Checking if a set is open and relatively open in the sphere ${x^2 + y^2 + z^2 = 1}$Is a sphere a closed...

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Checking if a set is open and relatively open in the sphere ${x^2 + y^2 + z^2 = 1}$


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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.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It seems OK.${}{}{}$
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 18:02










  • $begingroup$
    Thanks. But how can I formally prove that $B$ is not relatively open?
    $endgroup$
    – Gabi G
    Mar 9 at 18:16










  • $begingroup$
    You already did, I think it is enough.
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 19:20










  • $begingroup$
    I proved that A is. In the comment I am talking about B
    $endgroup$
    – Gabi G
    Mar 9 at 19:29


















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.










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It seems OK.${}{}{}$
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 18:02










  • $begingroup$
    Thanks. But how can I formally prove that $B$ is not relatively open?
    $endgroup$
    – Gabi G
    Mar 9 at 18:16










  • $begingroup$
    You already did, I think it is enough.
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 19:20










  • $begingroup$
    I proved that A is. In the comment I am talking about B
    $endgroup$
    – Gabi G
    Mar 9 at 19:29
















1












1








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.










share|cite|improve this question











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 9 at 22:20









Jean Marie

30.6k42154




30.6k42154










asked Mar 9 at 17:47









Gabi GGabi G

442110




442110








  • 1




    $begingroup$
    It seems OK.${}{}{}$
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 18:02










  • $begingroup$
    Thanks. But how can I formally prove that $B$ is not relatively open?
    $endgroup$
    – Gabi G
    Mar 9 at 18:16










  • $begingroup$
    You already did, I think it is enough.
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 19:20










  • $begingroup$
    I proved that A is. In the comment I am talking about B
    $endgroup$
    – Gabi G
    Mar 9 at 19:29
















  • 1




    $begingroup$
    It seems OK.${}{}{}$
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 18:02










  • $begingroup$
    Thanks. But how can I formally prove that $B$ is not relatively open?
    $endgroup$
    – Gabi G
    Mar 9 at 18:16










  • $begingroup$
    You already did, I think it is enough.
    $endgroup$
    – Giuseppe Negro
    Mar 9 at 19:20










  • $begingroup$
    I proved that A is. In the comment I am talking about B
    $endgroup$
    – Gabi G
    Mar 9 at 19:29










1




1




$begingroup$
It seems OK.${}{}{}$
$endgroup$
– Giuseppe Negro
Mar 9 at 18:02




$begingroup$
It seems OK.${}{}{}$
$endgroup$
– Giuseppe Negro
Mar 9 at 18:02












$begingroup$
Thanks. But how can I formally prove that $B$ is not relatively open?
$endgroup$
– Gabi G
Mar 9 at 18:16




$begingroup$
Thanks. But how can I formally prove that $B$ is not relatively open?
$endgroup$
– Gabi G
Mar 9 at 18:16












$begingroup$
You already did, I think it is enough.
$endgroup$
– Giuseppe Negro
Mar 9 at 19:20




$begingroup$
You already did, I think it is enough.
$endgroup$
– Giuseppe Negro
Mar 9 at 19:20












$begingroup$
I proved that A is. In the comment I am talking about B
$endgroup$
– Gabi G
Mar 9 at 19:29






$begingroup$
I proved that A is. In the comment I am talking about B
$endgroup$
– Gabi G
Mar 9 at 19:29












1 Answer
1






active

oldest

votes


















1












$begingroup$

Suppose $U$ is an open set in $mathbb{R}^3$ 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









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    $begingroup$

    Suppose $U$ is an open set in $mathbb{R}^3$ 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









    $endgroup$


















      1












      $begingroup$

      Suppose $U$ is an open set in $mathbb{R}^3$ 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









      $endgroup$
















        1












        1








        1





        $begingroup$

        Suppose $U$ is an open set in $mathbb{R}^3$ 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









        $endgroup$



        Suppose $U$ is an open set in $mathbb{R}^3$ 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












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 9 at 22:09









        PrototankPrototank

        1,254920




        1,254920






























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