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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074228%2fwhich-two-of-the-following-space-are-homeomorphic-to-each-other%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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
















    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






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074228%2fwhich-two-of-the-following-space-are-homeomorphic-to-each-other%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            What other Star Trek series did the main TNG cast show up in?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0; } 4 I'm referring to Picard, Riker, Data, Worf, Geordi, Dr. Crusher, Wesley Crusher, Chief O'Brien and family, Deanna Troi, Guinan. I know Picard and Worf show up on DS9, though I haven't seen it myself. I'm curious what happened to the characters after that last episode of TNG? star-trek star-trek-tng share | improve this question edited 10 hours ago Ham Sandwich 6,051 2 25 75
            Read more

            Berlina muro

            Image
            La berlina muro ĉe Bethaniendamm, 1986 La berlina muro aŭ simple la muro [1] , nomita ankaŭ "kontraŭfaŝisma remparo" en propaganda lingvo de la iama Germana Demokratia Respubliko, estis parto de la interngermana limo, kiu dividis de la 13-a de aŭgusto 1961 ĝis la 9-a de novembro 1989 okcidentan Berlinon de la orienta parto de la urbo kaj de la ĉirkaŭa tereno de la GDR. Ĝi estis unu el la plej famaj simboloj de la malvarma milito kaj de la disdivido de Germanio. Pli ol simpla muro, la berlina muro estis fakte tuta defendokonstruaĵo kun du muroj, ena malplena strio, rondirvojo, gardturoj kaj alarmsistemoj. Dum la provo transiri la severe garditan limon al okcidenta Berlino multaj homoj estis mortigitaj. La preciza nombro de viktimoj estas pridisputita kaj ne certa; la nombroj anoncataj varias inter 86 kaj 238 mortintoj. La malplifortiĝo de Sovetunio kaj la politiko de liberigo gvidita de Miĥail Gorbaĉov ebligis al la orientgermanoj faligi la 9-an de novembro 1989 l
            Read more

            Berlina aerponto

            Image
            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
            Read more