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
    Jan 15 at 9:14










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:16
















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
    Jan 15 at 9:14










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:16














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 Jan 15 at 9:29









José Carlos Santos

177k24138248




177k24138248










asked Jan 15 at 9:10









StammeringMathematicianStammeringMathematician

2,8321324




2,8321324








  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:14










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:16














  • 1




    $begingroup$
    $X_4$ is connected.
    $endgroup$
    – Kavi Rama Murthy
    Jan 15 at 9:14










  • $begingroup$
    @KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:16








1




1




$begingroup$
$X_4$ is connected.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 9:14




$begingroup$
$X_4$ is connected.
$endgroup$
– Kavi Rama Murthy
Jan 15 at 9:14












$begingroup$
@KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
$endgroup$
– StammeringMathematician
Jan 15 at 9:16




$begingroup$
@KaviRamaMurthy By mistake I wrote "connected", I meant homoeomorphic. Sorry
$endgroup$
– StammeringMathematician
Jan 15 at 9:16










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
    Jan 15 at 9:17










  • $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 9:19










  • $begingroup$
    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
    Jan 15 at 9:21










  • $begingroup$
    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
    Jan 15 at 9:28










  • $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:29



















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.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
      Jan 15 at 9:17










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      Jan 15 at 9:19










    • $begingroup$
      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
      Jan 15 at 9:21










    • $begingroup$
      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
      Jan 15 at 9:28










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:29
















    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
      Jan 15 at 9:17










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      Jan 15 at 9:19










    • $begingroup$
      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
      Jan 15 at 9:21










    • $begingroup$
      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
      Jan 15 at 9:28










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:29














    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 Jan 15 at 9:26

























    answered Jan 15 at 9:16









    José Carlos SantosJosé Carlos Santos

    177k24138248




    177k24138248












    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:17










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      Jan 15 at 9:19










    • $begingroup$
      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
      Jan 15 at 9:21










    • $begingroup$
      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
      Jan 15 at 9:28










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:29


















    • $begingroup$
      By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:17










    • $begingroup$
      I understood what you meant. What do you think about my answer?
      $endgroup$
      – José Carlos Santos
      Jan 15 at 9:19










    • $begingroup$
      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
      Jan 15 at 9:21










    • $begingroup$
      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
      Jan 15 at 9:28










    • $begingroup$
      Thanks for the quick reply.
      $endgroup$
      – StammeringMathematician
      Jan 15 at 9:29
















    $begingroup$
    By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:17




    $begingroup$
    By mistake I wrote "connected", I meant homoeomorphic. Sorry . I have edited the word in the post.
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:17












    $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 9:19




    $begingroup$
    I understood what you meant. What do you think about my answer?
    $endgroup$
    – José Carlos Santos
    Jan 15 at 9:19












    $begingroup$
    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
    Jan 15 at 9:21




    $begingroup$
    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
    Jan 15 at 9:21












    $begingroup$
    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
    Jan 15 at 9:28




    $begingroup$
    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
    Jan 15 at 9:28












    $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:29




    $begingroup$
    Thanks for the quick reply.
    $endgroup$
    – StammeringMathematician
    Jan 15 at 9:29











    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 Jan 15 at 9:29









        Mostafa AyazMostafa Ayaz

        18.1k31040




        18.1k31040






























            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

            Måne

            Image
            För jordens måne, se Månen. Några månar i solsystemet i skala till vår måne En måne , naturlig satellit eller drabant är en himlakropp som kretsar kring en planet eller en asteroid i ett solsystem. Ordet i bestämd form – månen – avser planeten jordens måne. Det finns inte i något fall belagt att någon måne har egna månar. De flesta stora månar i solsystemet är bundna i synkron rotation med en och samma sida vänd mot planeten. Ett exempel på en måne som inte beter sig på detta vis, är Hyperion som kaotiskt roterar kring Saturnus. En asteroidmåne är en måne som kretsar kring en asteroid. En måne kan ha bildats samtidigt som moderplaneten eller senare infångats av planetens gravitationskraft. I vårt eget solsystem känner vi i nuläget till 173 månar runt sex av de åtta planeterna [ 1 ] : Jorden – månen Mars – Phobos och Deimos Jupiter – Ganymedes (som är solsystemets största måne), Io, Europa, Callisto och 59 andra kända månar Saturnus – Titan (den enda måne...
            Read more

            Storängen

            Image
            För stadsdelen i Mariehamn, se Storängen, Åland. För ett naturreservat i Norrtälje kommun, se Storängens naturreservat. Storängen från luften i september 1967, vy mot väst. Till vänster syns Järlasjön och Saltsjöbanans spår. Till höger löper Värmdövägen och längst ut till höger syns anläggningsarbetena för Värmdöleden. Storängen är ett villaområde i centrala Nacka i Stockholms län. Området är beläget strax sydost om Nacka Forum vid Saltsjöbanan mellan Lillängen och Saltsjö-Duvnäs. Storängens villasamhälle grundades 1904 av "Tjänstemännens Egnahemsförening vid Storängen" och de första villorna började byggas samma år. Redan 1909 var 107 tomter bebyggda av totalt runt 160. [ 1 ] År 2013 fanns drygt 150 villor med lika många familjer i området. Egnahemsföreningen existerar fortfarande och fungerar idag som intresseförening för de boende i Storängen. Storängen anses vara ett av Sveriges bäst bevarade villasamhällen från 1900-talets början, tack vare sina tidstypis...
            Read more

            VLT Carioca

            Image
            VLT Carioca Informações Proprietário Prefeitura do Rio de Janeiro Local [[Rio de Janeiro, RJ]] País {{country data Brasil| country flag2| name = Brasil| variant = | size = }} Tipo de transporte Bonde (Veículo leve sobre trilhos) Número de linhas 2 (+1 em construção) Número de estações 26 (+3 em construção) Tráfego 80 mil passageiros/dia [ 1 ] (dezembro/2018) Website www.vltrio.com.br Funcionamento Início de funcionamento 5 de junho de 2016 (2 anos) Operadora(s) Concessionária do VLT Carioca S.A. Número de veículos 32 Alstom Citadis Comprimento dos veículos 44 m (144 ft) Dados técnicos Extensão do sistema 28 km (17,4 mi) [ 2 ] Frequência 5–10 minutos Bitola Bitola padrão/internacional 1 435 mm (4,71 ft) Eletrificação APS, 750V DC Velocidade média 15 km/h (9,32 mph) [ 2 ] Velocidade máxima 50 km/h (31,1 mph) ...
            Read more