$X$ is homeomorphic to $Xtimes X$ (TIFR GS $2014$)












9












$begingroup$


Question is :



Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:




  • $X$ is not connected.

  • $X$ is not compact

  • $X$ is not homeomorphic to a subset of $mathbb{R}$

  • None of the above.


I guess first two options are false.



We do have possibility that product of two connected spaces is connected.



So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.



We do have possibility that product of two compact spaces is compact.



So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.



I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.



So, only thing I have problem with is third option.



I could do nothing for that third option..



I would be thankful if some one can help me out to clear this.



Thank you :)






general-topology examples-counterexamples product-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    $mathbb N$ is a simple counterexample for the third one.
    $endgroup$
    – Dejan Govc
    Dec 10 '13 at 10:24










  • $begingroup$
    $[0,1]^infty{}{}$
    $endgroup$
    – user98602
    Dec 12 '15 at 7:06










  • $begingroup$
    @MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
    $endgroup$
    – Noah Schweber
    Dec 12 '15 at 7:08








  • 1




    $begingroup$
    @NoahSchweber: I had the advantage of not using words :)
    $endgroup$
    – user98602
    Dec 12 '15 at 7:08










  • $begingroup$
    Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
    $endgroup$
    – Henno Brandsma
    Dec 12 '15 at 8:37


















9












$begingroup$


Question is :



Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:




  • $X$ is not connected.

  • $X$ is not compact

  • $X$ is not homeomorphic to a subset of $mathbb{R}$

  • None of the above.


I guess first two options are false.



We do have possibility that product of two connected spaces is connected.



So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.



We do have possibility that product of two compact spaces is compact.



So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.



I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.



So, only thing I have problem with is third option.



I could do nothing for that third option..



I would be thankful if some one can help me out to clear this.



Thank you :)






general-topology examples-counterexamples product-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    $mathbb N$ is a simple counterexample for the third one.
    $endgroup$
    – Dejan Govc
    Dec 10 '13 at 10:24










  • $begingroup$
    $[0,1]^infty{}{}$
    $endgroup$
    – user98602
    Dec 12 '15 at 7:06










  • $begingroup$
    @MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
    $endgroup$
    – Noah Schweber
    Dec 12 '15 at 7:08








  • 1




    $begingroup$
    @NoahSchweber: I had the advantage of not using words :)
    $endgroup$
    – user98602
    Dec 12 '15 at 7:08










  • $begingroup$
    Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
    $endgroup$
    – Henno Brandsma
    Dec 12 '15 at 8:37
















9












9








9


5



$begingroup$


Question is :



Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:




  • $X$ is not connected.

  • $X$ is not compact

  • $X$ is not homeomorphic to a subset of $mathbb{R}$

  • None of the above.


I guess first two options are false.



We do have possibility that product of two connected spaces is connected.



So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.



We do have possibility that product of two compact spaces is compact.



So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.



I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.



So, only thing I have problem with is third option.



I could do nothing for that third option..



I would be thankful if some one can help me out to clear this.



Thank you :)






general-topology examples-counterexamples product-space







share|cite|improve this question











$endgroup$




Question is :



Suppose $X$ is a topological space of infinite cardinality which is homeomorphic to $Xtimes X$. Then which of the following is true:




  • $X$ is not connected.

  • $X$ is not compact

  • $X$ is not homeomorphic to a subset of $mathbb{R}$

  • None of the above.


I guess first two options are false.



We do have possibility that product of two connected spaces is connected.



So, $Xtimes X$ is connected if $X$ is connected. So I guess there is no problem.



We do have possibility that product of two compact spaces is compact.



So, $Xtimes X$ is compact if $X$ is compact. So I guess there is no problem.



I understand that this is not the proof to exclude first two options but I guess the chance is more for them to be false.



So, only thing I have problem with is third option.



I could do nothing for that third option..



I would be thankful if some one can help me out to clear this.



