Which two of the following space are homeomorphic to each other?












2












$begingroup$


Which two of the following space are homeomorphic to each other?



begin{align}
X_1&={(x,y)in mathbb{R}^2:xy=0}\
X_2&={(x,y)in mathbb{R}^2:xy=1}\
X_3&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=0}\
X_4&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=1}
end{align}



I want to use the contentedness property.



Efforts:



$X_1$ and $X_4$ are not homeomorphic as $X_1$ is connected and nd $X_2$ is disconnected(two connected component)



By same logic $X_1$ and $X_4$ are not homeomorphic.



If I remove the origin from $X_1$, there are 4 components, but if we remove the origin from $X_3$, there are three connected component. So $X_1$ and $X_3$ are not homeomorphic.



Am I going in the right direction?



How to proceed after this.






general-topology continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    1 hour ago
















2












$begingroup$


Which two of the following space are homeomorphic to each other?



begin{align}
X_1&={(x,y)in mathbb{R}^2:xy=0}\
X_2&={(x,y)in mathbb{R}^2:xy=1}\
X_3&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=0}\
X_4&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=1}
end{align}



I want to use the contentedness property.



Efforts:



$X_1$ and $X_4$ are not homeomorphic as $X_1$ is connected and nd $X_2$ is disconnected(two connected component)



By same logic $X_1$ and $X_4$ are not homeomorphic.



If I remove the origin from $X_1$, there are 4 components, but if we remove the origin from $X_3$, there are three connected component. So $X_1$ and $X_3$ are not homeomorphic.



Am I going in the right direction?



How to proceed after this.






general-topology continuity







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    1 hour ago














2












2








2





$begingroup$


Which two of the following space are homeomorphic to each other?



begin{align}
X_1&={(x,y)in mathbb{R}^2:xy=0}\
X_2&={(x,y)in mathbb{R}^2:xy=1}\
X_3&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=0}\
X_4&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=1}
end{align}



I want to use the contentedness property.



Efforts:



$X_1$ and $X_4$ are not homeomorphic as $X_1$ is connected and nd $X_2$ is disconnected(two connected component)



By same logic $X_1$ and $X_4$ are not homeomorphic.



If I remove the origin from $X_1$, there are 4 components, but if we remove the origin from $X_3$, there are three connected component. So $X_1$ and $X_3$ are not homeomorphic.



Am I going in the right direction?



How to proceed after this.






general-topology continuity







share|cite|improve this question











$endgroup$




Which two of the following space are homeomorphic to each other?



begin{align}
X_1&={(x,y)in mathbb{R}^2:xy=0}\
X_2&={(x,y)in mathbb{R}^2:xy=1}\
X_3&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=0}\
X_4&={(x,y)in mathbb{R}^2: x+ygeq 0 text{ and } xy=1}
end{align}



I want to use the contentedness property.



Efforts:



$X_1$ and $X_4$ are not homeomorphic as $X_1$ is connected and nd $X_2$ is disconnected(two connected component)



By same logic $X_1$ and $X_4$ are not homeomorphic.



If I remove the origin from $X_1$, there are 4 components, but if we remove the origin from $X_3$, there are three connected component. So $X_1$ and $X_3$ are not homeomorphic.



Am I going in the right direction?



How to proceed after this.






general-topology continuity




general-topology continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









José Carlos Santos

154k22123226




154k22123226










asked 1 hour ago









StammeringMathematicianStammeringMathematician

2,3041322




2,3041322








  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    1 hour ago














  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    1 hour ago










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    1 hour ago








1




1




$begingroup$
$X_4$ is connected.
$endgroup$
– Kavi Rama Murthy
1 hour ago




$begingroup$
$X_4$ is connected.
$endgroup$
– Kavi Rama Murthy
1 hour ago












$begingroup$
@KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
$endgroup$
– StammeringMathematician
1 hour ago




$begingroup$
@KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
$endgroup$
– StammeringMathematician
1 hour ago










2 Answers
2






active

oldest

votes


















3












$begingroup$

Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.



