando18, user10354138, Cesareo, KReiser Dec 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.
closed as off-topic by José Carlos Santos, Dando18, user10354138, Cesareo, KReiser Dec 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.
$begingroup$
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 $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
1
$begingroup$
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
1
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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
|
show 1 more comment
$begingroup$
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 $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
1
$begingroup$
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
1
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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
$begingroup$
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
$begingroup$
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 $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
$begingroup$
It can be closed when $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
1
1
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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
$begingroup$
@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
|
show 1 more comment
2 Answers
2
active
oldest
votes
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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 $R^{-1}(B_r(p)) = { te mid t > 0, e in E }$.
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
$endgroup$
add a comment |
$begingroup$
Recall that if $Asubseteqmathbb{R}^{n+1}$ and $tinmathbb{R}$ then $tA={ta | ain A}$.
Lemma 1. If $Csubseteq S^n$ is closed in $S^n$ then $V=bigcup_{tneq 0}tC$ is closed in $mathbb{R}^{n+1}backslash{0}$.
Proof. Assume that $(a_n)$ is a sequence in $V$ convergent to some $ainmathbb{R}^{n+1}backslash{0}$. Then $lVert a_nrVert$ converges to $lVert arVertneq 0$. Now note that $a_n/lVert a_nrVertin C$ and it is also convergent (as an image of $(a_n)$ via continuous function). Since $C$ is closed then it converges to some $cin C$. We know what that $c$ is: $c=a/lVert arVert$ again by continuity of $vmapsto v/lVert vrVert$. It follows that $a=lVert arVert c$ and so $ain V$. $Box$
Lemma 2. If $Usubseteq S^n$ is open then so is $V=bigcup_{tneq 0}tU$ in $mathbb{R}^{n+1}backslash{0}$.
Proof. By Lemma 1 $C'=bigcup_{tneq 0}t(S^nbackslash U)$ 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 $V=F$. $Box$
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $R^{-1}(B_r(p)) = { te mid t > 0, e in E }$.
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
$endgroup$
add a comment |
$begingroup$
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 $R^{-1}(B_r(p)) = { te mid t > 0, e in E }$.
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
$endgroup$
add a comment |
$begingroup$
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 $R^{-1}(B_r(p)) = { te mid t > 0, e in E }$.
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
$endgroup$
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 $R^{-1}(B_r(p)) = { te mid t > 0, e in E }$.
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
edited Dec 21 '18 at 16:36
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
answered Dec 21 '18 at 16:24
Paul FrostPaul Frost
11.3k3934
11.3k3934
add a comment |
add a comment |
$begingroup$
Recall that if $Asubseteqmathbb{R}^{n+1}$ and $tinmathbb{R}$ then $tA={ta | ain A}$.
Lemma 1. If $Csubseteq S^n$ is closed in $S^n$ then $V=bigcup_{tneq 0}tC$ is closed in $mathbb{R}^{n+1}backslash{0}$.
Proof. Assume that $(a_n)$ is a sequence in $V$ convergent to some $ainmathbb{R}^{n+1}backslash{0}$. Then $lVert a_nrVert$ converges to $lVert arVertneq 0$. Now note that $a_n/lVert a_nrVertin C$ and it is also convergent (as an image of $(a_n)$ via continuous function). Since $C$ is closed then it converges to some $cin C$. We know what that $c$ is: $c=a/lVert arVert$ again by continuity of $vmapsto v/lVert vrVert$. It follows that $a=lVert arVert c$ and so $ain V$. $Box$
Lemma 2. If $Usubseteq S^n$ is open then so is $V=bigcup_{tneq 0}tU$ in $mathbb{R}^{n+1}backslash{0}$.
Proof. By Lemma 1 $C'=bigcup_{tneq 0}t(S^nbackslash U)$ 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 $V=F$. $Box$
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
$endgroup$
add a comment |
$begingroup$
Recall that if $Asubseteqmathbb{R}^{n+1}$ and $tinmathbb{R}$ then $tA={ta | ain A}$.