Thank you :)






general-topology examples-counterexamples product-space




general-topology examples-counterexamples product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 19 '16 at 16:56









Martin Sleziak

44.8k10119273




44.8k10119273










asked Dec 10 '13 at 8:56







user87543



















  • $begingroup$
    $mathbb N$ is a simple counterexample for the third one.
    $endgroup$
    – Dejan Govc
    Dec 10 '13 at 10:24










  • $begingroup$
    $[0,1]^infty{}{}$
    $endgroup$
    – user98602
    Dec 12 '15 at 7:06










  • $begingroup$
    @MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
    $endgroup$
    – Noah Schweber
    Dec 12 '15 at 7:08








  • 1




    $begingroup$
    @NoahSchweber: I had the advantage of not using words :)
    $endgroup$
    – user98602
    Dec 12 '15 at 7:08










  • $begingroup$
    Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
    $endgroup$
    – Henno Brandsma
    Dec 12 '15 at 8:37




















  • $begingroup$
    $mathbb N$ is a simple counterexample for the third one.
    $endgroup$
    – Dejan Govc
    Dec 10 '13 at 10:24










  • $begingroup$
    $[0,1]^infty{}{}$
    $endgroup$
    – user98602
    Dec 12 '15 at 7:06










  • $begingroup$
    @MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
    $endgroup$
    – Noah Schweber
    Dec 12 '15 at 7:08








  • 1




    $begingroup$
    @NoahSchweber: I had the advantage of not using words :)
    $endgroup$
    – user98602
    Dec 12 '15 at 7:08










  • $begingroup$
    Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
    $endgroup$
    – Henno Brandsma
    Dec 12 '15 at 8:37


















$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24




$begingroup$
$mathbb N$ is a simple counterexample for the third one.
$endgroup$
– Dejan Govc
Dec 10 '13 at 10:24












$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06




$begingroup$
$[0,1]^infty{}{}$
$endgroup$
– user98602
Dec 12 '15 at 7:06












$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08






$begingroup$
@MikeMiller Drat, beat me to it! (As a good further exercise, it's easy to show that you can't get a simultaneous counterexample to $A$ and $C$.)
$endgroup$
– Noah Schweber
Dec 12 '15 at 7:08






1




1




$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08




$begingroup$
@NoahSchweber: I had the advantage of not using words :)
$endgroup$
– user98602
Dec 12 '15 at 7:08












$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37






$begingroup$
Why $mathbb{Q}$ in the discrete topology? The usual subspace topology, works, but it's not quite trivial to see that $mathbb{Q} times mathbb{Q} approx mathbb{Q}$, but $mathbb{Z}$ in the discrete = subspace topology is a better example (admittedly homeomorphic to the rationals in the discrete topology, but the latter does not directly embed into the reals). The irrationals or the Cantor set also work, etc.
$endgroup$
– Henno Brandsma
Dec 12 '15 at 8:37












4 Answers
4






active

oldest

votes


















7












$begingroup$

The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.



An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.






share|cite|improve this answer











$endgroup$









  • 2




    $begingroup$
    ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
    $endgroup$
    – martini
    Dec 10 '13 at 9:04










  • $begingroup$
    @martini : about the third statement? :(
    $endgroup$
    – user87543
    Dec 10 '13 at 9:06










  • $begingroup$
    As said by martini, my space must be of infinite cardinality...
    $endgroup$
    – user87543
    Dec 10 '13 at 9:06










  • $begingroup$
    @martini, thanks I had overlooked that, I will correct this.
    $endgroup$
    – Carsten S
    Dec 10 '13 at 9:07








  • 1




    $begingroup$
    You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
    $endgroup$
    – Carsten S
    Dec 10 '13 at 9:17



















8












$begingroup$

Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.



      Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.






      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%2f601113%2fx-is-homeomorphic-to-x-times-x-tifr-gs-2014%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown
























        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        7












        $begingroup$

        The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.



        An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
          $endgroup$
          – martini
          Dec 10 '13 at 9:04










        • $begingroup$
          @martini : about the third statement? :(
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          As said by martini, my space must be of infinite cardinality...
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          @martini, thanks I had overlooked that, I will correct this.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:07








        • 1




          $begingroup$
          You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:17
















        7












        $begingroup$

        The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.



        An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.






        share|cite|improve this answer











        $endgroup$









        • 2




          $begingroup$
          ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
          $endgroup$
          – martini
          Dec 10 '13 at 9:04










        • $begingroup$
          @martini : about the third statement? :(
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          As said by martini, my space must be of infinite cardinality...
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          @martini, thanks I had overlooked that, I will correct this.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:07








        • 1




          $begingroup$
          You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:17














        7












        7








        7





        $begingroup$

        The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.



        An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.






        share|cite|improve this answer











        $endgroup$



        The Cantor set is a counter-example to the second and third statement. Note that the Cantor set is homeomorphic to ${0,1}^{mathbb N}$, hence it is homeomorphic to the product with itself.



