$mathbb{R} times mathbb{R} times … times mathbb{R}$ ($n$ times) is homeomorphic to $mathbb{R}^{n}$











up vote
0
down vote

favorite












Here is a short sentence from the book "General Topology" of Stephen Willard :



"[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."



Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.



Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :




  • If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?

  • If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :


    • Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?

    • Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?




I hope it is clear enough. Thank you for your future answers.






general-topology







share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    Here is a short sentence from the book "General Topology" of Stephen Willard :



    "[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."



    Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.



    Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :




    • If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?

    • If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :


      • Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?

      • Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?




    I hope it is clear enough. Thank you for your future answers.






    general-topology







    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Here is a short sentence from the book "General Topology" of Stephen Willard :



      "[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."



      Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.



      Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :




      • If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?

      • If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :


        • Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?

        • Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?




      I hope it is clear enough. Thank you for your future answers.






      general-topology







      share|cite|improve this question













      Here is a short sentence from the book "General Topology" of Stephen Willard :



      "[...] In particular, $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology is homeomorphic to $mathbb{R}^{n}$."



      Recently, I asked a question about the fact that we don't have exactly $mathbb{R}^{n} = mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times). Thus, I know now that these two sets are in bijection.



      Now, my problem is the following : in the book in question, the author sometimes says things like "the usual topology on $mathbb{R}^{2}$" and I was wondering :




      • If we consider $mathbb{R}^{n}$, is the "usual topology" not the Tychonoff topology (product topology) ? Which is the same as the ones induced by any norm (considering $mathbb{R}^{n}$ as a vector space) or as the box topology ?

      • If we consider $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times), then :


        • Considering that what I said previously is correct (which is maybe not the case), why does the author say that we consider "$mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) with the product topology" ?

        • Again, considering that what I said is correct, if we assume that $mathbb{R}^{n}$ is endowed with its product topology, what is in fact the topology on the set $mathbb{R} times mathbb{R} times ... times mathbb{R}$ ($n$ times) that makes the author says that these two topological spaces are homeomorphic ?




      I hope it is clear enough. Thank you for your future answers.






      general-topology




      general-topology






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 22 at 14:32









      deeppinkwater

      618




      618






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          1
          down vote













          The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.



          Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.






          share|cite|improve this answer





















          • Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
            – deeppinkwater
            Nov 22 at 15:11












          • I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
            – Jean-Pierre de Villiers
            Nov 22 at 15:39













          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',
          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%2f3009207%2fmathbbr-times-mathbbr-times-times-mathbbr-n-times-is-homeo%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          1
          down vote













          The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.



          Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.






          share|cite|improve this answer





















          • Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
            – deeppinkwater
            Nov 22 at 15:11












          • I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
            – Jean-Pierre de Villiers
            Nov 22 at 15:39

















          up vote
          1
          down vote













          The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.



          Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.






          share|cite|improve this answer





















          • Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
            – deeppinkwater
            Nov 22 at 15:11












          • I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
            – Jean-Pierre de Villiers
            Nov 22 at 15:39















          up vote
          1
          down vote










          up vote
          1
          down vote









          The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.



          Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.






          share|cite|improve this answer












          The author is simply making a distinction between the topological product and the set product. By $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ he actually means the ( topological) product $(mathbb{R},tau_0)times(mathbb{R},tau_0)timesdotsbtimes(mathbb{R},tau_0)$ with $tau_0$ being the order topology on $mathbb{R}$. On the other hand by $mathbb{R}^n$ he means the $n$-dimensional topological space i.e. the topology $(mathbb{R}^n,tau_1)$ where $mathbb{R^n}$ is the set product $mathbb{R}timesmathbb{R}timesdotsbtimesmathbb{R}$ and $tau_1$ is the topology generated by the open balls.



          Author's like using these sorts of abbreviations but this can sometimes cause confusion the first time you see it. To avoid this, in fact, I've seen some authors (cannot immediately think of an example) that use $E_n$ to refer to the topological space $(mathbb{R}^n,tau_1)$ above.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 22 at 14:52









          Jean-Pierre de Villiers

          415




          415












          • Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
            – deeppinkwater
            Nov 22 at 15:11












          • I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
            – Jean-Pierre de Villiers
            Nov 22 at 15:39




















          • Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
            – deeppinkwater
            Nov 22 at 15:11












          • I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
            – Jean-Pierre de Villiers
            Nov 22 at 15:39


















          Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
          – deeppinkwater
          Nov 22 at 15:11






          Thank you for you answer. The product topology is over a set of the form $prod_{i in I} A_{i}$ right (where I suppose $I = {1, ..., n}$ finite) ? So I don't see why we endow $mathbb{R} times mathbb{R} times ... mathbb{R}$ with the product topology (because $prod_{i in I} A_{i} neq A_{1} times ... times A_{n}$). Or maybe I miss the point...
          – deeppinkwater
          Nov 22 at 15:11














          I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
          – Jean-Pierre de Villiers
          Nov 22 at 15:39






          I is indeed finite in this case. Well for one, given any set, we are free to force any topology that we please and thus one that suits our needs. In this case the author made it into the the product topology of the $1$-dimensional Euclidian space with itself $n$ times. He then proceeds to compare it with the $n$-dimensional Euclidian space and concludes that, though the two do not have the same underlying set, they are topologically the same structure. A similar sort of comparison can be made between say $(A,tau)$ and $({0}times A,{{0}times U:Uintau})$ for any set $A$.
          – Jean-Pierre de Villiers
          Nov 22 at 15:39




















          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3009207%2fmathbbr-times-mathbbr-times-times-mathbbr-n-times-is-homeo%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

          Probability when a professor distributes a quiz and homework assignment to a class of n students.

          Image
          0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
          Read more

          Aardman Animations

          Image
          Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
          Read more

          Are they similar matrix

          Image
          0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
          Read more