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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            Berlina aerponto

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            Flugkoridoroj dum la blokado de Berlino La Berlina aerponto estas la provizigo de okcidenta Berlino per aviadiloj fare de la aliancanoj inter la 23-a de junio 1948 kaj la 12-a de majo 1949 dum la blokado de Berlino fare de la sovetia armeo. Enhavo 1 Historio 1.1 La aerkoridoroj 1.2 Unuaj flugoj 1.3 Plibonigo de la aerponto 1.4 Fino de la aerponto 1.5 Rozinbombiloj 1.6 Bilanco 2 Flughavenoj 2.1 Berlino 2.2 Okcidenta Germanio 3 Aviadiloj de la berlina aerponto 4 15-a datreveno de la aerponto 5 Bibliografio 6 Referencoj 7 Eksteraj ligiloj Historio | General Lucius D. Clay, militguberniestro de la usona okupadzono Jam antaŭ la berlina aerponto estiĝis la malgranda aerponto . La estro de la sovetia militadministracio en Germanio, Vasili Daniloviĉ Sokolovski, ordonis la 1-an de aprilo 1948, kiel reago al la konferenco de Londono inter britoj, francoj kaj reprezentantoj de Benelukso pri aligo de Germa...
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