On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$begin{array}{ccc}X_4&longrightarrow&X_3\(x,y)&mapsto&begin{cases}(0,y-x)&text{ if }ygeqslant x\(x-y,0)&text{ otherwise.}end{cases}end{array}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
    $endgroup$
    – StammeringMathematician
    1 hour ago










  • $begingroup$
    I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
    $endgroup$
    – José Carlos Santos
    1 hour ago










  • $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    1 hour ago



















2












$begingroup$

Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.






share|cite|improve this answer









$endgroup$













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    2 Answers
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    2 Answers
    2






    active

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    active

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    3












    $begingroup$

    Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.



    On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$begin{array}{ccc}X_4&longrightarrow&X_3\(x,y)&mapsto&begin{cases}(0,y-x)&text{ if }ygeqslant x\(x-y,0)&text{ otherwise.}end{cases}end{array}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      1 hour ago
















    3












    $begingroup$

    Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.



    On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$begin{array}{ccc}X_4&longrightarrow&X_3\(x,y)&mapsto&begin{cases}(0,y-x)&text{ if }ygeqslant x\(x-y,0)&text{ otherwise.}end{cases}end{array}$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      1 hour ago














    3












    3








    3





    $begingroup$

    Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.



    On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$begin{array}{ccc}X_4&longrightarrow&X_3\(x,y)&mapsto&begin{cases}(0,y-x)&text{ if }ygeqslant x\(x-y,0)&text{ otherwise.}end{cases}end{array}$$






    share|cite|improve this answer











    $endgroup$



    Actually, $X_4$ is connected (it's a branch of a hyperbola) and therefore your argument fails. But if you remove $(0,0)$ from $X_1$, then what remains has four connected components. This proved that $X_1$ and $X_4$ are not homeomorphic.



    On the other hand, $X_3$ and $X_4$ are homeomorphic. Just consider the map$$begin{array}{ccc}X_4&longrightarrow&X_3\(x,y)&mapsto&begin{cases}(0,y-x)&text{ if }ygeqslant x\(x-y,0)&text{ otherwise.}end{cases}end{array}$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    José Carlos SantosJosé Carlos Santos

    154k22123226




    154k22123226












    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      1 hour ago


















    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
      $endgroup$
      – StammeringMathematician
      1 hour ago










    • $begingroup$
      I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
      $endgroup$
      – José Carlos Santos
      1 hour ago










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      1 hour ago
















    $begingroup$
    By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
    $endgroup$
    – StammeringMathematician
    1 hour ago




    $begingroup$
    By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
    $endgroup$
    – StammeringMathematician
    1 hour ago












    $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    1 hour ago




    $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    1 hour ago












    $begingroup$
    How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
    $endgroup$
    – StammeringMathematician
    1 hour ago




    $begingroup$
    How $X_3$ and $X_4$ are homeomorphic. Can you give some hint. And in the OP, am I right in deducing that removing the origin from $X_3$ leaves us with three connected component. Thanks
    $endgroup$
    – StammeringMathematician
    1 hour ago












    $begingroup$
    I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
    $endgroup$
    – José Carlos Santos
    1 hour ago




    $begingroup$
    I've added the homeomorphism that I had in mind. And if you remove a point from $X_3$, that will leave you with two connected components.
    $endgroup$
    – José Carlos Santos
    1 hour ago












    $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    1 hour ago




    $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    1 hour ago











    2












    $begingroup$

    Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.






        share|cite|improve this answer









        $endgroup$



        Yes you are right. In fact the only two homeomorphic sets are $X_3$ and $X_4$ since $X_4$ can be interpreted as a warped version of $X_3$. $X_3$, $X_4$ and $X_1$ are connected and $X_2$ is not. Therefore $X_2$ is not homeomorphic to any of the others. Also removing origin from $X_1$ leaves us with $4$ distinct parts in $X_1$ while leaving us $2$ distinct parts in $X_3$ and $X_4$. Therefore the only homeomorphic sets are $X_3$ and $X_4$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        Mostafa AyazMostafa Ayaz

        15.2k3939




        15.2k3939






























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            Gregoriusmis

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            Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
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