        An infinite set with the smallest topology (exactly two open sets) is a counter-example to the first statement. Martini gives a better counter-example in a comment.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 10 '13 at 9:10

























        answered Dec 10 '13 at 9:03









        Carsten SCarsten S

        7,26811436




        7,26811436








        • 2




          $begingroup$
          ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
          $endgroup$
          – martini
          Dec 10 '13 at 9:04










        • $begingroup$
          @martini : about the third statement? :(
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          As said by martini, my space must be of infinite cardinality...
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          @martini, thanks I had overlooked that, I will correct this.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:07








        • 1




          $begingroup$
          You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:17














        • 2




          $begingroup$
          ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
          $endgroup$
          – martini
          Dec 10 '13 at 9:04










        • $begingroup$
          @martini : about the third statement? :(
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          As said by martini, my space must be of infinite cardinality...
          $endgroup$
          – user87543
          Dec 10 '13 at 9:06










        • $begingroup$
          @martini, thanks I had overlooked that, I will correct this.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:07








        • 1




          $begingroup$
          You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
          $endgroup$
          – Carsten S
          Dec 10 '13 at 9:17








        2




        2




        $begingroup$
        ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
        $endgroup$
        – martini
        Dec 10 '13 at 9:04




        $begingroup$
        ... of infinite cardinality. But $[0,1]^{mathbb N}$ works as counterexample to the first and second statement.
        $endgroup$
        – martini
        Dec 10 '13 at 9:04












        $begingroup$
        @martini : about the third statement? :(
        $endgroup$
        – user87543
        Dec 10 '13 at 9:06




        $begingroup$
        @martini : about the third statement? :(
        $endgroup$
        – user87543
        Dec 10 '13 at 9:06












        $begingroup$
        As said by martini, my space must be of infinite cardinality...
        $endgroup$
        – user87543
        Dec 10 '13 at 9:06




        $begingroup$
        As said by martini, my space must be of infinite cardinality...
        $endgroup$
        – user87543
        Dec 10 '13 at 9:06












        $begingroup$
        @martini, thanks I had overlooked that, I will correct this.
        $endgroup$
        – Carsten S
        Dec 10 '13 at 9:07






        $begingroup$
        @martini, thanks I had overlooked that, I will correct this.
        $endgroup$
        – Carsten S
        Dec 10 '13 at 9:07






        1




        1




        $begingroup$
        You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
        $endgroup$
        – Carsten S
        Dec 10 '13 at 9:17




        $begingroup$
        You are welcome. Cantor space is the first space I think of when I see $Xtimes Xapprox X$.
        $endgroup$
        – Carsten S
        Dec 10 '13 at 9:17











        8












        $begingroup$

        Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.






        share|cite|improve this answer









        $endgroup$


















          8












          $begingroup$

          Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.






          share|cite|improve this answer









          $endgroup$
















            8












            8








            8





            $begingroup$

            Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.






            share|cite|improve this answer









            $endgroup$



            Fix your favorite compact connected space $X$ - then $X^omega$ is compact, connected, and homeomorphic to its own square.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 12 '15 at 7:07









            Noah SchweberNoah Schweber

            126k10151290




            126k10151290























                1












                $begingroup$

                Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.






                    share|cite|improve this answer









                    $endgroup$



                    Take the Cantor set. This gives a counterexample for B and C. For A, as Mike said in the comments, take the set $[0,1]^infty$.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 12 '15 at 7:08









                    seekerseeker

                    1,189414




                    1,189414























                        0












                        $begingroup$

                        Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.



                        Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.



                          Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.



                            Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.






                            share|cite|improve this answer









                            $endgroup$



                            Let $Q = prod_{n=1}^infty I_n$ where $I_n = [0,1]$. This space is known as the Hilbert cube. It is compact and connected and satisfies $Q times Q approx Q$.



                            Note that you cannot find a single counterexample $X$ for a) and c). Assume $X$ would be such a space. Then $X$ must be homeomorphic to a connected subset of $mathbb{R}$, that is, homeomorphic to an interval $J$. But $J times J$ is not homeomorphic to $J$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 6 '18 at 4:28









                            jasminejasmine

                            1,845418




                            1,845418






























                                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%2f601113%2fx-is-homeomorphic-to-x-times-x-tifr-gs-2014%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

                                mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                                Image
                                0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
                                Read more

                                How to change which sound is reproduced for terminal bell?

                                Image
                                3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                                Read more

                                Can I use Tabulator js library in my java Spring + Thymeleaf project?

                                Image
                                0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                                Read more