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1

$begingroup$

Let S be the sphere $(x,y,z) in mathbb R^3 mid x^2 + y^2 + z^2 = 1$.

First I want to check if the set $A = (cosphi sintheta, sinphi sintheta, costheta ) midphi in (0,pi), theta in (0,pi)$ is open in $mathbb R^3$.

I am pretty sure that it isn't open since it is half of the sphere S, and for the example for the point $(0,1,0) in A$ we can't find a ball in $A$ that surrounds it.

The second task is to find whether $A$ is relatively open in $S$.

Here I am not so sure but I think I can take the open set: $V = $ $B(0,2)$ with $y>0$ and then $S bigcap V = A$ which proves that it is relatively open in $S$.

However the set $B = (cosphi sintheta, sinphi sintheta, costheta ) midphi in [0,pi], theta in [0,pi]$ is not relatively open in $S$, right?

Am I correct here?

Help would be appreciated.

real-analysis calculus general-topology

share|cite|improve this question

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It seems OK.$$
  $endgroup$
  – Giuseppe Negro
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. But how can I formally prove that $B$ is not relatively open?
  $endgroup$
  – Gabi G
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You already did, I think it is enough.
  $endgroup$
  – Giuseppe Negro
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I proved that A is. In the comment I am talking about B
  $endgroup$
  – Gabi G
  2 days ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let S be the sphere $(x,y,z) in mathbb R^3 mid x^2 + y^2 + z^2 = 1$.

First I want to check if the set $A = (cosphi sintheta, sinphi sintheta, costheta ) midphi in (0,pi), theta in (0,pi)$ is open in $mathbb R^3$.

I am pretty sure that it isn't open since it is half of the sphere S, and for the example for the point $(0,1,0) in A$ we can't find a ball in $A$ that surrounds it.

The second task is to find whether $A$ is relatively open in $S$.

Here I am not so sure but I think I can take the open set: $V = $ $B(0,2)$ with $y>0$ and then $S bigcap V = A$ which proves that it is relatively open in $S$.

However the set $B = (cosphi sintheta, sinphi sintheta, costheta ) midphi in [0,pi], theta in [0,pi]$ is not relatively open in $S$, right?

Am I correct here?

Help would be appreciated.

real-analysis calculus general-topology

share|cite|improve this question

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It seems OK.$$
  $endgroup$
  – Giuseppe Negro
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. But how can I formally prove that $B$ is not relatively open?
  $endgroup$
  – Gabi G
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You already did, I think it is enough.
  $endgroup$
  – Giuseppe Negro
  2 days ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I proved that A is. In the comment I am talking about B
  $endgroup$
  – Gabi G
  2 days ago
  
  

  
  
  
  

add a comment | 

$begingroup$

Let S be the sphere $(x,y,z) in mathbb R^3 mid x^2 + y^2 + z^2 = 1$.

First I want to check if the set $A = (cosphi sintheta, sinphi sintheta, costheta ) midphi in (0,pi), theta in (0,pi)$ is open in $mathbb R^3$.

I am pretty sure that it isn't open since it is half of the sphere S, and for the example for the point $(0,1,0) in A$ we can't find a ball in $A$ that surrounds it.

The second task is to find whether $A$ is relatively open in $S$.

Here I am not so sure but I think I can take the open set: $V = $ $B(0,2)$ with $y>0$ and then $S bigcap V = A$ which proves that it is relatively open in $S$.

However the set $B = (cosphi sintheta, sinphi sintheta, costheta ) midphi in [0,pi], theta in [0,pi]$ is not relatively open in $S$, right?

Am I correct here?

Help would be appreciated.

real-analysis calculus general-topology

share|cite|improve this question

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

$endgroup$

share|cite|improve this question

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

share|cite|improve this question

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

edited 2 days ago

Jean Marie

30.6k42154

edited 2 days ago

Jean Marie

30.6k42154

asked 2 days ago

Gabi GGabi G

442110

asked 2 days ago

Gabi GGabi G

442110

1 Answer
1

active

oldest

votes

1

$begingroup$

Suppose $U$ is an open set in $mathbbR^3$ so that $Ucap S=B$. Then since $U$ is open, there is a ball $D$ around, say, the point $(0,0,1)$ with radius $epsilon$ that is contained inside $U$. But since $Ucap S=B$ and $Dsubset U$, it follows that $Dcap Ssubset B$. In other words, the small ball $D$ centered at the north pole intersects the sphere in some open set $Dcap S$ which will be contained in the closed half-sphere $B$. But this is impossible, you can find points in the intersection $Dcap S$ which have $theta's$ that fall outside of $0leqthetaleq pi$.

share|cite|improve this answer

answered 2 days ago

PrototankPrototank

1,229920

$endgroup$

add a comment | 

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1

$begingroup$

Suppose $U$ is an open set in $mathbbR^3$ so that $Ucap S=B$. Then since $U$ is open, there is a ball $D$ around, say, the point $(0,0,1)$ with radius $epsilon$ that is contained inside $U$. But since $Ucap S=B$ and $Dsubset U$, it follows that $Dcap Ssubset B$. In other words, the small ball $D$ centered at the north pole intersects the sphere in some open set $Dcap S$ which will be contained in the closed half-sphere $B$. But this is impossible, you can find points in the intersection $Dcap S$ which have $theta's$ that fall outside of $0leqthetaleq pi$.

share|cite|improve this answer

answered 2 days ago

PrototankPrototank

1,229920

$endgroup$

add a comment | 

1

$begingroup$

Suppose $U$ is an open set in $mathbbR^3$ so that $Ucap S=B$. Then since $U$ is open, there is a ball $D$ around, say, the point $(0,0,1)$ with radius $epsilon$ that is contained inside $U$. But since $Ucap S=B$ and $Dsubset U$, it follows that $Dcap Ssubset B$. In other words, the small ball $D$ centered at the north pole intersects the sphere in some open set $Dcap S$ which will be contained in the closed half-sphere $B$. But this is impossible, you can find points in the intersection $Dcap S$ which have $theta's$ that fall outside of $0leqthetaleq pi$.

share|cite|improve this answer

answered 2 days ago

PrototankPrototank

1,229920

$endgroup$

add a comment | 

$begingroup$

Suppose $U$ is an open set in $mathbbR^3$ so that $Ucap S=B$. Then since $U$ is open, there is a ball $D$ around, say, the point $(0,0,1)$ with radius $epsilon$ that is contained inside $U$. But since $Ucap S=B$ and $Dsubset U$, it follows that $Dcap Ssubset B$. In other words, the small ball $D$ centered at the north pole intersects the sphere in some open set $Dcap S$ which will be contained in the closed half-sphere $B$. But this is impossible, you can find points in the intersection $Dcap S$ which have $theta's$ that fall outside of $0leqthetaleq pi$.

share|cite|improve this answer

answered 2 days ago

PrototankPrototank

1,229920

$endgroup$

share|cite|improve this answer

answered 2 days ago

PrototankPrototank

1,229920

answered 2 days ago

PrototankPrototank

1,229920

answered 2 days ago

PrototankPrototank

1,229920