Lemma 1. If $Csubseteq S^n$ is closed in $S^n$ then $V=bigcup_{tneq 0}tC$ is closed in $mathbb{R}^{n+1}backslash{0}$.
Proof. Assume that $(a_n)$ is a sequence in $V$ convergent to some $ainmathbb{R}^{n+1}backslash{0}$. Then $lVert a_nrVert$ converges to $lVert arVertneq 0$. Now note that $a_n/lVert a_nrVertin C$ and it is also convergent (as an image of $(a_n)$ via continuous function). Since $C$ is closed then it converges to some $cin C$. We know what that $c$ is: $c=a/lVert arVert$ again by continuity of $vmapsto v/lVert vrVert$. It follows that $a=lVert arVert c$ and so $ain V$. $Box$
Lemma 2. If $Usubseteq S^n$ is open then so is $V=bigcup_{tneq 0}tU$ in $mathbb{R}^{n+1}backslash{0}$.
Proof. By Lemma 1 $C'=bigcup_{tneq 0}t(S^nbackslash U)$ 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 $V=F$. $Box$
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
$endgroup$
add a comment |
$begingroup$
Recall that if $Asubseteqmathbb{R}^{n+1}$ and $tinmathbb{R}$ then $tA={ta | ain A}$.
Lemma 1. If $Csubseteq S^n$ is closed in $S^n$ then $V=bigcup_{tneq 0}tC$ is closed in $mathbb{R}^{n+1}backslash{0}$.
Proof. Assume that $(a_n)$ is a sequence in $V$ convergent to some $ainmathbb{R}^{n+1}backslash{0}$. Then $lVert a_nrVert$ converges to $lVert arVertneq 0$. Now note that $a_n/lVert a_nrVertin C$ and it is also convergent (as an image of $(a_n)$ via continuous function). Since $C$ is closed then it converges to some $cin C$. We know what that $c$ is: $c=a/lVert arVert$ again by continuity of $vmapsto v/lVert vrVert$. It follows that $a=lVert arVert c$ and so $ain V$. $Box$
Lemma 2. If $Usubseteq S^n$ is open then so is $V=bigcup_{tneq 0}tU$ in $mathbb{R}^{n+1}backslash{0}$.
Proof. By Lemma 1 $C'=bigcup_{tneq 0}t(S^nbackslash U)$ 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 $V=F$. $Box$
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
$endgroup$
Recall that if $Asubseteqmathbb{R}^{n+1}$ and $tinmathbb{R}$ then $tA={ta | ain A}$.
Lemma 1. If $Csubseteq S^n$ is closed in $S^n$ then $V=bigcup_{tneq 0}tC$ is closed in $mathbb{R}^{n+1}backslash{0}$.
Proof. Assume that $(a_n)$ is a sequence in $V$ convergent to some $ainmathbb{R}^{n+1}backslash{0}$. Then $lVert a_nrVert$ converges to $lVert arVertneq 0$. Now note that $a_n/lVert a_nrVertin C$ and it is also convergent (as an image of $(a_n)$ via continuous function). Since $C$ is closed then it converges to some $cin C$. We know what that $c$ is: $c=a/lVert arVert$ again by continuity of $vmapsto v/lVert vrVert$. It follows that $a=lVert arVert c$ and so $ain V$. $Box$
Lemma 2. If $Usubseteq S^n$ is open then so is $V=bigcup_{tneq 0}tU$ in $mathbb{R}^{n+1}backslash{0}$.
Proof. By Lemma 1 $C'=bigcup_{tneq 0}t(S^nbackslash U)$ 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 $V=F$. $Box$
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
edited Dec 21 '18 at 15:55
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
answered Dec 21 '18 at 15:44
freakishfreakish
12.6k1630
12.6k1630
add a comment |
add a comment |
$begingroup$
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 $S_nsubset B_r(p)$
$endgroup$
– mm-crj
Dec 21 '18 at 15:08
1
$begingroup$
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
1
$begingroup$
So basically you take the quotient map $pi:mathbb{R}^{n+1}backslash{0}tomathbb{P}mathbb{R}^{n+1}$ 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