Example of a continuous function that don't have a continuous extensionExtending a continuous function...

Why, historically, did Gödel think CH was false?

Prove that NP is closed under karp reduction?

How to format long polynomial?

Is it tax fraud for an individual to declare non-taxable revenue as taxable income? (US tax laws)

How can I make my BBEG immortal short of making them a Lich or Vampire?

Do I have a twin with permutated remainders?

Why can't I see bouncing of a switch on an oscilloscope?

Why was the small council so happy for Tyrion to become the Master of Coin?

Why do falling prices hurt debtors?

Smoothness of finite-dimensional functional calculus

Is it possible to do 50 km distance without any previous training?

Example of a continuous function that don't have a continuous extension

The use of multiple foreign keys on same column in SQL Server

What are the differences between the usage of 'it' and 'they'?

Theorem, big Paralist and Amsart

Has the BBC provided arguments for saying Brexit being cancelled is unlikely?

What are these boxed doors outside store fronts in New York?

Is it unprofessional to ask if a job posting on GlassDoor is real?

How to test if a transaction is standard without spending real money?

If I cast Expeditious Retreat, can I Dash as a bonus action on the same turn?

Dragon forelimb placement

Arthur Somervell: 1000 Exercises - Meaning of this notation

Can a Warlock become Neutral Good?

Today is the Center



Example of a continuous function that don't have a continuous extension


Extending a continuous function defined on the rationalsLet $Asubset X$; let $f:Ato Y$ be continuous; let $Y$ be Hausdorff. Is there an example where there is no continuous function for $g$?Continuity of a product of two real valued continuous function.a counter example of extension of a continuous functionInverse of a continuous functionIf $Asubseteqmathbb R$ is closed and $f:Atomathbb R$ is right-continuous, is there a right-continuous extension of $f$ to $mathbb R$?Proving Topological Equivalence without finding a functionA function that can be continuously extended is continuousContinuous Extension of Densely Defined Continuous (but not Uniformly Continuous) Function.A topological space with the Universal Extension Property which is not homeomorphic to a retract of $mathbb{R}^J$?













3












$begingroup$



Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    2 hours ago
















3












$begingroup$



Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    2 hours ago














3












3








3





$begingroup$



Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.










share|cite|improve this question









$endgroup$





Give an example of a topological space $(X,tau)$, a subset $Asubset X$ that is dense in $X$ (i.e., $overline{A} = X$), and a continuous function $f:Atomathbb{R}$ that cannot be continually extended to $X$, that is, a $f$ for such do not exist a continuous function $g:Xto mathbb{R}$ such that $f(x) = g(x)$ for all $xin A$.




I just proved that if $f,g:Xtomathbb{R}$ are continuous and agree in a dense subset $Asubset X$ then they're equal.



I thought in $X=mathbb{R}$ with usual topology and $A = mathbb{R}-{0} =:mathbb{R}^* $, so I think $f:mathbb{R}^*tomathbb{R}, f(x) = x^{-1}$ is a continuous function that cannot be continually extended to $mathbb{R}$. I'm quite sure of this, but I'm stuck in proving it using the definition of continuity in general topological spaces.



Also, I'm quite confused on how this asked example is not a counterexample of what I proved.



Thanks in advance.







general-topology continuity






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 2 hours ago









AnalyticHarmonyAnalyticHarmony

689313




689313








  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    2 hours ago














  • 1




    $begingroup$
    The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
    $endgroup$
    – Melody
    2 hours ago








1




1




$begingroup$
The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
$endgroup$
– Melody
2 hours ago




$begingroup$
The open intervals form a basis for topology on the real line. A set is open if and only if it contains an open interval around each of these points. Using this definition of open sets you can show that the two different definitions of continuity are actually the same in this case. So you're example will work. And to show it will work you can show it using the usual definition of continuity you're used to in the real numbers.
$endgroup$
– Melody
2 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



    Now, drawing a picture will make the following obvious:



    Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



    If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






    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%2f3177651%2fexample-of-a-continuous-function-that-dont-have-a-continuous-extension%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









      2












      $begingroup$

      Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



      One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





      BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



        One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





        BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



          One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





          BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result






          share|cite|improve this answer











          $endgroup$



          Define $f(x)=1/x$ like you did, and assume you can find a continuous extension $g : mathbb{R}tomathbb{R}$. Well this $g $ takes a real numbered value at $0$, namely $-infty < g (0) < infty $, and it agrees with $f $ at non-zero values.



          One definition of continuity is that given a net of points in $X $ converging to $x_0$ and a function $g $, then the images converge to $g(x_0) $. Since $mathbb{R}$ is a metric space, we can use sequences instead of nets. But given a sequence of real numbers $(x_n )_{n=1}^{infty} $ converging to $0$, the sequence $(g (x_n))_{n=1}^{infty} $ converges to either positive or negative $infty $. So it does not converge to $g (0) $. So $g $ is not continuous





          BTW regarding your question on the results you proved. You proved a result about two functions that were continuous on the entire space, who agree on a dense subset. But the main question of your post is regarding a function who is not assumed to be continuous on the entire space, and comparing it to one that is continuous on the entire space. So the main example is not countering your original result







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 1 hour ago

























          answered 1 hour ago









          NazimJNazimJ

          77019




          77019























              1












              $begingroup$

              Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



              Now, drawing a picture will make the following obvious:



              Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



              If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                Now, drawing a picture will make the following obvious:



                Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                  Now, drawing a picture will make the following obvious:



                  Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                  If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$






                  share|cite|improve this answer









                  $endgroup$



                  Using sequences is the easiest way to go, but for a more "topological" proof, to show that no extension of $f$ is continuous at $x=0,$ we suppose there is one (we still call it $f$ for convenience), and we show that there is an $epsilon>0$ so that for any $delta >0$, there is an $xin (-delta,delta$), such that$f(x)>f(0)+epsilon$ (or that $f(x)<f(0)-epsilon$). Let's do the former.



                  Now, drawing a picture will make the following obvious:



                  Take $epsilon=1.$ Then, if $f(0)+1le 0$, then $text{any} xin (0,delta)$ will do because $f(x)=1/x>0.$



                  If $f(0)+1> 0$, all we need do is choose $x$ small enough so that $f(x)=1/x>f(0)+1,$ which is to say, choose $x<min{delta, frac{1}{f(0)+1}}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 45 mins ago









                  MatematletaMatematleta

                  12.1k21020




                  12.1k21020






























                      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%2f3177651%2fexample-of-a-continuous-function-that-dont-have-a-continuous-extension%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

                      Gersau Kjelder | Navigasjonsmeny46°59′0″N 8°31′0″E46°59′0″N...

                      Hestehale Innhaldsliste Hestehale på kvinner | Hestehale på menn | Galleri | Sjå òg |...

                      What is the “three and three hundred thousand syndrome”?Who wrote the book Arena?What five creatures were...